Solutions Chemistry Chapter 1: NCERT Solutions for Class 12

SimplyAcad is here to help students gear up for their approaching 12th Boards examinations, through providing the best answers of the Solutions class 12 chemistry NCERT solutions. The answers provided below contain in depth knowledge of the topics present in th chapter and will allow students to grasp them thoroughly. 

Solutions is an extremely crucial segment of the Chemistry syllabus prescribed by CBSE to teach students of the basic as well as advanced usage of the different types of solutions to perform chemical reactions. 

NCERT has prepared a unique set of questions in the exercises of the Chemistry textbook to check students’ understanding of the related topic.

Solutions class 12 chemistry NCERT solutions Chapter 1 Questions 1 to 5

Question 1:

Calculate the mass percentage of benzene (

$\text{C}_6\text{H}_6$

C6​H6​) and carbon tetrachloride (

$\text{CCl}_4$

CCl4​) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Solution:

Given:

$$\text{Mass of benzene (C}_6\text{H}_6) = 22 \, \text{g}$$

 

 

 

 

$$\text{Mass of carbon tetrachloride (CCl}_4) = 122 \, \text{g}$$

 

 

 

 

 

Total mass of the solution:

$$\text{Total mass} = \text{Mass of benzene} + \text{Mass of carbon tetrachloride} = 22 \, \text{g} + 122 \, \text{g} = 144 \, \text{g}$$

 

 

 

 

 

Mass percentage of a component is given by:

$$\text{Mass percentage} = \left(\frac{\text{Mass of component}}{\text{Total mass of solution}}\right) \times 100$$

 

 

 

 

 

Mass percentage of benzene:

$$\text{Mass percentage of benzene} = \left(\frac{22}{144}\right) \times 100 \approx 15.27\%$$

 

 

 

 

 

Mass percentage of carbon tetrachloride:

$$\text{Mass percentage of carbon tetrachloride} = \left(\frac{122}{144}\right) \times 100 \approx 84.72\%$$

 

 

 

 

 

Thus, the mass percentage of benzene is approximately 15.27%, and the mass percentage of carbon tetrachloride is approximately 84.72%.

---

Question 2:

Calculate the mole fraction of benzene in a solution containing 30% by mass of benzene in carbon tetrachloride.

Solution:

Given:

$$\text{Mass percentage of benzene} (\text{C}_6\text{H}_6) = 30\%$$

 

 

 

 

$$\text{Mass percentage of carbon tetrachloride} (\text{CCl}_4) = 70\%$$

 

 

 

 

 

Assume the total mass of the solution is 100 g:

$$\text{Mass of benzene} = 30 \, \text{g}$$

 

 

 

 

$$\text{Mass of carbon tetrachloride} = 70 \, \text{g}$$

 

 

 

 

 

Molar mass of benzene (

$\text{C}_6\text{H}_6$

C6​H6​):

$$6 \times 12.01 + 6 \times 1.01 = 78.12 \, \text{g/mol}$$

 

 

 

 

 

Molar mass of carbon tetrachloride (

$\text{CCl}_4$

CCl4​):

$$1 \times 12.01 + 4 \times 35.45 = 153.81 \, \text{g/mol}$$

 

 

 

 

 

Moles of benzene:

$$n_{\text{C}_6\text{H}_6} = \frac{30 \, \text{g}}{78.12 \, \text{g/mol}} \approx 0.384 \, \text{mol}$$

 

 

 

 

 

Moles of carbon tetrachloride:

$$n_{\text{CCl}_4} = \frac{70 \, \text{g}}{153.81 \, \text{g/mol}} \approx 0.455 \, \text{mol}$$

 

 

 

 

 

Mole fraction of benzene:

$$\chi_{\text{C}_6\text{H}_6} = \frac{n_{\text{C}_6\text{H}_6}}{n_{\text{C}_6\text{H}_6} + n_{\text{CCl}_4}} = \frac{0.384}{0.384 + 0.455} \approx 0.458$$

 

 

 

 

 

Mole fraction of carbon tetrachloride:

$$\chi_{\text{CCl}_4} = 1 – \chi_{\text{C}_6\text{H}_6} = 1 – 0.458 \approx 0.542$$

 

 

 

 

 

Thus, the mole fraction of benzene is approximately 0.458, and the mole fraction of carbon tetrachloride is approximately 0.542.

---

Question 3:

Calculate the molarity of each of the following solutions:

(A) 30 g of

$\text{Co}(\text{NO}_3)_2 \cdot 6\text{H}_2\text{O}$

Co(NO3​)2​⋅6H2​O in 4.3 L of solution.

(B) 30 ml of 0.5 M

$\text{H}_2\text{SO}_4$

H2​SO4​ diluted to 500 ml.

Solution:

(A) Molarity of $\text{Co}(\text{NO}_3)_2 \cdot 6\text{H}_2\text{O}$

 

 

 

 

Given:

$$\text{Mass of } \text{Co}(\text{NO}_3)_2 \cdot 6\text{H}_2\text{O} = 30 \, \text{g}$$

 

 

 

 

$$\text{Molecular mass of } \text{Co}(\text{NO}_3)_2 \cdot 6\text{H}_2\text{O} = 291 \, \text{g/mol}$$

 

 

 

 

 

Number of moles of

$\text{Co}(\text{NO}_3)_2 \cdot 6\text{H}_2\text{O}$

Co(NO3​)2​⋅6H2​O:

$$\text{Moles} = \frac{30 \, \text{g}}{291 \, \text{g/mol}} \approx 0.103 \, \text{mol}$$

 

 

 

 

 

Molarity (M) is defined as the number of moles of solute per liter of solution:

$$\text{Molarity} = \frac{0.103 \, \text{mol}}{4.3 \, \text{L}} \approx 0.024 \, \text{M}$$

 

 

 

 

 

(B) Molarity of diluted $\text{H}_2\text{SO}_4$

 

 

 

 

H2​SO4​

Using the dilution formula

$M_1V_1 = M_2V_2$

M1​V1​=M2​V2​:

$$M_1 = 0.5 \, \text{M} \quad \text{(initial molarity)}$$

 

 

 

 

$$V_1 = 30 \, \text{ml} \quad \text{(initial volume)}$$

 

 

 

 

$$V_2 = 500 \, \text{ml} \quad \text{(final volume)}$$

 

 

 

 

 

Let

$M_2$

M2​ be the final molarity after dilution:

$$0.5 \, \text{M} \times 30 \, \text{ml} = M_2 \times 500 \, \text{ml}$$

 

 

 

 

 

Solving for

$M_2$

M2​:

$$M_2 = \frac{15 \, \text{M}\cdot\text{ml}}{500 \, \text{ml}} = 0.03 \, \text{M}$$

 

 

 

 

 

Thus, the molarity of the diluted

$\text{H}_2\text{SO}_4$

H2​SO4​ solution is 0.03 M.

---

Question 4:

Calculate the mass of urea (

$\text{NH}_2\text{CONH}_2$

NH2​CONH2​) required to make 2.5 kg of a 0.25 molal aqueous solution.

Solution:

Given:

$$\text{Mass of solution} = 2.5 \, \text{kg}$$

 

 

 

 

$$\text{Molality of solution} = 0.25 \, \text{m}$$

 

 

 

 

 

Molar mass of urea (

$\text{NH}_2\text{CONH}_2$

NH2​CONH2​):

$$\text{Molar mass} = 2((1 \times 14) + (2 \times 1)) + 12 + 16 \, \text{g/mol} = 60 \, \text{g/mol}$$

 

 

 

 

 

A 0.25 molal (m) solution of urea means that 1000 g of water contains 0.25 mol of urea.

Mass of urea:

$$\text{Mass of urea} = \text{moles of urea} \times \text{molar mass of urea} = 0.25 \, \text{mol} \times 60 \, \text{g/mol} = 15 \, \text{g}$$

 

 

 

 

 

Total mass of solution:

$$\text{Total mass of solution} = 1000 \, \text{g (water)} + 15 \, \text{g (urea)} = 1015 \, \text{g}$$

 

 

 

 

 

For 2.5 kg (2500 g) of solution, the mass of urea is:

$$\text{Mass of urea} = \frac{15 \, \text{g}}{1015 \, \text{g}} \times 2500 \, \text{g} = 36.946 \, \text{g}$$

 

 

 

 

 

Thus, the mass of urea (

$\text{NH}_2\text{CONH}_2$

NH2​CONH2​) required is 36.946 g.

---

Question 5:

Calculate (a) molality, (b) molarity, and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g/mL.

Solution:

(a) Molality

Molar mass of KI:

$$M_{\text{KI}} = 39\ (\text{K}) + 127\ (\text{I}) = 166\ \text{g/mol}$$

 

 

 

 

 

Mass of KI and water in the solution:

$$20\% \, (\text{mass/mass}) \text{ aqueous solution means 20 g of KI in 100 g of solution.}$$

 

 

 

 

 

Mass of water = 100 g (total solution) – 20 g (KI) = 80 g of water.

Moles of KI:

$$\text{Moles of KI} = \frac{20\ \text{g}}{166\ \text{g/mol}} = 0.1205\ \text{mol}$$

 

 

 

 

 

Mass of water in kg:

$$80\ \text{g} = 0.080\ \text{kg}$$

 

 

 

 

 

Molality:

$$\text{Molality} = \frac{0.1205\ \text{mol}}{0.080\ \text{kg}} = 1.506\ \text{m}$$

 

 

 

 

 

Rounded to two decimal places, the molality is 1.51 m.

(b) Molarity

Density of solution:

$$\text{Given as 1.202 g/mL.}$$

 

 

 

 

 

Volume of 100 g solution:

$$\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100\ \text{g}}{1.202\ \text{g/mL}} = 83.19\ \text{mL}$$

 

 

 

 

 

Converting to liters:

$$83.19\ \text{mL} = 0.08319\ \text{L}$$

 

 

 

 

 

Molarity:

$$\text{Molarity} = \frac{0.1205\ \text{mol}}{0.08319\ \text{L}} = 1.45\ \text{M}$$

 

 

 

 

 

(c) Mole Fraction of KI

Moles of water (

$\text{H}_2\text{O}$

H2​O):

$$\text{Molar mass of water} = 18 \text{ g/mol.}$$

 

 

 

 

$$\text{Moles of H}_2\text{O} = \frac{80\ \text{g}}{18\ \text{g/mol}} = 4.444\ \text{mol}$$

 

 

 

 

 

Total moles in the solution:

$$\text{Total moles} = \text{Moles of KI} + \text{Moles of H}_2\text{O} = 0.1205\ \text{mol} + 4.444\ \text{mol} = 4.5645\ \text{mol}$$

 

 

 

 

 

Mole fraction of KI:

$$\text{Mole fraction of KI}=\frac{\text{Moles of KI}}{\text{Total moles}}=\frac{0.1205}{4.5645}=0.0264$$

 

 

 

 

Rounded to four decimal places, the mole fraction of KI is 0.0263.

Solutions class 12 chemistry NCERT solutions Chapter 1 Questions 6 to 10

Question 6:

H

$_2$

2​S is a toxic gas with a rotten egg-like smell and is used for qualitative analysis. If the solubility of H

$_2$

2​S in water is 0.195 m, calculate the Henry’s law constant.

Solution:

Henry’s law states:

 

$$C=k_H\cdot P$$

 

where:

• 
  $C$ 

  $C$ is the concentration of the gas in the solution (mol/L),

• 
  $k_H$ 

  kH​ is the Henry’s law constant (mol/(L·atm)),

• 
  $P$ 

  $P$ is the partial pressure of the gas above the solution (atm).

Given:

• The solubility (concentration) of H
  $_2$ 

  2​S in water is

  $C = 0.195 \, \text{m}$ 

  $C=0.195\text{m}$ (molality). Since molality and molarity can be considered approximately equal for dilute aqueous solutions, we use

  $C \approx 0.195 \, \text{mol/L}$ 

  $C\approx 0.195\text{mol/L}$ (molarity).

• Assuming the partial pressure of H
  $_2$ 

  2​S

  $P = 1 \, \text{atm}$ 

  $P=1\text{atm}$ (standard conditions),

 

$$k_H=\frac{C}{P}=\frac{0.195\text{mol/L}}{1\text{atm<span style="font-family: Georgia, 'Times New Roman', 'Bitstream Charter', Times, serif;"></span>}}$$

 

Therefore,

 

$$k_H=0.195\text{mol/(L}\text{cdotp}\text{atm)}$$

 

---

Question 7:

Henry’s law constant for CO

$_2$

2​ in water is

$1.67 \times 10^8$

1.67×108 Pa at 298 K. Calculate the quantity of CO

$_2$

2​ in 500 mL of soda water when packed under 2.5 atm CO

$_2$

2​ pressure at 298 K.

Solution:

According to Henry’s law:

 

$$p=K_H\cdot x$$

 

where:

• 
  $p$ 

  $p$ is the partial pressure of the gas (Pa),

• 
  $K_H$ 

  KH​ is the Henry’s law constant (Pa),

• 
  $x$ 

  $x$ is the mole fraction of the gas in the solution.

Given:

• 
  $K_H = 1.67 \times 10^8 \, \text{Pa}$ 

  KH​=1.67×108Pa

• 
  $P_{\text{CO}_2} = 2.5 \, \text{atm} = 2.5 \times 1.01325 \times 10^5 \, \text{Pa} = 2.533 \times 10^5 \, \text{Pa}$ 

  PCO2​​=2.5atm=2.5×1.01325×105Pa=2.533×105Pa

To find the mole fraction of CO

$_2$

2​,

$x$

x:

 

$$x = \frac{P_{\text{CO}_2}}{K_H} = \frac{2.533 \times 10^5 \, \text{Pa}}{1.67 \times 10^8 \, \text{Pa}} = 0.00152$$

The mole fraction

$x$

x of CO

$_2$

2​ is given by:

 

$$x = \frac{n_{\text{CO}_2}}{n_{\text{CO}_2} + n_{\text{H}_2\text{O}}}$$

​​​

To find the moles of water:

 

$$\text{Volume of water}, V = 500 \, \text{mL} = 0.5 \, \text{L}$$

 

$$\text{Density of water} = 1 \, \text{g/mL}, \text{ so mass of water} = 500 \, \text{g}$$

 

$$\text{Molar mass of water} = 18 \, \text{g/mol}$$

 

$$n_{\text{H}_2\text{O}} = \frac{500 \, \text{g}}{18 \, \text{g/mol}} = 27.78 \, \text{mol}$$

 

Let

$n_{\text{CO}_2}$

nCO2​​ be the number of moles of CO

$_2$

2​:

 

$$x = \frac{n_{\text{CO}_2}}{n_{\text{CO}_2} + 27.78}$$

​​​

$$0.00152 = \frac{n_{\text{CO}_2}}{n_{\text{CO}_2} + 27.78}$$

​​​

$$n_{\text{CO}_2} = 0.00152 \times (n_{\text{CO}_2} + 27.78)$$

 

$$n_{\text{CO}_2} \approx 0.042 \, \text{mol}$$

 

Finally, the mass of CO

$_2$

2​ is:

 

$$\text{Mass of CO}_2 = \text{moles} \times \text{molar mass}$$

 

$$= 0.042 \, \text{mol} \times 44 \, \text{g/mol} = 1.848 \, \text{g}$$

 

---

Question 8:

The vapour pressure of pure liquids A and B are 450 mm Hg and 700 mm Hg respectively at 350 K. Find the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.

Solution:

Given:

 

$$\text{Vapour pressure of pure liquid A}, P_A^0 = 450 \text{ mm Hg}$$

 

$$\text{Vapour pressure of pure liquid B}, P_B^0 = 700 \text{ mm Hg}$$

 

$$\text{Total vapour pressure of the mixture}, P_{\text{total}} = 600 \text{ mm Hg}$$

 

Using Raoult’s law for a binary mixture:

 

$$P_{\text{total}} = P_A^0 \cdot X_A + P_B^0 \cdot X_B$$

​

Since

$X_A + X_B = 1$

(where

$X_A$

XA​ and

$X_B$

XB​ are the mole fractions of liquids A and B in the mixture, respectively), we have:

 

$$P_{\text{total}} = P_A^0 \cdot X_A + P_B^0 \cdot (1 – X_A)$$

 

Substituting the given values:

 

$$600 = 450 \cdot X_A + 700 \cdot (1 – X_A)$$

 

$$600 = 450X_A + 700 – 700X_A$$

​

$$600 = 700 – 250X_A$$

​

$$250X_A = 700 – 600$$

 

$$250X_A = 100$$

 

$$X_A = \frac{100}{250} = 0.4$$

 

Thus, the mole fraction of liquid A in the mixture is

$X_A = 0.4$

XA​=0.4. The mole fraction of liquid B is:

 

$$X_B = 1 – X_A = 1 – 0.4 = 0.6$$

 

To find the composition of the vapour phase:

 

$$P_A = P_A^0 \times X_A = 450 \times 0.4 = 180 \, \text{mm Hg}$$

 

$$P_B = P_B^0 \times X_B = 700 \times 0.6 = 420 \, \text{mm Hg}$$

 

The mole fraction of A in the vapour phase,

$Y_A$

YA​, is:

 

$$Y_A = \frac{P_A}{P_A + P_B} = \frac{180}{180 + 420} = \frac{180}{600} = 0.30$$

 

The mole fraction of B in the vapour phase,

$Y_B$

YB​, is:

 

$$Y_B = 1 – Y_A = 1 – 0.30 = 0.70$$

 

---

Question 9:

The vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH$_2$CONH$_2$) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Solution:

Given:

 

$$\text{Mass of urea (W}_B\text{)} = 50 \text{ g}$$

 

$$\text{Mass of water (W}_A\text{)} = 850 \text{ g}$$

 

$$\text{Vapour pressure of pure water (P}^0_A\text{)} = 23.8 \text{ mm Hg}$$

 

Molar mass of urea (

$\text{NH}_2\text{CONH}_2$

​):

 

$$M_B = 14 + 1 \times 2 + 12 + 16 + 14 + 1 \times 2 = 60 \, \text{g/mol}$$

 

Molar mass of water (

$\text{M}_A$

) = 18 g/mol

Moles of Compounds

To calculate the moles (

$n$

n), we use the formula:

 

$$\text{moles} (n) = \frac{\text{mass}}{\text{molar mass}}$$

​

Number of moles of urea (

$n_{\text{urea}}$

​):

 

$$n_{\text{urea}} = \frac{W_B}{M_B} = \frac{50}{60} = 0.83 \, \text{mol}$$

 

Number of moles of water (

$n_{\text{H}_2\text{O}}$

​):

 

$$n_{{\text{H}}_2\text{O}}=\frac{W_A}{M_A}=\frac{850}{18}=47.2\text{mol}$$

 

Mole Fraction of Urea

The mole fraction of urea (

$\chi_{\text{urea}}$

χurea​) is calculated as:

 

$${\chi }_{\text{urea}}=\frac{n_{\text{urea}}}{n_{\text{urea}}+n_{{\text{H}}_2\text{O}}}=\frac{0.83}{0.83+47.2}=0.017$$

 

Relative Lowering in Vapour Pressure

The relative lowering of vapour pressure is equal to the mole fraction of the solute (urea):

 

$${\chi }_{\text{urea}}=\frac{P_A^0-P_A}{P_A^0}=0.017$$

 

Thus, the relative lowering of vapour pressure of the solution is 0.017.

Vapour Pressure of Water in Solution

We can find the vapour pressure of water in the solution (

$P_A$

PA​) using the formula:

 

$$\chi_{\text{urea}} = \frac{P^0_A – P_A}{P^0_A}$$

 

$$0.017 = \frac{23.8 – P_A}{23.8}$$

​​

$$P_A = 23.8 – (0.017 \times 23.8)$$

 

$$P_A=23.40\text{mm Hg}$$

 

Thus, the vapour pressure of water in this solution is 23.40 mm Hg, and its relative lowering is 0.017.

---

Question 10:

The boiling point of water at 750 mm Hg is 99.63°C. How much sucrose should be added to 500 g of water so that it boils at 100°C? The molal elevation constant for water is 0.52 K kg mol$^{-1}$.

Solution:

Given:

 

$$\Delta T_b = (100 + 273) – (99.63 + 273) = 0.37 \, \text{K}$$

 

$$\text{Mass of water}, W_1 = 500 \, \text{g}$$

 

$$\text{Molar mass of sucrose} (\text{C}_{12}\text{H}_{22}\text{O}_{11}), M_2 = 12 \times 12 + 22 \times 1 + 11 \times 16 = 342 \, \text{g/mol}$$

 

$$\text{Molal elevation constant}, K_b = 0.52 \, \text{K kg mol}^{-1}$$

 

Using the formula for the elevation of boiling point:

 

$$\Delta T_b = K_b \cdot m$$

 

where

$m$

is the molality of the solution.

First, calculate the molality

$m$

:

 

$$m = \frac{\Delta T_b}{K_b} = \frac{0.37}{0.52} \approx 0.71 \, \text{mol/kg}$$

 

To find the mass of sucrose

$W_2$

​, use the formula for molality:

 

$$m = \frac{n_2}{W_1}$$

​​

where

$n_2$

is the number of moles of sucrose.

Rearranging for

$n_2$

:

 

$$n_2 = m \times W_1 = 0.71 \times 0.5 = 0.355 \, \text{mol}$$

 

Now, convert moles of sucrose to grams:

 

$$W_2=n_2\times M_2=0.355\times 342\text{g}\approx 121.67\text{g}$$

 

Thus, approximately 121.67 g of sucrose needs to be added to the water.

Solutions class 12 chemistry NCERT solutions Chapter 1 Questions 11 to 15

Question 11

Calculate the mass of ascorbic acid (Vitamin C, C

$_6$

6​H

$_8$

8​O

$_6$

6​) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C.

$K_f = 3.9 \, \text{K kg mol}^{-1}$

Kf​=3.9K kg mol−1.

Solution
Given:

• Depression in freezing point,
  $\Delta T_f = 1.5^\circ\text{C}$ 

   

• Cryoscopic constant (
  $K_f$ 

  ​) = 3.9 K kg mol

  $^{-1}$ 

  −1

• Mass of acetic acid (
  $w_A$ 

  ​) = 75 g

• Molar mass of ascorbic acid (
  $M_B$ 

  ​) = 176 g/mol

The formula for depression in freezing point is:

$$\Delta T_f = K_f \times \text{molality} (m)$$

 

 

 

 

 

Molality (

$m$

) is defined as:

$$m=\frac{\text{mass of solute }(w_B)}{\text{molar mass of solute }(M_B)}\times \frac{1}{\text{mass of solvent }(w_A)\text{ in kg<span style="font-family: Georgia, 'Times New Roman', 'Bitstream Charter', Times, serif; font-size: 16px;"></span>}}$$

 

 

 

 

Rearranging for the mass of solute (

$w_B$

):

$$\Delta T_f = K_f \times \frac{w_B}{M_B} \times \frac{1}{w_A}$$

 

 

 

 

​ $$w_B = \frac{\Delta T_f \times M_B \times w_A}{K_f}$$

 

 

 

 

​​

Substituting the known values:

$$w_B = \frac{1.5 \, \text{K} \times 176 \, \text{g/mol} \times 75 \times 10^{-3} \, \text{kg}}{3.9 \, \text{K kg mol}^{-1}}$$

 

 

 

 

​

Calculating the mass of ascorbic acid:

$$w_B = \frac{1.5 \times 176 \times 75 \times 10^{-3}}{3.9} \, \text{g}$$

 

 

 

 

$$w_B \approx 5.077 \, \text{g}$$

 

 

 

 

 

Hence, the required mass of ascorbic acid is approximately 5.077 g.

---

Question 12

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

Solution
Given:

• Mass of polymer (
  $W_B$ 

  ​) = 1 g

• Molar mass of polymer (
  $M_B$ 

  ) = 185,000 g/mol

• Volume of water (
  $V$ 

  ) = 450 mL = 0.45 L

• Temperature (
  $T$ 

  ) = 37°C = 310 K (since

  $T = 37 + 273$ 

  $T$)

The formula for osmotic pressure (

$\pi$

) is:

$$\pi =CRT$$

 

 

 

 

Where:

• 
  $C$ 

  is the molarity of the solution,

• 
  $R$ 

  is the gas constant,

• 
  $T$ 

  is the temperature in Kelvin.

First, calculate the molarity

$C$

of the solution:

$$C = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1}{\text{volume of solution (L)}}$$

 

 

 

 

​ $$C = \frac{1 \, \text{g}}{185,000 \, \text{g/mol}} \times \frac{1}{0.45 \, \text{L}}$$

 

 

 

 

​ $$C \approx 1.204 \times 10^{-5} \, \text{mol/L}$$

 

 

 

 

 

Using the value of the gas constant

$R = 8.314 \times 10^3 \, \text{Pa L K}^{-1} \text{mol}^{-1}$

the osmotic pressure

$\pi$

is given by:

$$\pi = C \times R \times T$$

 

 

 

 

$$\pi = 1.204 \times 10^{-5} \, \text{mol/L} \times 8.314 \times 10^3 \, \text{Pa L K}^{-1} \text{mol}^{-1} \times 310 \, \text{K}$$

 

 

 

 

$$\pi \approx 30.96 \, \text{Pa}$$

 

 

 

 

 

Thus, the osmotic pressure is approximately 30.96 Pa.

---

Question 13

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Solution
Homogeneous mixtures of two or more than two components are known as solutions.

There are three types of solutions:

1. Gaseous solution:
   The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.
2. Liquid solution:
   The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid. For example, a solution of ethanol in water is a liquid solution.
3. Solid solution:
   The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid, or solid. For example, a solution of copper in gold is a solid solution.

---

Question 14

Give an example of a solid solution in which the solute is a gas.

Solution
In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.

---

Question 15

Define the following terms:

(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage

Solution
(i) Mole fraction:
The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture. Mole fraction is denoted by ‘x’.

(ii) Molality:
Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as:

$$\text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent (kg)}}$$

 

 

 

 

​

(iii) Molarity:
Molarity (M) is defined as the number of moles of the solute dissolved in one liter of the solution. It is expressed as:

$$\text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution (L)}}$$

 

 

 

 

​

(iv) Mass percentage:
The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution. It is expressed as:

$$\text{Mass \% of a component} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$$

 

 

 

 

Solutions class 12 chemistry NCERT solutions Chapter 1 Questions 16 to 20

Question 16

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g/mL?

Solution
Given:

• Mass percentage of nitric acid = 68%
• Density of the solution = 1.504 g/mL
• Molar mass of nitric acid (HNO
  $_3$ 

  ) = 63 g/mol

Step 1: Calculate the mass of nitric acid in 100 g of the solution:
68% by mass means that 100 g of the solution contains 68 g of nitric acid.

Step 2: Calculate the number of moles of nitric acid:
Number of moles of nitric acid:

 

$$\text{Moles of HNO}_3 = \frac{\text{Mass of HNO}_3}{\text{Molar mass of HNO}_3}$$

 

$${\text{Moles of HNO}}_3=\frac{68\text{g}}{63\text{g/mol}}=1.079\text{mol}$$

 

Step 3: Calculate the volume of the solution:
Given the density of the solution, the volume of 100 g of solution can be calculated as:

 

$$\text{Volume} = \frac{\text{Mass}}{\text{Density}}$$

​

$$\text{Volume}=\frac{100\text{g}}{1.504\text{g/mL}}\approx 66.5\text{mL}=0.0665\text{L}$$

 

Step 4: Calculate the molarity of the solution:
Molarity (M) is defined as the number of moles of solute per liter of solution:

 

$$\text{Molarity} = \frac{\text{Moles of HNO}_3}{\text{Volume of solution (L)}}$$

​​

$$\text{Molarity} = \frac{1.079 \, \text{mol}}{0.0665 \, \text{L}} \approx 16.22 \, \text{M}$$

 

Thus, the molarity of the concentrated nitric acid solution is approximately 16.22 M.

---

Question 17

A solution of glucose in water is labeled as 10% w/w. What would be the molality and mole fraction of each component in the solution? If the density of the solution is 1.2 g/mL, then what shall be the molarity of the solution?

Solution
Given:

• 10% w/w solution of glucose means 10 g of glucose is present in 100 g of the solution.
• Density of the solution = 1.2 g/mL
• Molar mass of glucose (C
  $_6$ 

  6​H

  $_{12}$ 

  12​O

  $_6$ 

  6​) = 180 g/mol

Step 1: Calculating the mass of water:
Since 10 g of glucose is present in 100 g of the solution, the mass of water is:

 

$$\text{Mass of water} = 100 \, \text{g} – 10 \, \text{g} = 90 \, \text{g}$$

 

Step 2: Calculating the number of moles of glucose:

 

$$\text{Number of moles of glucose} = \frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}$$

​

$$\text{Number of moles of glucose}=\frac{10\text{g}}{180\text{g/mol}}=0.056\text{mol}$$

 

Step 3: Calculating the number of moles of water:
Molar mass of water (H

$_2$

2​O) = 18 g/mol

 

$$\text{Number of moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}}$$

 

$$\text{Number of moles of water}=\frac{90\text{g}}{18\text{g/mol}}=5\text{mol}$$

 

Step 4: Calculating the molality of the solution:
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

 

$$\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$$

​

$$\text{Molality} = \frac{0.056 \, \text{mol}}{0.090 \, \text{kg}} = 0.62 \, \text{m}$$

 

Step 5: Calculating the mole fraction of glucose:

 

$$\text{Mole fraction of glucose} = \frac{\text{Moles of glucose}}{\text{Total moles}}$$

Mole fraction of glucose=Total molesMoles of glucose​

$$\text{Total moles} = \text{Moles of glucose} + \text{Moles of water} = 0.056 + 5 = 5.056$$

Total moles=Moles of glucose+Moles of water=0.056+5=5.056

$$\text{Mole fraction of glucose} = \frac{0.056}{5.056} \approx 0.011$$

 

Step 6: Calculating the mole fraction of water:

 

$$\text{Mole fraction of water}=1-\text{Mole fraction of glucose}=1-0.011=0.989$$

 

Step 7: Calculating the molarity of the solution:
The density of the solution is 1.2 g/mL, and the total mass of the solution is 100 g, so the volume of the solution is:

 

$$\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \, \text{g}}{1.2 \, \text{g/mL}} \approx 83.33 \, \text{mL}$$

 

$$\text{Volume in liters} = 0.08333 \, \text{L}$$

 

$$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}=\frac{0.056\text{mol}}{0.08333\text{L}}\approx 0.67\text{M}$$

 

Final Answer

• Molality of the solution: 0.62 m
• Mole fraction of glucose: 0.011
• Mole fraction of water: 0.989
• Molarity of the solution: 0.67 M

---

Question 18

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na

$_2$

2​CO

$_3$

3​ and NaHCO

$_3$

3​ containing equimolar amounts of both?

Solution
Let the amount of Na

$_2$

2​CO

$_3$

in the mixture be

$x$

x g.

Then, the amount of NaHCO

$_3$

in the mixture is

$1 − x$

g.

Molar mass of Na

$_2$

2​CO

$_3$

​ = 106 g/mol
Molar mass of NaHCO

$_3$

​ = 84 g/mol

Number of moles of Na

$_2$

2​CO

$_3$

:

 

$$\text{Moles of Na}_2\text{CO}_3 = \frac{x}{106} \, \text{mol}$$

 

Number of moles of NaHCO

$_3$

3:

 

$${\text{Moles of NaHCO}}_3=\frac{1-x}{84}\text{mol}$$

 

According to the question:

 

$$\frac{x}{106}=\frac{1-x}{\mathrm{84<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;"></span>}}$$

 

Solving:

 

$$84x = 106 – 106x$$

 

$$190x = 106$$

 

$$x = \frac{106}{190} \approx 0.5579 \, \text{g}$$

 

Thus, the number of moles of Na

$_2$

2​CO

$_3$

:

 

$$\text{Moles of Na}_2\text{CO}_3 = \frac{0.5579}{106} \approx 0.0053 \, \text{mol}$$

 

The number of moles of NaHCO

$_3$

:

 

$${\text{Moles of NaHCO}}_3=\frac{1-0.5579}{84}\approx 0.0053\text{mol}$$

 

Reactions with HCl:

1 mol of Na

$_2$

2​CO

$_3$

3​ reacts with 2 mol of HCl. Therefore, 0.0053 mol of Na

$_2$

2​CO

$_3$

3​ reacts with 2

$\times$

× 0.0053 mol = 0.0106 mol.

1 mol of NaHCO

$_3$

3​ reacts with 1 mol of HCl. Therefore, 0.0053 mol of NaHCO

$_3$

3​ reacts with 0.0053 mol of HCl.

Total moles of HCl required = 0.0106 mol + 0.0053 mol = 0.0159 mol.

Given that 0.1 M HCl contains 0.1 mol of HCl in 1000 mL of solution:

Therefore, the volume of 0.1 M HCl required for 0.0159 mol of HCl:

 

$$\text{Volume of HCl}=\frac{0.0159\text{mol}}{0.1\text{mol/L}}\times 1000\text{mL}\approx 159\text{mL}$$

 

Hence, 159 mL of 0.1 M HCl is required to react completely with 1 g mixture of Na

$_2$

2​CO

$_3$

3​ and NaHCO

$_3$

3​, containing equimolar amounts of both.

---

Question 19

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Solution
Total amount of solute present in the mixture:

 

$$\text{Mass of solute in 25\% solution} = 0.25 \times 300 = 75 \, \text{g}$$

 

$$\text{Mass of solute in 40\% solution} = 0.4 \times 400 = 160 \, \text{g}$$

 

$$\text{Total mass of solute} = 75 \, \text{g} + 160 \, \text{g} = 235 \, \text{g}$$

 

 

$$\text{Total mass of solution} = 300 \, \text{g} + 400 \, \text{g} = 700 \, \text{g}$$

 

 

$$\text{Mass \% of solute} = \frac{235 \, \text{g}}{700 \, \text{g}} \times 100 \approx 33.57\%$$

 

 

$$\text{Mass \% of solvent}=100-33.57\approx 66.43\mathrm{\%}$$

 

---

Question 20

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C

$_2$

2​H

$_6$

6​O

$_2$

2​) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g/mL, then what shall be the molarity of the solution?

Solution

Step 1: Calculating the Molality of the Solution

1. Molar Mass of Ethylene Glycol:
   $$\text{Molar mass of ethylene glycol (C}_2\text{H}_6\text{O}_2) = 62 \, \text{g/mol}$$ 
2. Number of Moles of Ethylene Glycol:
   $$\text{Moles of ethylene glycol}=\frac{222.6\text{g}}{62\text{g/mol}}\approx 3.59\text{mol}$$ 
3. Molality Calculation: Molality (m) is defined as the number of moles of solute per kilogram of solvent.
   $$\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$$​

   $$\text{Molality}=\frac{3.59\text{mol}}{0.200\text{kg}}=17.95\text{m}$$ 

Step 2: Calculating the Molarity of the Solution

1. Total Mass of Solution:
   $$\text{Total mass of solution} = \text{Mass of ethylene glycol} + \text{Mass of water}$$ 

   $$\text{Total mass of solution} = 222.6 \, \text{g} + 200 \, \text{g} = 422.6 \, \text{g}$$ 

2. Volume of Solution:
   $$\text{Volume of solution} = \frac{\text{Total mass of solution}}{\text{Density}}$$​

   $$\text{Volume of solution}=\frac{422.6\text{g}}{1.072\text{g/mL}}\approx 394\text{mL}=0.394\text{L}$$ 

3. Molarity Calculation: Molarity (M) is defined as the number of moles of solute per liter of solution.
   $$\text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution (L)}}$$​

   $$\text{Molarity}=\frac{3.59\text{mol}}{0.394\text{L}}\approx 9.11\text{M}$$ 

Final Answer

• Molality of the solution: 17.95 m
• Molarity of the solution: 9.11 M

 

Solutions class 12 chemistry NCERT solutions Chapter 1 Questions 21 to 30

Question 21:

What role does the molecular interaction play in a solution of alcohol and water?

Solution:

In pure alcohol and water, the molecules are held tightly by strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol−alcohol and water−water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker, and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

---

Question 22:

Why do gases always tend to be less soluble in liquids as the temperature is raised?

Solution:

The solubility of gases in liquids decreases with an increase in temperature. This is because the dissolution of gases in liquids is an exothermic process. When the temperature is increased, heat is supplied, causing the equilibrium to shift backward, thereby decreasing the solubility of gases.

---

Question 23:

State Henry’s law and mention some important applications.

Solution:

Henry’s law states that the partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. Mathematically, it can be expressed as:

 

$$p=K_H\cdot x$$

 

Where:

• 
  $p$ 

  is the partial pressure of the gas,

• 
  $K_H$ 

  is Henry’s law constant,

• 
  $x$ 

  s the mole fraction of the gas in the solution.

Applications of Henry’s Law:

1. Carbonated Beverages: Bottles are sealed under high pressure to increase the solubility of CO
   $_2$ 

   2​ in soft drinks and soda water.

2. Scuba Diving: The increased pressure underwater causes more nitrogen to dissolve in the blood of divers. As they ascend, reduced pressure causes nitrogen to come out of solution, potentially leading to decompression sickness, or “the bends.” Scuba tanks are often filled with a mixture of oxygen and helium to mitigate this.
3. High Altitude Physiology: At high altitudes, the partial pressure of oxygen decreases, leading to lower oxygen levels in the blood and tissues. This can cause symptoms such as weakness and impaired thinking, known as anoxia.

---

Question 24:

The partial pressure of ethane over a solution containing

$6.56 \times 10^{-3} \, \text{g}$

6.56×10−3g of ethane is 1 bar. If the solution contains

$5.00 \times 10^{-2} \, \text{g}$

5.00×10−2g of ethane, what will be the partial pressure of the gas?

Solution:

According to Henry’s law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. This relationship can be expressed as:

 

$$W \cdot P’ = W’ \cdot P$$

 

Where:

• 
  $W$ 

  is the mass of ethane in the first solution,

• 
  $P$ 

  is the partial pressure of ethane in the first solution,

• 
  $W’$ 

  is the mass of ethane in the second solution,

• 
  $P’$ 

  is the partial pressure of ethane in the second solution.

Given:

• 
  $W = 6.56 \times 10^{-3} \, \text{g}$ 

   

• 
  $P = 1 \, \text{bar}$ 

   

• 
  $W’ = 5.00 \times 10^{-2} \, \text{g}$ 

   

We need to find

$P’$

:

 

$$6.56\times 10^{-3}\text{g}\times P^{\mathrm{\prime }}=5.00\times 10^{-2}\text{g}\times 1\text{bar}$$

 

Solving for

$P’$

:

 

$$P^{\mathrm{\prime }}=\frac{5.00\times 10^{-2}}{6.56\times 10^{-3}}\approx 7.62\text{bar}$$

 

Final Answer: The partial pressure of ethane in the solution containing

$5.00 \times 10^{-2} \, \text{g}$

of ethane will be approximately 7.6 bar.

---

Question 25:

What is meant by positive and negative deviations from Raoult’s law and how is the sign of

$\Delta_{\text{mix}} H$

related to positive and negative deviations from Raoult’s law?

Solution:

Raoult’s law states that the partial vapour pressure of each component in an ideal solution is directly proportional to its mole fraction. For an ideal solution, the total vapour pressure is given by:

 

$$P_{\text{total}} = P_A + P_B = P_A^0 X_A + P_B^0 X_B$$

​

Where:

• 
  $P_A^0$ 

  ​ and

  $P_B^0$ 

  are the vapour pressures of pure components A and B, respectively,

• 
  $X_A$ 

  and

  $X_B$ 

  are the mole fractions of components A and B, respectively.

Positive Deviation from Raoult’s Law:

Positive deviation occurs when the interactions between unlike molecules (A and B) are weaker than those between like molecules (A-A and B-B). This results in higher vapour pressure than predicted by Raoult’s law, indicating that the solution is less stable and has a greater tendency to vaporize.

 

$$P_{\text{total}} = P_A + P_B > P_A^0 X_A + P_B^0 X_B$$

​

In such cases, the enthalpy of mixing (

$\Delta_{\text{mix}} H$

Δmix​H) is positive, indicating that the mixing process is endothermic (absorbs heat).

Negative Deviation from Raoult’s Law:

Negative deviation occurs when the interactions between unlike molecules (A and B) are stronger than those between like molecules (A-A and B-B). This results in lower vapour pressure than predicted by Raoult’s law, indicating that the solution is more stable and has a lower tendency to vaporize.

 

$$P_{\text{total}}=P_A+P_B<P_A^0{X}_A+P_B^0{X}_{\mathrm{B<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;\; font-size:\; 16px;"></span>}}$$

 

In these cases, the enthalpy of mixing (

$\Delta_{\text{mix}} H$

) is negative, indicating that the mixing process is exothermic (releases heat).

Summary:

• Positive deviation: Higher vapour pressure than ideal, weaker A-B interactions, positive
  $\Delta_{\text{mix}} H$ 

  $.$

• Negative deviation: Lower vapour pressure than ideal, stronger A-B interactions, negative
  $\Delta_{\text{mix}} H$ 

  $.$

---

Question 26:

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Solution:

Given:

• Vapour pressure of the solution at normal boiling point (
  $p_1$ 

  ) = 1.004 bar

• Vapour pressure of pure water at normal boiling point (
  $p_0$ 

  ​) = 1.013 bar

• Mass of solute (
  $w_2$ 

  ​) = 2 g

• Mass of solvent (water) (
  $w_1$ 

  ) = 98 g

• Molar mass of solvent (water) (
  $M_1$ 

  M1​) = 18 g/mol

Using Raoult’s law, the relative lowering of vapour pressure is:

 

$$\frac{p_0 – p_1}{p_0} = \frac{w_2 \cdot M_1}{w_1 \cdot M_2}$$

​​

Substituting the values:

 

$$\frac{1.013 – 1.004}{1.013} = \frac{2 \cdot 18}{98 \cdot M_2}$$

​

Solving for

$M_2$

 

 

$$M_2\approx 41.35\text{g/mol}$$

 

Final Answer: The molar mass of the solute is approximately 41.35 g/mol.

---

Question 27:

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa, respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Solution:

To find the total vapour pressure of the solution, we use Raoult’s law, which states that the partial vapour pressure of each component in an ideal solution is proportional to its mole fraction and the vapour pressure of the pure component.

Given:

• Vapour pressure of pure heptane (
  $P_{\text{heptane}}^0$ 

  ) = 105.2 kPa

• Vapour pressure of pure octane (
  $P_{\text{octane}}^0$ 

  ) = 46.8 kPa

• Mass of heptane = 26.0 g
• Mass of octane = 35.0 g

1. Molar Masses:

 

$$\text{Molar mass of heptane (C}_7\text{H}_{16}\text{): } 7 \times 12 + 16 \times 1 = 100 \text{ g/mol}$$

 

$${\text{Molar mass of octane (C}}_8{\text{H}}_{18}\text{): }8\times 12+18\times 1=114\text{g/mol}$$

 

2. Number of Moles:

 

$$\text{Number of moles of heptane (}{n}_{\text{heptane}}\text{)}=\frac{26.0\text{ g}}{100\text{ g/mol}}=0.26\text{mol}$$

 

$$\text{Number of moles of octane (}{n}_{\text{octane}}\text{)}=\frac{35.0\text{ g}}{114\text{ g/mol}}=0.31\text{mol}$$

 

3. Mole Fractions:

 

$$\text{Total moles}=n_{\text{heptane}}+n_{\text{octane}}=0.26+0.31=0.57\text{mol}$$

 

$$\text{Mole fraction of heptane (}x_{\text{heptane}}\text{)} = \frac{0.26}{0.57} \approx 0.456$$

 

$$\text{Mole fraction of octane (}x_{\text{octane}}\text{)} = 1 – 0.456 \approx 0.544$$

 

4. Partial Pressures:

 

$$\text{Partial pressure of heptane (}P_{\text{heptane}}\text{)} = x_{\text{heptane}} \times P_{\text{heptane}}^0 = 0.456 \times 105.2 \text{ kPa} \approx 47.97 \text{ kPa}$$

 

$$\text{Partial pressure of octane (}P_{\text{octane}}\text{)} = x_{\text{octane}} \times P_{\text{octane}}^0 = 0.544 \times 46.8 \text{ kPa} \approx 25.46 \text{ kPa}$$

 

5. Total Vapour Pressure:

 

$$P_{\text{total}} = P_{\text{heptane}} + P_{\text{octane}} = 47.97 \text{ kPa} + 25.46 \text{ kPa} \approx 73.43 \text{ kPa}$$

 

Final Answer: The total vapour pressure of the mixture is approximately 73.43 kPa.

---

Question 28:

The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a 1 molal solution of a non-volatile solute in it.

Solution:

To find the vapour pressure of a 1 molal solution of a non-volatile solute in water, we use Raoult’s law, which states that the vapour pressure of the solvent above a solution (

$P_1$

P1​) is equal to the vapour pressure of the pure solvent (

$P^0$

) multiplied by the mole fraction of the solvent in the solution.

Given:

• Vapour pressure of pure water (
  $P^0$ 

  ) at 300 K = 12.3 kPa

• 1 molal solution means 1 mole of solute in 1000 g of solvent (water).

1. Moles of Water:

 

$$\text{Molar mass of water (H}_2\text{O)} = 18 \text{ g/mol}$$

 

$$\text{Mass of water} = 1000 \text{ g}$$

 

$$\text{Moles of water} = \frac{1000 \text{ g}}{18 \text{ g/mol}} = 55.56 \text{ mol}$$

 

2. Mole Fraction of Solvent (Water):

 

$$\text{Moles of solute} = 1 \text{ mol}$$

 

$$\text{Total moles} = \text{Moles of solute} + \text{Moles of solvent} = 1 + 55.56 = 56.56 \text{ mol}$$

 

$$\text{Mole fraction of water (}x_{\text{water}}\text{)} = \frac{\text{Moles of water}}{\text{Total moles}} = \frac{55.56}{56.56}$$

​

3. Vapour Pressure of Solution:

Using Raoult’s law:

 

$$P_1 = x_{\text{water}} \times P^0$$

 

$$P_1 = \left( \frac{55.56}{56.56} \right) \times 12.3 \text{ kPa}$$

 

$$P_1 \approx \left( 0.9823 \right) \times 12.3 \text{ kPa}$$

 

$$P_1 \approx 12.08 \text{ kPa}$$

 

Final Answer: The vapour pressure of the 1 molal solution of the non-volatile solute in water is approximately 12.08 kPa.

---

Question 29:

Calculate the mass of a non-volatile solute (molar mass 40 g/mol) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Solution:

Let

$p$

p be the vapour pressure of pure octane. The vapour pressure of the solution is given as 80% of

$p$

p, which can be written as:

 

$$p’ = 0.8p$$

 

The molar mass of the solute (M) and octane (m) are 40 g/mol and 114 g/mol, respectively. The mass of octane,

$w$

w, is 114 g. According to Raoult’s law, the relative lowering of vapour pressure is proportional to the mole fraction of the solute in the solution:

 

$$\frac{p – p’}{p} = \text{mole fraction of solute}$$

 

Substituting the known values:

 

$$\frac{p – 0.8p}{p} = 0.2$$

 

This gives:

 

$$0.2 = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$$

​​

Where:

• 
  $n_{\text{solute}}$ 

  = moles of solute =

  $\frac{W}{M}$ 

  ​

• 
  $n_{\text{solvent}}$ 

  = moles of octane =

  $\frac{w}{m}$ 

  ​

Substituting values:

 

$$0.2 = \frac{\frac{W}{40}}{\frac{W}{40} + \frac{114}{114}}$$

​​

$$0.2 = \frac{\frac{W}{40}}{\frac{W}{40} + 1}$$

0.2=40W​+140W​​

 

 

$$0.2\left(\frac{W}{40} + 1\right) = \frac{W}{40}$$

 

$$0.2 = \frac{W}{40} – 0.2 \cdot \frac{W}{40}$$

​

$$0.2 = 0.8 \cdot \frac{W}{40}$$

​

$$W = \frac{0.2 \times 40}{0.8} = 10 \, \text{g}$$

 

Final Answer: 10 g of the non-volatile solute is required.

---

Question 30:

A solution containing 30 g of non-volatile solute is dissolved in 90 g of water, resulting in a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is added to the solution, increasing the vapour pressure to 2.9 kPa at 298 K. Calculate: (i) The molar mass of the solute (ii) The vapour pressure of water at 298 K.

Solution:

Let the molar mass of the solute be

$M_B$

and the vapour pressure of pure water at 298 K be

$P^0_A$

 

Initial Solution:

• Weight of solute,
  $W_B = 30$ 

   

• Weight of water,
  $W_A = 90$ 

  $,$

• Vapour pressure of solution,
  $P_A = 2.8$ 

According to Raoult’s law:

 

$$\frac{P^0_A – P_A}{P^0_A} = X_B \approx \frac{W_B \cdot M_A}{M_B \cdot W_A}$$

​​

$$\frac{P^0_A – 2.8}{P^0_A} = \frac{30 \cdot 18}{M_B \cdot 90}$$

​

$$2.8P^0_A = M_B – 6M_B \quad \dots \quad (1)$$

 

After Adding Water:

• Weight of solute,
  $W_B = 30$ 
• Weight of water,
  $W_A = 90 + 18 = 108$ 
• Vapour pressure of solution,
  $P_A = 2.9$ 

  PA​

According to Raoult’s law:

 

$$\frac{P^0_A – P_A}{P^0_A} = X_B \approx \frac{W_B \cdot M_A}{M_B \cdot W_A}$$

​​

$$\frac{P^0_A – 2.9}{P^0_A} = \frac{30 \cdot 18}{M_B \cdot 108}$$

​

$$2.9P^0_A = M_B – 5M_B \quad \dots \quad (2)$$

 

Divide equation (1) by equation (2):

 

$$\frac{2.8}{2.9} = \frac{M_B – 6}{M_B – 5}$$

​

Solving for

$M_B$

MB​:

 

$$M_B = 34 \, \text{g/mol}$$

 

Substituting the value of

$M_B$

MB​ in equation (1):

 

$$2.8P^0_A = \frac{34 – 6}{34}$$

​

$$P_A^0=3.4\text{kPa}$$

 

Final Answer: The molar mass of the solute is 34 g/mol, and the vapour pressure of water at 298 K is 3.4 kPa.

Solutions class 12 chemistry NCERT solutions Chapter 1 Questions 31 to 40

Question 31:

A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of a 5% glucose solution in water if the freezing point of pure water is 273.15 K.

Solution: A 5% solution by mass of cane sugar means 100 g of solution contains 5 g of cane sugar. The molecular weight of cane sugar (

$\text{C}_{12}\text{H}_{22}\text{O}_{11}$

​) is 342 g/mol.

The number of moles of cane sugar is:

 

$$\text{Number of moles of cane sugar}=\frac{5\text{g}}{342\text{g/mol}}=0.0146\text{mol}$$

 

The mass of water in the solution is:

 

$$\text{Mass of water}=100\text{g}-5\text{g}=95\text{g}=0.095\text{kg}$$

 

The molality

$m$

m of the solution is:

 

$$m=\frac{0.0146\text{mol}}{0.095\text{kg}}=0.154\text{mol/kg}$$

 

The depression in the freezing point

$\Delta T_f$

is:

 

$$\Delta T_f = 273.15 \, \text{K} – 271 \, \text{K} = 2.15 \, \text{K}$$

Using the formula

$\Delta T_f = K_f \cdot m$

, we can find the cryoscopic constant

$K_f$

:

 

$$K_f = \frac{\Delta T_f}{m} = \frac{2.15 \, \text{K}}{0.154 \, \text{mol/kg}} = 13.97 \, \text{K} \cdot \text{kg/mol}$$

Now, for the glucose solution:

A 5% by mass solution of glucose means 100 g of solution contains 5 g of glucose. The molecular weight of glucose (

$\text{C}_6\text{H}_{12}\text{O}_6$

) is 180 g/mol.

The number of moles of glucose is:

 

$$\text{Number of moles of glucose}=\frac{5\text{g}}{180\text{g/mol}}=0.0277\text{mol}$$

 

The mass of water in the solution is:

 

$$\text{Mass of water}=100\text{g}-5\text{g}=95\text{g}=0.095\text{kg}$$

 

The molality

$m$

m of the glucose solution is:

 

$$m=\frac{0.0277\text{mol}}{0.095\text{kg}}=0.292\text{mol/kg}$$

 

The depression in the freezing point for the glucose solution is:

 

$$\mathrm{\Delta }{T}_f=K_f\cdot m=13.97\text{K}\cdot \text{kg/mol}\times 0.292\text{mol/kg}=4.09\text{K}$$

 

The freezing point of the glucose solution is:

 

$$$$

 

---

Question 32:

Two elements A and B form compounds having formula AB

$_2$

and AB

$_4$

. When dissolved in 20 g of benzene (

$\text{C}_6\text{H}_6$

), 1 g of AB

$_2$

lowers the freezing point by 2.3 K whereas 1.0 g of AB

$_4$

​ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol

$^{-1}$

−1. Calculate the atomic masses of A and B.

Solution: We know that,

 

$$\mathrm{\Delta }{T}_f=K_f\cdot m$$

 

For AB

$_2$

:

 

$$\mathrm{\Delta }{T}_f=2.3\text{K},K_f=5.1{\text{K kg mol}}^{-1}$$

 

$$m = \frac{1 \, \text{g}}{\text{Molar mass of AB}_2} \times \frac{1}{0.02 \, \text{kg}}$$

​

$$2.3 = 5.1 \times \frac{1}{\text{Molar mass of AB}_2} \times \frac{1}{0.02}$$

​

$$\text{Molar mass of AB}_2 = 110.87 \, \text{g/mol}$$

For AB

$_4$

:

$$m = \frac{1 \, \text{g}}{\text{Molar mass of AB}_4} \times \frac{1}{0.02 \, \text{kg}}$$

 

$$1.3 = 5.1 \times \frac{1}{\text{Molar mass of AB}_4} \times \frac{1}{0.02}$$

​

$${\text{Molar mass of AB}}_4=196.15\text{g/mol}$$

 

Let the atomic masses of A and B be

$x$

and

$y$

respectively.

For AB

$_2$

:

 

$$x+2y=110.87\dots (i)$$

 

For AB

$_4$

​:

 

$$x+4y=196.15\dots (ii)$$

 

Subtracting equation (i) from (ii), we have

 

$$2y = 85.28$$

 

$$y = 42.64$$

 

Putting the value of

$y$

y in equation (i), we have

 

$$x + 2 \times 42.64 = 110.87$$

 

$$x = 25.59$$

 

Final Answer: The atomic masses of A and B are 25.59 u and 42.64 u, respectively.

---

Question 33:

At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Solution: Given:

• 
  $T = 300 \, \text{K}$ 

  $T$

• 
  $\pi = 1.52 \, \text{bar}$ 
• 
  $R = 0.083 \, \text{bar L K}^{-1} \text{mol}^{-1}$ 

Using the relation:

 

$$\pi = CRT$$

 

$$C = \frac{\pi}{RT} = \frac{1.52 \, \text{bar}}{0.083 \, \text{bar L K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}} = 0.061 \, \text{mol/L}$$

 

Final Answer: The concentration of the solution is 0.061 M.

---

Question 34:

Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n-hexane and n-octane
(ii)

$\text{I}_2$

​ and

$\text{CCl}_4$

​
(iii)

$\text{NaClO}_4$

and water
(iv) methanol and acetone
(v) acetonitrile (

$\text{CH}_3\text{CN}$

) and acetone (

$\text{C}_3\text{H}_6\text{O}$

).

Solution:

(i) Van der Waals forces of attraction.
(ii) Van der Waals forces of attraction.
(iii) Ion-dipole interaction.
(iv) Dipole-dipole interaction.
(v) Dipole-dipole interaction.

---

Question 35:

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain.

Cyclohexane, KCl, CH

$_3$

​OH, CH

$_3$

CN.

Solution: n-Octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in n-octane.

The order of increasing polarity is:

Cyclohexane < CH

$_3$

CN < CH

$_3$

​OH < KCl

Therefore, the order of increasing solubility is:

KCl < CH

$_3$

OH < CH

$_3$

​CN < Cyclohexane

---

Question 36:

Amongst the following compounds, identify which are insoluble, partially soluble, and highly soluble in water?

(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.

Solution:

(i) Phenol (C

$_6$

H

$_5$

​OH) has the polar group –OH and non-polar group –C

$_6$

H

$_5$

​. Thus, phenol is partially soluble in water.
(ii) Toluene (C

$_6$

H

$_5$

​–CH

$_3$

) has no polar groups. Thus, toluene is insoluble in water.
(iii) Formic acid (HCOOH) has the polar group –OH and can form H-bond with water. Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol has polar –OH group and can form H–bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi) Pentanol (C

$_5$

5​H

$_{11}$

OH) has polar –OH group, but it also contains a very bulky non-polar –C

$_5$

5​H

$_{11}$

​ group. Thus, pentanol is partially soluble in water.

---

Question 37:

If the density of some lake water is 1.25 g/mL and contains 92 g of Na

$^+$

+ ions per kg of water, calculate the molarity of Na

$^+$

+ ions in the lake.

Solution:

Step 1:

Mass of sodium ions = 92 g
Molar mass of sodium ions = 23 g/mol

Number of moles of sodium ions:

 

$$\text{Number of moles of sodium ions} = \frac{92 \, \text{g}}{23 \, \text{g/mol}} = 4 \, \text{mol}$$

 

Mass of water = 1 kg
Volume of water:

 

$$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{1000 \, \text{g}}{1.25 \, \text{g/mL}} = 800 \, \text{mL} = 0.8 \, \text{L}$$

 

Step 2:

As we know, molarity (\textit{M}) is defined as:

 

$$\text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$$

​

$$\text{Molarity of } \text{Na}^+ \text{ ions (M)} = \frac{\text{Number of moles of Na}^+ \text{ ions}}{\text{Volume of solution in litres}}$$

​

$$\text{Molarity}=\frac{4\text{mol}}{0.8\text{L}}=5\text{mol/L or }5\text{M}$$

 

---

Question 38:

If the solubility product of CuS is

$6 \times 10^{-16}$

6×10−16, calculate the maximum molarity of CuS in aqueous solution.

Solution:

Solubility product of CuS,

$K_{sp}$

​ =

$6 \times 10^{-16}$

 

Let

$s$

s be the solubility of CuS in mol L

$^{-1}$

 

Now,

 

$$K_{sp} = s^2$$

 

$$s = \sqrt{K_{sp}} = \sqrt{6 \times 10^{-16}} = 2.45 \times 10^{-8} \, \text{mol/L}$$

 

Final Answer: The maximum molarity of CuS in an aqueous solution is

$2.45 \times 10^{-8}$

2.45×10−8 mol L

$^{-1}$

−1.

---

Question 39:

Calculate the mass percentage of aspirin (

$\text{C}_9\text{H}_8\text{O}_4$

​) in acetonitrile (

$\text{CH}_3\text{CN}$

) when 6.5 g of

$\text{C}_9\text{H}_8\text{O}_4$

is dissolved in 450 g of

$\text{CH}_3\text{CN}$

.

Solution:

Given:
6.5 g of

$\text{C}_9\text{H}_8\text{O}_4$

is dissolved in 450 g of

$\text{CH}_3\text{CN}$

.

Then, the total mass of the solution:

 

$$\text{Total mass of the solution} = 6.5 \, \text{g} + 450 \, \text{g} = 456.5 \, \text{g}$$

 

Mass percentage of

$\text{C}_9\text{H}_8\text{O}_4$

C9​H8​O4​:

 

$$\text{Mass percentage of }{\text{C}}_9{\text{H}}_8{\text{O}}_4=\frac{6.5\text{g}}{456.5\text{g}}\times 100=1.424\mathrm{\%}$$

 

Final Answer: The mass percentage of aspirin in acetonitrile is 1.424%.

---

Question 40:

Nalorphene (C

$_{19}$

​H

$_{21}$

​NO

$_3$

), similar to morphine, is used to combat withdrawal symptoms in narcotic users. A dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5

$\times$

× 10

$^{-3}$

−3 m aqueous solution required for the above dose.

Solution:

The molar mass of nalorphene (C

$_{19}$

H

$_{21}$

​NO

$_3$

​) is given as:

In 1.5

$\times$

× 10

$^{-3}$

−3 m aqueous solution of nalorphene, 1 kg (1000 g) of water contains 1.5

$\times$

× 10

$^{-3}$

−3 mol of nalorphene.

The molar mass of nalorphene (C

$_{19}$

​H

$_{21}$

​NO

$_3$

​) can be calculated as follows:

 

$$\text{Molar mass of nalorphene}=19\times 12+21\times 1+1\times 14+3\times 16=311\text{g/mol}$$

 

Therefore, the mass of nalorphene in the solution is:

 

$$\text{Mass of nalorphene} = 1.5 \times 10^{-3} \, \text{mol} \times 311 \, \text{g/mol} = 0.4665 \, \text{g}$$

 

The total mass of the solution is the mass of the water plus the mass of the nalorphene:

 

$$\text{Total mass of the solution} = 1000 \, \text{g (water)} + 0.4665 \, \text{g (nalorphene)} = 1000.4665 \, \text{g}$$

 

To find the mass of the solution containing 1.5 mg of nalorphene, we use the proportion:

 

$$\text{Mass of the solution}=\frac{1000.4665\text{g}}{0.4665\text{g}}\times 1.5\text{mg}=3.22\text{g}$$

 

Final Answer: The mass of the aqueous solution required is 3.22 g.

Note: There is a slight variation in this answer and the one given in the NCERT textbook.

Solutions class 12 chemistry NCERT solutions Chapter 1 Questions 41 to 52

Question 41:

Calculate the amount of benzoic acid (C

$_6$

​H

$_5$

COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Solution: A 0.15 M solution of benzoic acid in methanol means:

1000 mL of solution contains 0.15 mol of benzoic acid.

Therefore, 250 mL of solution contains:

 

$$\text{Moles of benzoic acid}=0.15\text{mol/L}\times \frac{250\text{mL}}{1000\text{mL/L}}=0.0375\text{mol}$$

 

Molar mass of benzoic acid (C

$_6$

​H

$_5$

​COOH):

 

$$\text{Molar mass} = 7 \times 12 + 6 \times 1 + 2 \times 16 = 122 \, \text{g/mol}$$

 

Hence, required benzoic acid:

 

$$\text{Mass of benzoic acid}=0.0375\text{mol}\times 122\text{g/mol}=4.575\text{g}$$

 

---

Question 42:

The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid, and trifluoroacetic acid increases in the order given above. Explain briefly.

Solution: The observed depression in the freezing point is related to the number of particles in the solution, which is influenced by the degree of dissociation of the acids. When strongly electron-withdrawing groups are present on the α-carbon atom of acetic acid, the acid strength and the degree of dissociation increase. This results in a higher van’t Hoff factor,

$i$

i, and consequently a greater depression in the freezing point.

• Acetic acid (
  ${\text{CH}}_3\text{COOH<span style="font-family: Georgia, 'Times New Roman', 'Bitstream Charter', Times, serif;">) has no electron-withdrawing groups on the α-carbon, resulting in the lowest degree of dissociation among the three acids.</span>}$ 
• Trichloroacetic acid (
  $\text{CCl}_3\text{COOH}$ 

  ) contains three chlorine atoms, which are electron-withdrawing groups. This increases the acid’s strength and degree of dissociation compared to acetic acid, leading to a greater depression in the freezing point.

• Trifluoroacetic acid (
  $\text{CF}_3\text{COOH}$ 

  ) contains three fluorine atoms, which are even more strongly electron-withdrawing than chlorine atoms. This makes trifluoroacetic acid the most acidic of the three, with the highest degree of dissociation. Therefore, it has the maximum depression in the freezing point.

Thus, the depression in the freezing point increases in the order: acetic acid < trichloroacetic acid < trifluoroacetic acid, due to the increasing electron-withdrawing effect and acid strength, which enhances the degree of dissociation and the van’t Hoff factor,

$i$

i.

---

Question 43:

Calculate the depression in the freezing point of water when 10 g of CH

$_3$

​CH

$_2$

CHClCOOH is added to 250 g of water. Given:

$K_a = 1.4 \times 10^{-3}$

,

$K_f = 1.86 \, \text{K kg mol}^{-1}$

 

Solution: Molar mass of CH

$_3$

3​CH

$_2$

CHClCOOH:

 

$$\text{Molar mass}=(4\times 12.01)+(7\times 1.01)+(1\times 35.45)+(2\times 16.00)=122.56\text{g/mol}$$

 

Number of moles of CH

$_3$

CH

$_2$

CHClCOOH:

 

$$\text{Number of moles}=\frac{10\text{g}}{122.56\text{g/mol}}\approx 0.0816\text{mol}$$

 

Molality of the solution:

 

$$\text{Molality} (m) = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.0816 \, \text{mol}}{0.250 \, \text{kg}} = 0.3264 \, \text{mol/kg}$$

 

Let

$\alpha$

be the degree of dissociation of CH

$_3$

​CH

$_2$

CHClCOOH. The dissociation equation is:

 

$$\text{CH}_3\text{CH}_2\text{CHClCOOH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{CHClCOO}^- + \text{H}^+$$

 

The expression for the dissociation constant,

$K_a$

, is:

 

$$K_a=\frac{[{\text{CH}}_3{\text{CH}}_2{\text{CHClCOO}}^{-}][{\text{H}}^{+}]}{[{\text{CH}}_3{\text{CH}}_2\text{CHClCOOH}]<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;"></span>}$$

 

Assuming

$\alpha$

is small,

$1 – \alpha \approx 1$

 

 

$$K_a \approx \alpha^2 C$$

 

Where

$C$

is the initial concentration of the solute:

 

$$C = 0.3264 \, \text{mol/kg}$$

 

Solving for

$\alpha$

:

 

$$\alpha =\sqrt{\frac{K_a}{C}}=\sqrt{\frac{1.4\times 10^{-3}}{0.3264}}\approx 0.0655$$

 

Total moles at equilibrium:

 

$$1-\alpha +\alpha +\alpha =1+\alpha =1+0.0655=1.0655$$

 

Van’t Hoff factor,

$i = 1 + \alpha = 1.0655$

i

Depression in the freezing point:

 

$$\mathrm{\Delta }{T}_f=i\cdot K_f\cdot m=1.0655\times 1.86\times 0.3264\approx 0.646\text{K}$$

 

Therefore, the depression in the freezing point of water is approximately 0.646 K.

---

Question 44:

Calculate the depression in the freezing point of water when 10 g of CH

$_3$

​CH

$_2$

CHClCOOH is added to 250 g of water. Given:

$K_a = 1.4 \times 10^{-3}$

,

$K_f = 1.86 \, \text{K kg mol}^{-1}$

 

Solution: Molar mass of CH

$_3$

​CH

$_2$

CHClCOOH:

 

$$\text{Molar mass} = (4 \times 12.01) + (7 \times 1.01) + (1 \times 35.45) + (2 \times 16.00) = 122.56 \, \text{g/mol}$$

Number of moles of CH

$_3$

CH

$_2$

CHClCOOH:

 

$$\text{Number of moles}=\frac{10\text{g}}{122.56\text{g/mol}}\approx 0.0816\text{mol}$$

 

Molality of the solution:

 

$$\text{Molality} (m) = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.0816 \, \text{mol}}{0.250 \, \text{kg}} = 0.3264 \, \text{mol/kg}$$

 

Let

$\alpha$

be the degree of dissociation of CH

$_3$

CH

$_2$

CHClCOOH. The dissociation equation is:

 

$${\text{CH}}_3{\text{CH}}_2\text{CHClCOOH}\rightleftharpoons {\text{CH}}_3{\text{CH}}_2{\text{CHClCOO}}^{-}+{\text{H}}^{+}$$

 

The expression for the dissociation constant,

$K_a$

is:

 

$$K_a = \frac{[\text{CH}_3\text{CH}_2\text{CHClCOO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{CHClCOOH}]}$$

​

Assuming

$\alpha$

is small,

$1 – \alpha \approx 1$

 

 

$$K_a \approx \alpha^2 C$$

 

Where

$C$

is the initial concentration of the solute:

 

$$C = 0.3264 \, \text{mol/kg}$$

 

Solving for

$\alpha$

:

 

$$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.4 \times 10^{-3}}{0.3264}} \approx 0.0655$$

 

Total moles at equilibrium:

 

$$1 – \alpha + \alpha + \alpha = 1 + \alpha = 1 + 0.0655 = 1.0655$$

 

Van’t Hoff factor,

$i = 1 + \alpha = 1.0655$

Depression in the freezing point:

 

$$\mathrm{\Delta }{T}_f=i\cdot K_f\cdot m=1.0655\times 1.86\times 0.3264\approx 0.646\text{K}$$

 

Therefore, the depression in the freezing point of water is approximately 0.646 K.

---

Question 45:

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Solution: Vapour pressure of pure water,

$P^0_{\text{water}} = 17.535$

mm Hg
Mass of glucose,

$w_2$

=25 g
Mass of water,

$w_1$

​=450 g

Molar masses:

 

$${\text{Molar mass of glucose (C}}_6{\text{H}}_{12}{\text{O}}_6\text{)},M_2=6\times 12+12\times 1+6\times 16=180\text{g/mol}$$

 

$$\text{Molar mass of water,} M_1 = 18 \text{ g/mol}$$

 

 

 

$$n_2 = \frac{w_2}{M_2} = \frac{25}{180} \approx 0.139 \text{ mol}$$

 

Number of moles of water:

 

$$n_1 = \frac{w_1}{M_1} = \frac{450}{18} = 25 \text{ mol}$$

 

Using Raoult’s law, the relative lowering of vapour pressure is given by:

 

$$\frac{P^0_{\text{water}} – P_{\text{solution}}}{P^0_{\text{water}}} = \frac{n_2}{n_1 + n_2}$$

​​

$$\frac{17.535 – P_{\text{solution}}}{17.535} = \frac{0.139}{25 + 0.139}$$

​

$$17.535 – P_{\text{solution}} = 17.535 \times \frac{0.139}{25.139} \approx 0.097$$

 

$$P_{\text{solution}} = 17.535 – 0.097 = 17.438 \text{ mm Hg}$$

 

Final Answer: Hence, the vapour pressure of the solution is approximately 17.44 mm Hg.

---

Question 46:

Henry’s law constant for the molality of methane in benzene at 298 K is

$4.27 \times 10^5$

mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Solution: Given:

 

$$p = 760 \text{ mm Hg}, \quad k_H = 4.27 \times 10^5 \text{ mm Hg}$$

 

According to Henry’s law:

 

$$p = k_H \cdot x$$

 

Where

$x$

is the mole fraction of methane in benzene.

 

$$x = \frac{p}{k_H} = \frac{760 \text{ mm Hg}}{4.27 \times 10^5 \text{ mm Hg}} \approx 1.78 \times 10^{-3}$$

 

Final Answer: Hence, the mole fraction of methane in benzene is approximately $1.78 \times 10^{-3}$

---

Question 47:

100 g of liquid A (molar mass 140 g/mol) was dissolved in 1000 g of liquid B (molar mass 180 g/mol). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Solution: Number of moles:

For liquid A:

 

$$n_A = \frac{100 \text{ g}}{140 \text{ g/mol}} = 0.714 \text{ mol}$$

 

For liquid B:

 

$$n_B = \frac{1000 \text{ g}}{180 \text{ g/mol}} = 5.556 \text{ mol}$$

 

Mole fractions:

The mole fraction of A:

 

$$X_A = \frac{n_A}{n_A + n_B} = \frac{0.714}{0.714 + 5.556} \approx 0.114$$

 

The mole fraction of B:

 

$$X_B = 1 – X_A = 1 – 0.114 = 0.886$$

 

Total vapour pressure:

Given:

 

$$P_{\text{total}} = 475 \text{ torr}, \quad P^0_B = 500 \text{ torr}$$

 

Using Raoult’s law:

 

$$P_{\text{total}} = P_A + P_B = P^0_A \cdot X_A + P^0_B \cdot X_B$$

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Substituting the known values:

 

$$475 = P^0_A \times 0.114 + 500 \times 0.886$$

 

$$475 = 0.114 P^0_A + 443$$

 

$$0.114 P^0_A = 475 – 443 = 32$$

 

$$P^0_A = \frac{32}{0.114} \approx 280.7 \text{ torr}$$

 

Vapour pressure of A in the solution:

 

$$P_A = P^0_A \times X_A = 280.7 \times 0.114 \approx 32 \text{ torr}$$

 

Final Answer: The vapour pressure of pure liquid A is approximately 280.7 torr, and its vapour pressure in the solution is approximately 32 torr.