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JEE Main 20127 MayOnline
Exam
JEE Main
Year
2012
Date
7 May
Mode
Online
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2012 · 7 May · Online · Chemistry · Some Basic Concepts of Chemistry

The concentrated sulphuric acid that is peddled commercial is 95 \%...

The concentrated sulphuric acid that is peddled commercial is 95%H2SO495 \% \mathrm{H}_2 \mathrm{SO}_4 by weight. If the density of this commercial acid is 1.834 g cm−31.834 \mathrm{~g} \mathrm{~cm}^{-3}, the molarity of this solution is

Options

  1. A17.8M17.8 \mathrm{M}
  2. B12.0M12.0 \mathrm{M}
  3. C10.5M10.5 \mathrm{M}
  4. D15.7M15.7 \mathrm{M}

Correct answer

A. 17.8M17.8 \mathrm{M}

Step-by-step solution

95%H2SO495 \% \mathrm{H}_2 \mathrm{SO}_4 by weight means 100 gH2SO4100 \mathrm{~g} \mathrm{H}_2 \mathrm{SO}_4 solution contains 95 gH2SO495 \mathrm{~g} \mathrm{H}_2 \mathrm{SO}_4 by mass. Molar mass of H2SO4=98 g mol−1\mathrm{H}_2 \mathrm{SO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1} Moles in 95 g=9598=0.96995 \mathrm{~g}=\frac{95}{98}=0.969 mole Volume of 100 gH2SO4100 \mathrm{~g} \mathrm{H}_2 \mathrm{SO}_4= mass  density =100 g1.834 g cm−3=54.52 cm3=54.52×10−3 L\begin{aligned}& =\frac{\text{ mass }}{\text{ density }}=\frac{100 \mathrm{~g}}{1.834 \mathrm{~g} \mathrm{~cm}^{-3}}\\ & =54.52 \mathrm{~cm}^3=54.52 \times10^{-3} \mathrm{~L} \end{aligned} Molarity = Moles of solute  Volme of solute in L=0.96954.52×10−3=17.8M\begin{aligned} \text{ Molarity }& =\frac{\text{ Moles of solute }}{\text{ Volme of solute in } \mathrm{L}}\\ & =\frac{0.969}{54.52 \times 10^{-3}}=17.8 \mathrm{M} \end{aligned}$$
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