CBSE Class 11 Chemistry HOTs Some Basic Concepts of Chemistry Set 02

Higher Order Thinking Questions

Question 1. What is the difference between 160 cm and 160.0 cm in a measurement?
Answer:
160 cm: zeros at the end of a number without a decimal point are ambiguous. The number may be read as having two or three significant figures, and it is usually taken as three when the length is known to the nearest centimetre.
160.0 cm has four significant figures, because the zero after the decimal point is significant and leaves no doubt.
Hence 160.0 cm shows a more precise measurement: the length is known to the nearest 0.1 cm, not just the nearest 1 cm.
In simple words: More significant figures mean a finer measuring scale was used.

Teacher's Note:
1. Zeros after a decimal point show the precision of the instrument.
2. Do not drop a trailing zero from a measured value.

Question 2. In the combustion of methane in air, which is the limiting reagent and why?
Answer:
CH₄ + 2O₂ → CO₂ + 2H₂O.
Methane is the limiting reagent. The other reactant, oxygen of the air, is always present in large excess.
So the amounts of CO₂ and H₂O formed depend on the amount of methane burnt.
In simple words: When one reactant is freely available (air), the other one decides the amount of product.

Teacher's Note:
1. The limiting reagent is the one that gets used up first.
2. Burning in open air means oxygen is effectively unlimited.

Question 3. A compound of two elements A and B contains A = 70% and B = 30%. Their relative numbers of moles in the compound are 1.25 and 1.88. Calculate (a) the atomic masses of A and B and (b) the molecular formula of the compound if its molecular mass is 160.
Answer:
(a) Relative number of moles = % ÷ atomic mass, so atomic mass = % ÷ relative moles.
Atomic mass of A = 70 ÷ 1.25 = 56; atomic mass of B = 30 ÷ 1.88 = 15.96 ≈ 16.
(b) Ratio A : B = 1.25 : 1.88 = 1 : 1.5 = 2 : 3, so the empirical formula is A₂B₃.
Empirical formula mass = 2 × 56 + 3 × 16 = 160, so n = 160 ÷ 160 = 1 and the molecular formula is A₂B₃ (this fits Fe₂O₃).
In simple words: Reverse the usual mole calculation: atomic mass = percentage ÷ number of moles.

Teacher's Note:
1. Clear the decimal by multiplying the ratio 1 : 1.5 by 2.
2. Compare with known formulae such as Fe₂O₃ to check.

Question 4. Assertion (A): The mass of a body is constant, but its weight may change from place to place.
Reason (R): Weight is the force exerted by gravity on a body, and gravity varies from place to place.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Mass is the amount of matter in a body and does not change. Weight is the force of gravity acting on that mass, and it changes with the value of gravity, which differs on the earth, the moon and so on.
So R correctly explains A.
In simple words: Mass stays the same everywhere, but weight depends on gravity.

Teacher's Note:
1. SI unit of mass is kg, and of weight (a force) is newton.
2. Mass is measured with a balance and weight with a spring balance.

Question 5. Assertion (A): Molality of a solution does not change with temperature, but molarity does.
Reason (R): Molality is defined using the mass of the solvent, while molarity is defined using the volume of the solution.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Mass of the solvent does not change on heating or cooling, so molality (mol per kg of solvent) is constant.
The volume of a solution expands or contracts with temperature, so molarity (mol per litre of solution) changes. R gives exactly this reason.
In simple words: Mass is temperature-proof; volume is not.

Teacher's Note:
1. Use molality when working at different temperatures.
2. Molarity needs volume of solution in litres.

Question 6. Assertion (A): One mole of hydrogen gas (H₂) and one mole of helium (He) contain the same number of atoms.
Reason (R): One mole of any substance contains 6.022 × 10²³ elementary entities.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (d) A is false: 1 mol of H₂ has 6.022 × 10²³ molecules, i.e. 2 × 6.022 × 10²³ atoms, while 1 mol of He has only 6.022 × 10²³ atoms (it is monoatomic).
R is a true statement of the definition of the mole.
In simple words: A mole counts the entities named, molecules for H₂ and atoms for He.

Teacher's Note:
1. Always state whether atoms or molecules are being counted.
2. Atoms in one mole of a molecule = atomicity × 6.022 × 10²³.

Question 7. Assertion (A): When 28 g of N₂ and 28 g of H₂ are mixed to form ammonia, N₂ is the limiting reagent.
Reason (R): 28 g of N₂ requires only 6 g of H₂ for complete reaction.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) N₂ + 3H₂ → 2NH₃. 28 g (1 mol) of N₂ reacts with 3 mol (6 g) of H₂.
Since 28 g of H₂ (14 mol) is far more than 6 g, N₂ is used up first and is the limiting reagent. R explains A.
In simple words: Only 6 g of H₂ is needed for 28 g of N₂, so hydrogen is in large excess.

Teacher's Note:
1. Compare moles available with the ratio from the balanced equation.
2. Mass ratio is not the same as mole ratio.

Question 8. Which of the following statements are correct for 0.5 mol of methane (CH₄)? (More than one may be correct.)
(i) Its mass is 8 g.
(ii) It contains 3.011 × 10²³ molecules.
(iii) It occupies 11.2 L at STP.
(iv) It contains 5 × 6.022 × 10²³ atoms.
(a) (i) and (ii) only
(b) (i), (ii) and (iii) only
(c) (ii) and (iv) only
(d) (i), (ii), (iii) and (iv)
Answer:
Ans. (b)
(i) 0.5 × 16 = 8 g: correct. (ii) 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules: correct. (iii) 0.5 × 22.4 = 11.2 L at STP: correct.
(iv) Each molecule has 5 atoms, so atoms = 0.5 × 5 × 6.022 × 10²³ = 2.5 × 6.022 × 10²³: incorrect.
In simple words: Work out mass, molecules, volume and atoms separately from 0.5 mol.

Teacher's Note:
1. Multiply by the atomicity (5 for CH₄) to count atoms.
2. Check every statement; do not stop at the first correct one.

Question 9. Why is the atomic mass of chlorine taken as 35.5 u when no single chlorine atom has this mass?
Answer:
Chlorine occurs naturally as two isotopes, Cl-35 and Cl-37, in the approximate ratio 3 : 1.
Average atomic mass = (35 × 3 + 37 × 1) ÷ 4 = 142 ÷ 4 = 35.5 u.
So the atomic mass given in the periodic table is the weighted average of the masses of all naturally occurring isotopes.
In simple words: 35.5 u is an average over Cl-35 and Cl-37 atoms.

Teacher's Note:
1. Atomic masses are averages when isotopes exist.
2. The abundance of each isotope is used as the weight.

Question 10. A student prepares a solution by dissolving 4 g of NaOH in 1 L of water and says its molarity is 0.1 M. Is the statement correct? Explain.
Answer:
Molarity is defined as moles of solute per litre of solution, not per litre of solvent.
Dissolving 4 g of NaOH in 1 L of water gives a solution whose volume is not exactly 1 L (it changes slightly on dissolving and, for NaOH, the solution warms up), so the molarity is only approximately 0.1 M.
To get exactly 0.1 M, dissolve 4 g of NaOH in a little water and then make up the volume to exactly 1 L in a volumetric flask.
In simple words: Molarity needs the final volume of solution, so make up to the mark in a flask.

Teacher's Note:
1. Moles of NaOH = 4 ÷ 40 = 0.1 mol.
2. Molality, in contrast, uses the mass of the solvent only.

Question 11. How much water should be added to 100 mL of 0.5 M NaCl solution to make it 0.2 M? (Assume the volumes are additive.)
Answer:
Using M₁V₁ = M₂V₂: 0.5 × 100 = 0.2 × V₂, so V₂ = 250 mL.
Volume of water to be added = 250 − 100 = 150 mL.
Check: moles of NaCl = 0.5 × 0.100 = 0.05 mol, and 0.05 ÷ 0.250 L = 0.2 M.
In simple words: Diluting changes the volume and the molarity but not the number of moles of solute.

Teacher's Note:
1. Moles of solute stay constant on dilution.
2. Water added = final volume − initial volume.

Question 12. 12 g of carbon is burnt in 48 g of oxygen to form carbon dioxide. Which reactant is in excess, how much of it is left, and how does the result illustrate the law of conservation of mass? (C = 12, O = 16)
Answer:
C + O₂ → CO₂. Moles of C = 12 ÷ 12 = 1 mol; moles of O₂ = 48 ÷ 32 = 1.5 mol.
1 mol of C needs only 1 mol (32 g) of O₂, so carbon is the limiting reagent and oxygen is in excess.
O₂ left unreacted = 48 − 32 = 16 g; CO₂ formed = 44 g.
Mass before the reaction = 12 + 48 = 60 g; mass after = 44 + 16 = 60 g. The total mass is unchanged, which is the law of conservation of mass.
In simple words: Carbon finishes first, so 16 g of oxygen is left over; the total mass stays 60 g.

Teacher's Note:
1. Mass is conserved even when a reactant is left over, if you count it.
2. The limiting reagent gives the amount of product.

Question 13. Haemoglobin contains 0.33% iron by mass. Its molar mass is about 67200 g mol⁻¹. How many iron atoms are present in one molecule of haemoglobin? (Fe = 56)
Answer:
Mass of iron in 67200 g (1 mol) of haemoglobin = 0.33 ÷ 100 × 67200 = 221.8 g.
Moles of Fe atoms = 221.8 ÷ 56 = 3.96 ≈ 4.
So one molecule of haemoglobin has 4 iron atoms (the number of atoms must be a whole number).
In simple words: Find the mass of iron in one mole of haemoglobin, then convert to moles of atoms.

Teacher's Note:
1. One mole of haemoglobin means one mole of molecules.
2. Round to the nearest whole number because atoms cannot be fractions.