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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2019 · 11 January · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

25 ~mL of the given HCl solution requires 30 ~mL of 0.1 M sodium...

25 mL25 \mathrm{~mL} of the given HCl\mathrm{HCl} solution requires 30 mL30 \mathrm{~mL} of 0.1M0.1 \mathrm{M} sodium carbonate solution. What is the volume of this HCl\mathrm{HCl} solution required to titrate 30 mL30 \mathrm{~mL} of 0.2M0.2 \mathrm{M} aqueous NaOH\mathrm{NaOH} solution?

Options

  1. A25 mL25 \mathrm{~mL}
  2. B75 mL75 \mathrm{~mL}
  3. C50 mL50 \mathrm{~mL}
  4. D12.5 mL12.5 \mathrm{~mL}

Correct answer

A. 25 mL25 \mathrm{~mL}

Step-by-step solution

25 \mathrm{ mL}ofof\mathrm{HCl}solutionrequiressolution requires30 \mathrm{ mL}ofof0.1 \mathrm{M}\mathrm{Na}_{2} \mathrm{CO}_{3}solution.solution.∵N1V1=N2V2∴25×N1=30×0.2(0.1MNa2CO3=0.2NNa2CO3)N1=625=0.24N\begin{array}{l} \because\mathrm{N}_{1} \mathrm{ V}_{1}=\mathrm{N}_{2} \mathrm{ V}_{2}\\ \therefore 25 \times \mathrm{N}_{1}=30 \times0.2\left(0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}=0.2 \mathrm{ N}\right.\\ \left.\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\\ \mathrm{N}_{1}=\frac{6}{25}=0.24 \mathrm{ N} \end{array}Now,HClsolutionistitratedwithNow, HCl solution is titrated with\mathrm{NaOH}solution.solution.M1V1=M2V2;0.24NHCl=0.24MHCl∴V×0.24×1=30×0.2×1⇒V=25mL\begin{array}{l}\mathrm{M}_{1}\mathrm{ V}_{1}=\mathrm{M}_{2}\mathrm{ V}_{2}; 0.24 \mathrm{NHCl}=0.24 \mathrm{M} \mathrm{HCl}\\ \therefore \mathrm{V} \times0.24 \times1=30 \times0.2 \times1 \Rightarrow \mathrm{V}=25 \mathrm{ mL} \end{array}$
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