Question

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

• A. $70\text{g}conc.HN{O}_3$
• B. $54\text{g}conc.HN{O}_3$
• C. $45\text{g}conc.HN{O}_3$
• D. $90\text{g}conc.HN{O}_3$

Answer 1

The correct option is C $45\text{g}conc.HN{O}_3$

$\text{Molarity}=\frac{w_{HNO3}\times 1000}{M_{HN{O}_3}\times V_{sol}\left(mL\right)}$
$\Rightarrow 2=\frac{w_{HN{O}_3}}{63}\times \frac{1000}{250}\Rightarrow w_{HN{O}_3}=\frac{63}{2}\text{g}$
$\text{Mass of acid}\times \frac{70}{100}=\frac{63}{2}$
$\text{Mass of acid}=45\text{g}$