Question

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% $HN{O}_3$.

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

Answer 1

The correct option is A 45.0 g conc. $HN{O}_3$

The number of moles of nitric acid can be obtained by multiplying the molarity with volume

$M\times V=$Moles of $HN{O}_3=\frac{250\times 2}{1000}=0.5$

The mass of nitric acid can be obtained by multiplying the number of mols with molar mass and dividing with percentage concentration.
$\therefore HN{O}_3$ required $=0.5\times 63\times \frac{100}{70}=45g$

Thus 45.0 g conc. $HN{O}_3$ of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M $HN{O}_3$.

Hence, the correct option is $A$.