NCERT Exemplar for Class 11 Chemistry Chapter 1

SimplyAcad has provided the NCERT Exemplar for Class 11 Chemistry below to help students learn about the different topics in a detailed manner. The exemplar will allow students to gain deep insights of all the sections and prepares you better for the upcoming examination. Chemistry is an interesting subject which requires attention to minor details, hence, completing exemplars will be an effective way to increase your marks and boost confidence in you. The given exemplar contains MCQs of two different types, Short, Matching, Assertion and Reason, and Long Answer Type Questions, there are a total of 45 questions. Students can access the NCERT exemplar for class 11 chemistry, Chapter 1: Some Basic Concepts of Chemistry by scrolling below. Along with this, there are several NCERT exemplar for class 11 science of all the chapters provided on this platform.

Access the Solutions of NCERT Exemplar Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

NCERT Exemplar Class 11 Chemistry Chapter 1 : Multiple Choice Questions (Type-I)

Question 1

Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below. The correct reading of mass is 3.0 g. Based on given data, mark the correct option out of the following statements. Student Readings

	(i)	(ii)
A	3.01	2.99
B	3.05	2.95

(i) Results of both the students are neither accurate nor precise.

(ii) Results of student A are both precise and accurate.

(iii) Results of student B are neither precise nor accurate.

(iv) Results of student B are both precise and accurate.

Solution:

Option (ii) is the answer.

Question 2

A measured temperature on the Fahrenheit scale is 200 °F. What will this reading

be on a Celsius scale?

(i) 40 °C

(ii) 94 °C

(iii) 93.3 °C

(iv) 30 °C

Solution:

Option (iii) is the answer.

Question 3

What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per

500 mL?

(i) 4 mol L-1

(ii) 20 molL-1

(iii) 0.2 molL-1

(iv) 2molL-1

Solution:

Option (iii) is the answer.

Question 4

If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of

the solution obtained?

(i) 1.5 M

(ii) 1.66 M

(iii) 0.017 M

(iv) 1.59 M

Solution:

Option (ii) is the answer.

Question 5

The number of atoms present in one mole of an element is equal to Avogadro

number. Which of the following element contains the greatest number of

atoms?

(i) 4g He

(ii) 46g Na

(iii) 0.40g Ca

(iv) 12g He

Solution:

Option (iv) is the answer.

Question 6

If the concentration of glucose (C6H12O6) in the blood is 0.9 g L-1, what will be the

molarity of glucose in the blood?

(i) 5 M

(ii) 50 M

(iii) 0.005 M

(iv) 0.5 M

Solution:

Option (iii) is the answer.

Question 7

What will be the molality of the solution containing 18.25 g of HCl gas in

500 g of water?

(i) 0.1 m

(ii) 1 M

(iii) 0.5 m

(iv) 1 m

Solution:

Option (iv) is the answer.

Question 8

One mole of any substance contains 6.022 × 1023 atoms/molecules. Number

of molecules of H2SO4 present in 100 mL of 0.02M H2SO4 solution is ______.

(i) 12.044 × 1020 molecules

(ii) 6.022 × 1023 molecules

(iii) 1 × 1023 molecules

(iv) 12.044 × 1023molecules

Solution:

Option (i) is the answer.

Question 9

What is the mass per cent of carbon in carbon dioxide?

(i) 0.034%

(ii) 27.27%

(iii) 3.4%

(iv) 28.7%

Solution:

Option (ii) is the answer.

Question 10

The empirical formula and molecular mass of a compound are CH2O and

180 g respectively. What will be the molecular formula of the compound?

(i) C9H18O9

(ii) CH2O

(iii) C6H12O6

(iv) C2H4O2

Solution:

Option (iii) is the answer.

Question 11

If the density of a solution is 3.12 g mL-1, the mass of 1.5 mL solution in

significant figures is _______.

(i) 4.7g

(ii) 4680 × 10 -3g

(iii) 4.680g

(iv) 46.80g

Solution:

Option (i) is the answer

Question 12

Which of the following statements about a compound is incorrect?

(i) A molecule of a compound has atoms of different elements.

(ii) A compound cannot be separated into its constituent elements by

physical methods of separation.

(iii) A compound retains the physical properties of its constituent elements.

(iv) The ratio of atoms of different elements in a compound is fixed.

Solution:

Option (iii) is the answer.

Question 13

 Which of the following statements is correct about the reaction given below:

4Fe(s) + 3O2(g) → 2Fe2O3(g)

(i) The total mass of iron and oxygen in reactants = total mass of iron and

oxygen in product therefore it follows the law of conservation of mass.

(ii) The total mass of reactants = total mass of product; therefore, the law of multiple

proportions is followed.

(iii) Amount of Fe2O3 can be increased by taking any one of the reactants

(iron or oxygen) in excess.

(iv) Amount of Fe2O3 produced will decrease if the amount of any one of the

reactants (iron or oxygen) is taken in excess.

Solution:

Option (i) is the answer.

Question 14

Which of the following reactions is not correct according to the law of

conservation of mass.

(i) 2Mg(s) + O2(g) →2MgO(s)

(ii) C3H8(g) + O2(g) → CO2(g) + H2O(g)

(iii) P4(s) + 5O2(g) → P4O10(s)

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O (g)

Solution:

Option (ii) is the answer.

Question 15

Which of the following statements indicates that the law of multiple proportions is

being followed.

(i) Sample of carbon dioxide taken from any source will always have carbon

and oxygen in the ratio 1:2.

(ii) Carbon forms two oxides namely CO2 and CO, where masses of oxygen

which combine with a fixed mass of carbon are in the simple ratio 2:1.

(iii) When magnesium burns in oxygen, the amount of magnesium taken

for the reaction is equal to the amount of magnesium in magnesium

oxide formed.

(iv) At constant temperature and pressure, 200 mL of hydrogen will combine

with 100 mL oxygen to produce 200 mL of water vapour.

Solution:

Option (ii) is the answer.

NCERT Exemplar Class 11 Chemistry Chapter 1 : Multiple Choice Questions (Type-II)

In the following questions, two or more options may be correct.

Question 16

 One mole of oxygen gas at STP is equal to _______.

(i) 6.022 × 1023 molecules of oxygen

(ii) 6.022 × 1023 atoms of oxygen

(iii) 16 g of oxygen

(iv) 32 g of oxygen

Solution:

Option (i) and (iv) are the answers.

Question 17

Sulphuric acid reacts with sodium hydroxide as follows:

H2SO4 + 2NaOH → Na2SO4+ 2H2O

When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M

sodium hydroxide solution, the amount of sodium sulphate formed and its

molarity in the solution obtained is

(i) 0.1 mol L-1

(ii) 7.10 g

(iii) 0.025 mol L-1

(iv) 3.55 g

Solution:

Option (ii) and (iii) are the answers.

Question 18

Which of the following pairs have the same number of atoms?

(i) 16 g of O2(g) and 4 g of H2(g)

(ii) 16 g of O2 and 44 g of CO2

(iii) 28 g of N2 and 32 g of O2

(iv) 12 g of C(s) and 23 g of Na(s)

Solution:

Option (iii) and (iv) are the answers.

Question 19

Which of the following solutions have the same concentration?

(i) 20 g of NaOH in 200 mL of solution

(ii) 0.5 mol of KCl in 200 mL of solution

(iii) 40 g of NaOH in 100 mL of solution

(iv) 20 g of KOH in 200 mL of solution

Solution:

Option (i) and (ii) are the answers.

Question 20

16 g of oxygen has the same number of molecules as in

(i) 16 g of CO

(ii) 28 g of N2

(iii) 14 g of N2

(iv) 1.0 g of H2

Solution:

Option (iii) and (iv) are the answers.

Question 21

Which of the following terms is unitless?

(i) Molality

(ii) Molarity

(iii) Mole fraction

(iv) Mass per cent

Solution:

Option (iii) and (iv) are the answers.

Question 22

One of the statements of Dalton’s atomic theory is given below:

“Compounds are formed when atoms of different elements combine in a fixed

ratio”

Which of the following laws is not related to this statement?

(i) Law of conservation of mass

(ii) Law of definite proportions

(iii) Law of multiple proportions

Solution:

Option (i) and (iv)

NCERT Exemplar Class 11 Chemistry Chapter 1 : Short Answer Type

Question 23

What will be the mass of one atom of C-12 in grams?

Solution:

1 mole of carbon atom = 12g= 6.022 × 1023 atoms.

Question 24

How many significant figures should be present in the answer to the following

calculations?

2.5 1.25 3.5/2.01

Solution:

Two significant figures should be present in this.

Since the least number of significant figures from the given figure is 2 (in 2.5 and 3.5).

Question 25

What is the symbol for the SI unit of a mole? How is the mole defined?

Solution:

The symbol for the SI unit of the mole is mol. A mole is defined as the amount of substance that contains as many entities as there are atoms in 12g carbon.

Question 26

What is the difference between molality and molarity?

Solution:

Molarity is the number of moles of solute dissolved in 1 litre of the solution. Molality is the number of moles of solute present in 1kg of the solvent.

Question 27

Calculate the mass percent of calcium, phosphorus and oxygen in calcium

phosphate Ca3(PO4)

Solution:

Molecular mass of Ca3(PO4) = 3*40+2*31+8*16 =310

the Mass per cent of Ca = 3*40/310*100 = 38.71%

Mass per cent of P = 2*31/310*100 = 20%

and Mass per cent of O = 8*16/310 = 41.29%

Question 28

45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous

oxide was formed. The reaction is given below:

2N2(g) + O2(g) → 2N2O(g)

Which law is being obeyed in this experiment? Write the statement of the law?

Solution:

The above experiment proves Gay-Lussac’s law which states that gases combine or produced in a chemical reaction in a simple whole-number ratio by volume provided that all gases are the same temperature and pressure.

Question 29

 If two elements can combine to form more than one compound, the masses of

one element that combine with a fixed mass of the other element, are in whole-number ratio.

(a) Is this statement true?

(b) If yes, according to which law?

(c) Give one example related to this law.

Solution:

(a) Yes, the statement is true.

(b) According to the law of multiple proportions

(c), Hydrogen and oxygen react to form water and hydrogen peroxide

H2 + 1/2O2 → H2O

H2 + O2 → H2O2

Masses of oxygen which combine the fixed mass of hydrogen are in the ratio 16:32 or 1

Question 30

Calculate the average atomic mass of hydrogen using the following data :

Isotope %	Natural abundance	Molar mass
1H	99.985	1
2H	0.015	2

Solution:

Average atomic mass = 99.985*1+0.015*2/100

=099.985*1+0.015*2/100

=1.00015u

Question 31

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with

granulated zinc. Following reaction takes place.

Zn + 2HCl → ZnCl2 + H2

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc

reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of

Zn = 65.3 u.

Solution:

1 mol of gas occupies = 22.7L Volume at STP atomic mass of Zn = 65.3u

From the above equation,

65.3g of Zn when reacts with HCl produces = 22.7L H2 at STP

Therefore, 32.65g of Zn when reacts with HCl will produce = 22.7 * 32.65/65.3 =11.35L of H2 at STP

Question 32

 The density of 3 molal solutions of NaOH is 1.110 g mL–1. Calculate the molarity

of the solution.

Solution:

3 molal solution of NaOH = 3 moles of NaOH dissolved in 1000g water

3 mole of NaOH = 3*40g = 120g

Density of solution = 1.110gmL-1

Volume = mass/density = 1120g/1.110gmL-1 =1.009L

Molarity of the solution = 3/1.009 = 2.97M

Question 33

 The volume of a solution changes with change in temperature, then, will the molality

of the solution be affected by temperature? Give a reason for your answer.

Solution:

Mass does not change as the temperature changes. Therefore, the molality of a solution does not change.

Molality = moles of solute/ weight of solvent (in g) *1000

Question 34

 If 4 g of NaOH dissolves in 36 g of H2O, calculate the mole fraction of each

component in the solution. Also, determine the molarity of the solution (specific

gravity of solution is 1g mL–1).

Solution:

Mole fraction of H2O = No; of moles of H2O/ Total no: of moles (H2O+NaOH)

No: of moles of H2O = 36/18=2moles

No: of moles of NaOH = 4/40=0.1mol

Total no: of moles = 2+0.1= 2.1

Mole fraction of H2O = 2/2.1 = 0.952

Mole fraction of NaOH = 0.1/2.1 = 0.048

Mass of solution = Mass of H2O + Mass of NaOH = 36+4=40G

Volume of the solution = 40/1 = 40mL

Molarity = 0.1/0.04 = 2.5M

Question 35

The reactant which is entirely consumed in the reaction is known as limiting reagent.

In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B,

then

(i) which is the limiting reagent?

(ii) calculate the amount of C formed?

Solution;

(i) B will be the limiting reagent as it gives a lesser amount of product.

(ii) Let B is completely consumed

4 mol B gives 3 mol C

6 mol B will give 3/4 *6 mol C =4.5 mol C

NCERT Exemplar Class 11 Chemistry Chapter 1 : Match The Following Type

Question 36

(i) 88 g of CO2 (ii) 6.022 ×1023 molecules of H2O (iii) 5.6 litres of O2 at STP (iv) 96 g of O2 (v) 1 mol of any gas	(a) 0.25 mol (b) 2 mol (c) 1 mol (d) 6.022 × 1023 molecules (e) 3 mol

Solution:

A → b

B → c

C → a

D → e

E → d

Question 37

 Match the following

Physical quantity	Unit
(i) Molarity (ii) Mole fraction (iii) Mole (iv) Molality (v) Pressure (vi) Luminous intensity (vii) Density (viii) Mass	(a) g mL–1 (b) mol (c) Pascal (d) Unitless (e) mol L–1 (f) Candela (g) mol kg–1 (h) Nm–1 (i) kg

Solution:

(i → e)

(ii → d)

(iii → b)

(iv → g)

(v → c)

(vi → f)

(vii → a)

(viii → i)

NCERT Exemplar Class 11 Chemistry Chapter 1 : Assertion and Reason Type

In the following questions, a statement of Assertion (A) followed by a

statement of Reason (R) is given. Choose the correct option out of the

choices are given below each question.

Question 38

Assertion (A): The empirical mass of ethene is half of its molecular mass.

Reason (R): The empirical formula represents the simplest whole-number

the ratio of various atoms present in a compound.

(i) Both A and R are true and R is the correct explanation of A.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

Solution:

Option (i) is correct.

Question 39

 Assertion (A): One atomic mass unit is defined as one-twelfth of the mass of

one carbon-12 atom.

Reason (R): Carbon-12 isotope is the most abundant isotope of carbon

and has been chosen as the standard.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Solution:

Option (ii) is correct. Carbon-12 is considered a standard for defining the atomic and molecular mass.

Question 40

 Assertion (A): Significant figures for 0.200 is 3 whereas for 200 it is 1.

Reason (R): Zero at the end or right of a number are significantly provided

they are not on the right side of the decimal point.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not a correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Solution:

Option (iii) is correct. Significant figures for 0.200 = 3 and for 200 =1

Zero at the end of a number without decimal point may or may not be significant depending on the accuracy of the measurement.

Question 41

Assertion (A): Combustion of 16 g of methane gives 18 g of water.

Reason (R): In the combustion of methane, water is one of the products.

(i) Both A and R are true but R is not the correct explanation of A.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

Solution:

Option (iii) is correct.

16g of CH4 on complete combustion will give 36g of water.

NCERT Exemplar Class 11 Chemistry Chapter 1 : Long Answer Type Question

Question 42

A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The

gas is now transferred to another vessel at a constant temperature, where

the pressure becomes half of the original pressure. Calculate

(i) the volume of the new vessel.

(ii) a number of molecules of dioxygen.

Solution:

(i) Moles of oxygen = 1.6/32 = 0.05mol

At STP, 1 mol of O2 = 22.4L

Then volume of O2 = 22.4 × 0.05 = 1.12L

V1 = 1.12L

V2 =?

P1 = 1atm

P2 = ½ = 0.5atm

According to Boyle’s law, p1V1 = p2V2

Substituting the values

V2 = 1 × 1.12/0.5 = 2.24L

(ii) No of molecules in 1.6g or 0.005mol = 6.022 × 1023 × 0.05 = 3.011 × 1022

Question 43

Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according

to the reaction given below:

CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)

What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with

1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles

of CaCl2 formed in the reaction.

Solution:

No: of moles of HCl taken = MV/1000 = 0.76*250/1000 = 0.19

No: of moles of CaCO3 = Mass/Molar mass = 1000/100 = 10

1. When CaCO3 is completely consumed

1 mol of CaCO3 = 1 mol CaCl2

10 mol CaCO3 = 10mol CaCl2

2. When HCl is completely consumed.

2 mol HCl = 1 mol CaCl2

0.19mol HCl = ½ × 0.19mol CaCl2 = 0.095 mol CaCl2

HCl will be the limiting reagent and the number of moles of CaCl2 formed will be 0.095mol

Question 44

Define the law of multiple proportions. Explain it with two examples. How

does this law point to the existence of atoms?

Solution:

When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other bear a simple ratio to one another is the law of multiple proportions.

For example, carbon combines with oxygen to form two compounds they are carbon dioxide and carbon monoxide

The masses of oxygen which combine with a fixed mass of carbon in carbon dioxide and carbon monoxide are 32 and 16. Therefore oxygen bears: 32:16 ratio or 2:1

Example 2: Sulphur combines with oxygen to form sulphur trioxide and sulphur dioxide

The masses of oxygen which combine with a fixed mass of sulphur in SO3 and SO2 are 48 and 32. Therefore oxygen bears a ratio of 48:32 or 3:2

Question 45

A box contains some identical red coloured balls, labelled as A, each weighing

2 grams. Another box contains identical blue coloured balls, labelled as B,

each weighing 5 grams. Consider the combinations AB, AB2, A2B and A2B3

and show that the law of multiple proportions is applicable.

Solution:

	AB	ab2	A,B	A2B3
Mass of A (in g)	2	2	4	415
Mass of B (in g)	5	10	5

Masses of B which combine with a fixed mass of A are

10g, 20g, 5g, 15g

2 : 4 : 1 : 3

This is the simple whole-number ratio.

NCERT Exemplar For Class 11 Science

The NCERT exemplars are an effective study material for scoring higher marks in the examination paper. Students must practise these additional questions for their own benefits, as these are curated by the best subject-matter experts. Students can easily access the ncert exemplar for class 11 science by visiting our website SimplyAcad and solve all the questions listed to secure maximum marks.

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