The molarity of $H N O_{3}$ in a sample which has density 1.4 g/mL and mass percentage of 63% is (Molecular weight of $H N O_{3} = 63) .$

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Solution

$% \frac{w}{w} = 63 %$
$ρ = 1.4 g / m L$
$M = \frac{(63 \times ρ)}{M M \times 100} \times 1000 m o l / L$
$M = \frac{(63 \times 1.4 \times 10)}{63}$
M = 14 mol/L

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