Home/
Competitive Exams/
JEE Main & Advanced/
Chemistry/
Mole Concept and Stoichiometry/
Question 600012
JEE Chemistry · Medium

Mole Concept and Stoichiometry: Mole–Mass Conversion MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

Timer-based practiceDetailed answer reviewMobile-friendly flow
Mole–Mass ConversionMediumQuestion 600012

Question

How many grams of CaCO₃ (molar mass 100 g/mol) contain 0.5 mol of CO₃²⁻ ions?
  1. A
    25 g
  2. B
    50 g
    Correct
  3. C
    100 g
  4. D
    200 g

Correct answer

50 g

Explanation

1 mol CaCO₃ gives 1 mol CO₃²⁻. So 0.5 mol CO₃²⁻ requires 0.5 mol CaCO₃ = 0.5 × 100 = 50 g.

Review Mole Concept and Stoichiometry notesStart timed chapter practice