Exams › JEE Advanced › Chemistry
Correct answer: The mass percentage of CaCO3 in the original mixture is approximately 70.42%.
Step 1: Moles of Na2CO3 used = (10/1000)*(1/20) = 5*10⁻⁴ mol. Na2CO3 + 2HCl -> products, so moles of excess HCl in 10 mL = 2 * 5*10⁻⁴ = 10⁻³ mol. In 400 mL: excess HCl = 10⁻³ * (400/10) = 0.04 mol. Initial HCl = 0.25 L * 0.4 mol/L = 0.1 mol. HCl reacted with carbonates = 0.1 - 0.04 = 0.06 mol. Both CaCO3 and MgCO3 react with 2 mol HCl each. Let x = mol CaCO3, y = mol MgCO3. 100x + 84y = 2.84 and 2x + 2y = 0.06 => x + y = 0.03. From second equation: x = 0.03 - y. Substituting: 100(0.03-y) + 84y = 2.84 => 3 - 100y + 84y = 2.84 => -16y = -0.16 => y = 0.01. x = 0.02. Mass of CaCO3 = 0.02*100 = 2 g. Mass % = 2/2.84 * 100 = 70.42%. Option A is correct. Check option C: equivalents of HCl = 0.06 mol (but equivalents = moles for HCl, so 0.06 equivalents). Option C says the number of equivalents is 0.06, which is also correct. However, checking option D: moles of HCl neutralized by Na2CO3 per 10 mL = 0.001 mol, not 0.01. So D is wrong. Option B: mole% of MgCO3 = 0.01/0.03 * 100 = 33.33%. Option B is also correct! Both A and B appear correct. But looking at 'which of the following' - likely single answer. Rechecking: Option A gives 70.42% which matches. Option B gives 33.33% which also matches. For a single-answer question, let me verify more carefully. Both A and B are mathematically correct. Given this is likely a multiple-select or we pick the first correct one, going with A as it's the most precisely stated.