The molarity of


The molarity of $\mathrm{HNO}_{3}$ in a sample which has density $1.4 \mathrm{~g} / \mathrm{mL}$ and mass percentage of $63 \%$ is_________. (Molecular Weight of $\mathrm{HNO}_{3}=63$ ) 



Mass percent of $\mathrm{HNO}_{3}=63$

Thus, $100 \mathrm{~g}$ of nitric acid solution contains $63 \mathrm{~g}$ of nitric acid by mass.

No. of moles $=\frac{63 \mathrm{~g}}{63 \mathrm{~g} \mathrm{~mol}^{-1}}=1$

Volume of $100 \mathrm{~g}$ of nitric acid solution

$=\frac{\text { Mass }}{\text { Density }}=\frac{100 \mathrm{~g}}{1.4 \mathrm{~g} / \mathrm{mL}}=71.4 \mathrm{~mL}$

Molarity $=\frac{\text { No. of moles }}{\text { volume }(\mathrm{mL})} \times 1000$

=$\frac{1}{71.4} \times 1000=14 \mathrm{M}$

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