Mole Concept and Stoichiometry: Back Titration MCQ

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Back TitrationHardQuestion 600044

Question

0.5 g of impure CaCO₃ is treated with 50 mL of 0.2 M HCl. Excess HCl required 10 mL of 0.1 M NaOH to neutralise. Mass of pure CaCO₃ (M = 100) is:

A
0.45 g
Correct
B
0.50 g
C
0.40 g
D
0.48 g

Correct answer

0.45 g

Explanation

Moles HCl = 0.05 × 0.2 = 0.01 mol. Moles NaOH (back) = 0.01 × 0.1 = 0.001 mol = moles excess HCl. Moles HCl reacted with CaCO₃ = 0.01 − 0.001 = 0.009 mol. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Moles CaCO₃ = 0.009/2 = 0.0045. Mass = 0.0045 × 100 = 0.45 g.

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