Question

How many grams of concentrated nitric acid solution should be used to prepare $250mL$ of $2.0MHN{O}_3$ ? The concentrated acid is $70\%HN{O}_3$.

• A. $45.0g$ conc. $HN{O}_3$
• B. $90.0g$ conc. $HN{O}_3$
• C. $70.0g$ conc. $HN{O}_3$
• D. $54.0g$ conc. $HN{O}_3$

Answer 1

Answer: (A)

Molarity
$=\frac{\text{ Weight of }HN{O}_3}{\text{ Molecular mass of }HN{O}_3\times \text{ Volume of solution (in L) }}$
$\therefore$ Weight of $HN{O}_3=\text{ Molarity }\times \text{ Molecular mass }\times \text{ Volume (in L)}$
$=2\times 63\times \frac{250}{1000}=31.5g$
It is the weight of $100\%HN{O}_3$.
But the given acid is $70\%HN{O}_3$
$\therefore$ Its weight $=31.5\times \frac{100}{70}g=45.0g$ conc. $HN{O}_3$