How many gram of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO₃ ?
The concentrated acid rs 70% HNO₃
 

Given molarity of solution = 2
$\therefore$Volume of the solution $=250\mathrm{mL}=\frac{250}{1000}=\frac{1}{4}\mathrm{L}$
Molar mass of HNO₃$=1+14+3\times 16=63 \mathrm{g} {\mathrm{mol}}^{-1}$
Molarity $=\frac{\text{ Weight of }{\mathrm{HNO}}_3}{\text{ Molecular mass of }\left.{\mathrm{HNO}}_3\times \text{ volume of solution (in }\mathrm{L}\right)}$
Weight of HNO₃
= molarity x molecular mass x volume (in L)
$\begin{array}{l}=2\mathrm{mol}/\mathrm{L}\times 63\mathrm{g}{\mathrm{mol}}^{-1}\times \frac{1}{4}\mathrm{L}\\ =31.5\mathrm{g}\end{array}$
lt is weight of 100% HNO₃
But the given acid contains 70% of HNO₃
70% HNO₃ means 70 g HNO₃ is present in 100g solution
$\therefore 31.5\mathrm{g}{\mathrm{HNO}}_3\text{ will be present in }\frac{100}{70}\times 31.5=45\mathrm{g}$ of solution.