How many grams of concentrated nitric acid solution should be used to prepare 250 ~mL of 2.0 M HNO_3 ? The concentrated acid is 70 % HNO_3.

Given, molarity of solution $=2$=2 Volume of solution $=250 \mathrm{~mL}=\frac{250}{1000}=\frac{1}{4} \mathrm{~L}$=250 mL=1000250​=41​ L Molar mass of $$\mathrm{HNO}_3=1+14+3 \times 16=63 \mathrm{~g} \mathrm{~mol}^{-1}$$HNO3​=1+14+3×16=63 g mol−1 $\because$∵ Molarity $$\begin{aligned} & =\frac{\text { weight of } \mathrm{HNO}_3}{\text { mass of } \left.\mathrm{HNO}_3 \times \text { volume of solution ( } L\right)} \\ & \therefore \text { Weight of } \mathrm{HNO}_3=\text { molarity } \times \text { mol. mass } \\ & \times \text { volume }(\mathrm{L}) \\ & =2 \times 63 \times \frac{1}{4} \mathrm{~g}=31.5 \mathrm{~g} \\ & \end{aligned}$$​= mass of HNO3​× volume of solution ( L) weight of HNO3​​∴ Weight of HNO3​= molarity × mol. mass × volume (L)=2×63×41​ g=31.5 g​ It is the weight of $100 \% \mathrm{HNO}_3$100%HNO3​. But the given acid is $70 \% \mathrm{HNO}_3$70%HNO3​. $$\therefore \text { Its weight }=31.5 \times \frac{100}{70} \mathrm{~g}=45 \mathrm{~g}$$∴ Its weight =31.5×70100​ g=45 g