Home/
Competitive Exams/
JEE Main & Advanced/
Chemistry/
Mole Concept and Stoichiometry/
Question 600067
JEE Chemistry · Hard

Mole Concept and Stoichiometry: Stoichiometry MCQ

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

Timer-based practiceDetailed answer reviewMobile-friendly flow
StoichiometryHardQuestion 600067

Question

X g of hydrated oxalic acid (H₂C₂O₄·2H₂O, M=126) dissolves to give 100 mL of solution. 10 mL of this neutralises 15 mL of 0.1 M NaOH. X is:
  1. A
    0.473 g
  2. B
    0.945 g
    Correct
  3. C
    0.315 g
  4. D
    0.630 g

Correct answer

0.945 g

Explanation

Moles NaOH = 0.015×0.1 = 0.0015 mol. H₂C₂O₄ has 2 acidic protons → moles H₂C₂O₄ in 10 mL = 0.0015/2 = 0.00075 mol. In 100 mL: 0.0075 mol. X = 0.0075×126 = 0.945 g.

Review Mole Concept and Stoichiometry notesStart timed chapter practice