Question

How many grams of concentrated nitric acid solution should be used to prepare $250\, mL$ of $2 \cdot 0\, M HNO _{3}$ ? The concentrated acid is $70 \% HNO _{3}$.

• A. $90.0\, g$ conc. $HNO_3$
• B. $70.0\, g$ conc. $HNO_3$
• C. $54.0\, g$ conc. $HNO_3$
• D. $45.0\, g$ conc. $HNO_3$

Answer 1

Answer: (D)

$45.0\, g$ conc. $HNO_3$

Answer 2

The number of moles of nitric acid can be obtained by multiplying the molarity with volume
$M \times V =$ Moles of $HNO _{3}=\frac{250 \times 2}{1000}=0.5$
The mass of nitric acid can be obtained by multiplying the number of mols with molar mass and dividing with percentage concentration.
$\therefore HNO _{3}$ required $=0.5 \times 63 \times \frac{100}{70}=45\, g$
Thus $45.0\, g$ conc. $HNO _{3}$ of concentrated nitric acid solution should be used to prepare $250\, mL$ of $2.0\, M\, HNO _{3}$.