The molartiry of  ${\mathrm{HNO}}_3$  in a sample which has density 1.4 g/mL and mass percentage of 63% is ______ ( Molecular weight of${\mathrm{HNO}}_3=63$  )

Mass percentage is 63% which means that 63 g of solute present in 100 g of solution.

$$\begin{gathered} \mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Volume}} \\ 1.4 \mathrm{g}/\mathrm{mL}=\frac{100}{\mathrm{volume}} \\ \mathrm{volume}=71.428 \mathrm{mL} \end{gathered}$$

Molarity is calculated as:

$$\begin{gathered} \mathrm{molarity}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}\times \frac{1000}{\mathrm{volume} \mathrm{in} \mathrm{mL}} \\ \mathrm{molarity}=\frac{63\times 1000}{63\times 71.428}=14 \mathrm{M} \end{gathered}$$