The density of nitric acid solution is $1.5 \mathrm{~g} \mathrm{~mL}^{-1}$. Its weight percentage is 68 .…
The density of nitric acid solution is $1.5 \mathrm{~g} \mathrm{~mL}^{-1}$. Its weight percentage is 68 . What is the approximate concentration (in $\left.\mathrm{mol} \mathrm{L}^{-1}\right)$ of nitric acid? $(\mathrm{N}=14 u ; \mathrm{O}=16 u ; \mathrm{H}=1 u)$
$14.2$
$11.6$
$18.2$
$16.2$
Solution
$\begin{aligned} & \text { Given }- \text { Density }(\delta)=1.5 \mathrm{~g} / \mathrm{ml} \\ & \text { Volume }(\mathrm{v})=1000 \mathrm{ml} \\ & \text { Weight percentage }=68 \\ & \text { Mass of solution }=\text { density } \times \text { volume } \\ & 1.5 \mathrm{~g} / \mathrm{ml} \times 1000 \mathrm{ml}=1500 \mathrm{~g} \\ & \text { Mass of Nitric Acid }=\text { weight percentage } \times \text { Total mass of } \\ & \text { solution. }\end{aligned}$
Mass of Nitric acid $=0.68 \times 1500 \mathrm{~g}=1020 \mathrm{~g}$ Molar mass of Nitric Acid $\left(\mathrm{HNO}_2\right)$
$=1(\mathrm{H})+14(\mathrm{~N})+3 \times 16(0)=63 \mathrm{~g} / \mathrm{mol}$
Moles of Nitric acid $=\frac{\text { mass }}{\text { molar mass }}$
Moles of Nitric acid $=\frac{1020 \mathrm{~g}}{63 \mathrm{~g} / \mathrm{mol}}=16.19 \mathrm{~mol}$
$\operatorname{Molarity}(\mathrm{m})=\frac{\text { moles of solute }}{\text { Volume of solution in litres }}$
Molarity $=\frac{16.19 \mathrm{~mol}}{\mathrm{LL}}=19.19 \mathrm{~m} \approx 16.2$