Stoichiometry & Calculations — IISER Chemistry MCQs with Solutions
Free IISER Chemistry Stoichiometry & Calculations MCQs with step-by-step solutions (7 questions). Part of Some Basic Concepts of Chemistry. Practise online on Prepizo — no login needed.
▶ Practise Stoichiometry & Calculations online (free)
Questions with solutions
Q1 — Stoichiometry & Calculations · medium · theory
Assertion (A): The molarity of a given solution decreases when its temperature is raised.
Reason (R): On heating, the volume of the solution increases while the number of moles of solute stays the same.
Choose the correct option.
A. A is true but R is false
B. Both A and R are true and R is the correct explanation of A ✓ Correct
C. Both A and R are true but R is NOT the correct explanation of A
D. A is false but R is true
Solution: Molarity = moles of solute ÷ volume of solution in litres. Heating expands the solution, so the volume in the denominator rises while the moles of solute are unchanged; the ratio therefore falls, so A is true. R states exactly this cause, so R is true and is the correct explanation of A.
Q2 — Stoichiometry & Calculations · medium · theory
Assertion (A): Molarity is preferred over molality when a reaction is studied at several different temperatures.
Reason (R): Molality is defined per kilogram of solvent, and mass does not change with temperature.
Choose the correct option.
A. Both A and R are true but R is NOT the correct explanation of A
B. A is true but R is false
C. Both A and R are true and R is the correct explanation of A
D. A is false but R is true ✓ Correct
Solution: Molality uses the mass of solvent, which is temperature-independent, whereas molarity uses volume, which changes with temperature. So for work over a range of temperatures, molality (not molarity) is preferred — this makes A false. R correctly states why molality is temperature-independent, so R is true: the correct choice is "A is false but R is true".
Q3 — Stoichiometry & Calculations · medium · numerical
For the reaction A + B₂ → AB₂, a mixture contains 5 mol of A and 2.5 mol of B₂. Identify the limiting reagent and the amount of AB₂ formed.
A. B₂ is limiting; 5 mol AB₂
B. No limiting reagent; 7.5 mol AB₂
C. B₂ is limiting; 2.5 mol AB₂ ✓ Correct
D. A is limiting; 5 mol AB₂
Solution: The equation needs A and B₂ in a 1 : 1 ratio. To react all 5 mol A would need 5 mol B₂, but only 2.5 mol B₂ are present, so B₂ is the limiting reagent. It fixes the product: 2.5 mol B₂ gives 2.5 mol AB₂ (leaving 2.5 mol A unreacted).
Q4 — Stoichiometry & Calculations · hard · numerical
Taking 1 g of each substance, which one contains the largest number of atoms? (H = 1, C = 12, O = 16, Na = 23, Ca = 40, Cl = 35.5)
A. NaCl
B. CaCO₃
C. H₂O ✓ Correct
D. CO₂
Solution: Atoms ∝ (mass ÷ molar mass) × (atoms per formula unit). H₂O: (1 ÷ 18) × 3 = 0.167; CO₂: (1 ÷ 44) × 3 = 0.068; CaCO₃: (1 ÷ 100) × 5 = 0.050; NaCl: (1 ÷ 58.5) × 2 = 0.034 (all × 6.022 × 10²³). Water gives the largest value because its low molar mass combines with 3 atoms per molecule.
Q5 — Stoichiometry & Calculations · hard · numerical
A gaseous hydrocarbon CₓHᵧ is burned in oxygen. 10 mL of the hydrocarbon produces 30 mL of CO₂ and 40 mL of H₂O vapour, all volumes measured at the same temperature and pressure. The hydrocarbon is:
A. C₃H₄
B. C₃H₈ ✓ Correct
C. C₄H₁₀
D. C₃H₆
Solution: At constant temperature and pressure, gas volumes are proportional to moles (Avogadro law). Per 1 volume of hydrocarbon: CO₂ = 30 ÷ 10 = 3, so x = 3. Water = 40 ÷ 10 = 4, and each hydrocarbon supplies y/2 water molecules, so y/2 = 4 → y = 8. The formula is C₃H₈ (propane).
Q6 — Stoichiometry & Calculations · medium · numerical
For Mg + 2HCl → MgCl₂ + H₂, what volume of H₂ measured at STP is produced when 4.8 g of magnesium reacts with excess HCl? (Mg = 24, molar volume at STP = 22.7 L mol⁻¹)
A. 4.48 L
B. 9.08 L
C. 2.27 L
D. 4.54 L ✓ Correct
Solution: Moles of Mg = 4.8 ÷ 24 = 0.2 mol. From the equation, 1 mol Mg gives 1 mol H₂, so 0.2 mol H₂ forms. Volume at STP = 0.2 × 22.7 = 4.54 L. Using the old 22.4 L mol⁻¹ gives the trap 4.48 L.
Q7 — Stoichiometry & Calculations · hard · numerical
A 1.0 mol kg⁻¹ (molal) aqueous solution of glucose (molar mass 180 g mol⁻¹) has a density of 1.10 g mL⁻¹. Its molarity is:
A. 1.00 mol L⁻¹
B. 1.10 mol L⁻¹
C. 0.93 mol L⁻¹ ✓ Correct
D. 0.85 mol L⁻¹
Solution: 1.0 molal means 1 mol glucose per 1 kg (1000 g) of water. Mass of solution = 1000 g water + 1 × 180 g glucose = 1180 g. Volume = 1180 ÷ 1.10 = 1072.7 mL = 1.0727 L. Molarity = 1 ÷ 1.0727 = 0.93 mol L⁻¹. Assuming molarity equals molality gives the wrong 1.00 mol L⁻¹.