How many grams of concentrated nitric acid solution should be used to prepare $$250 mL$$  of $$2.0\,M\,HN{O_3}?$$    The concentrated acid is $$70\% \,HN{O_3}.$$

Solution :
$$\eqalign{ & {\text{Given, molarity of solution = 2}} \cr & {\text{Volume of solution}} \cr & = 250\,mL \cr & = \frac{{250}}{{1000}} \cr & = \frac{1}{4}L \cr & {\text{Molar mass of}} \cr & HN{O_3} = 1 + 14 + 3 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 63\,g\,mo{l^{ - 1}} \cr & \because \,{\text{Molarity}} \cr & = \frac{{{\text{Weight of}}\,HN{O_3}}}{{{\text{Molecular mass of}}\,HN{O_3} \times {\text{volume of solution}}\,\left( L \right)}} \cr} $$

$$\therefore \,\,{\text{Weight of}}\,HN{O_3}$$     $$ = {\text{molarity}} \times {\text{molecular mass}} \times {\text{volume }}(L)$$
$$\eqalign{ & = 2 \times 63 \times \frac{1}{4}g \cr & = 31.5\,g \cr & {\text{It is the weight of }}100\% {\text{ }}HN{O_3} \cr & {\text{But the given acid is }}70\% {\text{ }}HN{O_3} \cr & \therefore \,\,{\text{Its weight}}\,{\text{ = }}\,{\text{31}}{\text{.5}} \times \frac{{100}}{{70}}\,g \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45\,g \cr} $$