NCERT Exemplar Solutions-Basic Concepts of Chemistry for Class 11

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NCERT Exemplar Solutions for Class 11 Chemistry Chapter Some Basic Concepts of Chemistry

Multiple Choice Questions (Type-1)

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) 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

Multiple Choice Questions (Type-11)

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

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.

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.

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.

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.

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.

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.

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)

III. Short Answer Type

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.

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).

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.

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.

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

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

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

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

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.

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

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

Solution:

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

=099.985*1+0.015*2/100

=1.00015u

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

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

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

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=40 gm/mol

Volume of the solution = 40/1 = 40mL

Molarity = 0.1/0.04 = 2.5M

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

Match The Following Type

36.

Solution:

A → b

B → c

C → a

D → e

E → d

37. Match the following

Solution:

(i → e)

(ii → d)

(iii → b)

(iv → g)

(v → c)

(vi → f)

(vii → a)

(viii → i)

V. 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.

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.

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.

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.

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.

Long Answer Type 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

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

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 bear: 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 bear a ratio of 48:32 or 3:2

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:

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.

Important Topics of Chemistry Chapter 1 Some Basic Concepts of Chemistry

• Importance of Chemistry
• Nature of Matter
• Properties of Matter and Their Measurement
  • The International System of Units (Si)
  • Mass and Weight
• Uncertainty in Measurement
  • Scientific Notation
  • Significant Figures
  • Dimensional Analysis
• Laws of Chemical Combinations
  • Law of Conservation of Mass
  • Law of Definite Proportions
  • Law of Multiple Proportions
  • Gay Lussac’s Law of Gaseous Volumes
  • Avogadro Law
• Dalton’s Atomic Theory
• Atomic and Molecular Masses
  • Atomic Mass
  • Average Atomic Mass
  • Molecular Mass
  • Formula Mass
• Mole Concept and Molar Masses
• Percentage Composition
• Empirical Formula for Molecular Formula
• Stoichiometry and Stoichiometric Calculations
  • Limiting Reagent
  • Reactions in Solutions