Question

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 $70 \% HNO _{3}$.

• A. $45.0\, g$ conc. $HNO _{3}$
• B. $90.0\,g$ conc. $HNO _{3}$
• C. $70.0\, g$ conc. $HNO _{3}$
• D. $54.0\,g$ conc. $HNO _{3}$

Answer 1

Answer: (A)

Molarity
$=\frac{\text { Weight of } HNO _{3}}{\text { Molecular mass of } HNO _{3} \times \text { Volume of solution (in L) }}$
$\therefore$ Weight of $HNO _{3}=\text { Molarity } \times \text { Molecular mass } \times \text { Volume (in L)}$
$=2 \times 63 \times \frac{250}{1000}=31.5\, g$
It is the weight of $100 \% HNO _{3}$.
But the given acid is $70 \% HNO _{3}$
$\therefore$ Its weight $=31.5 \times \frac{100}{70} g =45.0\, g$ conc. $HNO _{3}$