The molarity of $1 \mathrm{~L}$ orthophosphoric acid $\left(\mathrm{H}_3 \mathrm{PO}_4\right)$ having $70 \%$ purity by weight (specific gravity $1.54 \mathrm{~g} \mathrm{~cm}^{-3}$) is __________ $\mathrm{M}$.
(Molar mass of $\mathrm{H}_3 \mathrm{PO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1}$)
Answer (integer)
11
Solution
<p>Specific gravity (density) $=1.54 \mathrm{~g} / \mathrm{cc}$.</p>
<p>Volume $=1 \mathrm{~L}=1000 \mathrm{~ml}$</p>
<p>Mass of solution $=1.54 \times 1000$</p>
<p>$=1540 \mathrm{~g}$</p>
<p>$\%$ purity of $\mathrm{H}_2 \mathrm{SO}_4$ is $70 \%$</p>
<p>So weight of $\mathrm{H}_3 \mathrm{PO}_4=0.7 \times 1540=1078 \mathrm{~g}$</p>
<p>Mole of $\mathrm{H}_3 \mathrm{PO}_4=\frac{1078}{98}=11$</p>
<p>Molarity $=\frac{11}{1 \mathrm{~L}}=11$</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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