Home/
Competitive Exams/
JEE Main & Advanced/
Chemistry/
Mole Concept and Stoichiometry/
Question 600045
JEE Chemistry · Hard

Mole Concept and Stoichiometry: Mixture Stoichiometry MCQ

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

Timer-based practiceDetailed answer reviewMobile-friendly flow
Mixture StoichiometryHardQuestion 600045

Question

A mixture of NaOH and Na₂CO₃ is titrated with HCl. In the first titration (phenolphthalein), 25 mL of 0.1 M HCl is used. In the second (methyl orange), 15 mL more is needed. Moles of NaOH and Na₂CO₃ are:
  1. A
    NaOH: 0.001, Na₂CO₃: 0.0015
    Correct
  2. B
    NaOH: 0.001, Na₂CO₃: 0.0025
  3. C
    NaOH: 0.0025, Na₂CO₃: 0.0015
  4. D
    NaOH: 0.0015, Na₂CO₃: 0.0025

Correct answer

NaOH: 0.001, Na₂CO₃: 0.0015

Explanation

First end-point: neutralises NaOH + converts Na₂CO₃ → NaHCO₃. Second: converts NaHCO₃ → CO₂. Let n(NaOH)=x, n(Na₂CO₃)=y. V₁: x + y = 0.025×0.1 = 0.0025. V₂: y = 0.015×0.1 = 0.0015. So x = 0.0025−0.0015 = 0.001. NaOH: 0.001 mol, Na₂CO₃: 0.0015 mol.

Review Mole Concept and Stoichiometry notesStart timed chapter practice