How many grams of concentrated nitric acid solution should be used to prepare $250 \mathrm{~mL}$ of $2.0…
How many grams of concentrated nitric acid solution should be used to prepare $250 \mathrm{~mL}$ of $2.0 \mathrm{M} \mathrm{HNO}_3$ ? The concentrated acid is $70 \% \mathrm{HNO}_3$.
$45.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$
$90.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$
$70.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$
$54.0 \mathrm{~g}$ conc. $\mathrm{HNO}_3$
Solution
Given, molarity of solution $=2$
Volume of solution $=250 \mathrm{~mL}=\frac{250}{1000}=\frac{1}{4} \mathrm{~L}$
Molar mass of
$\mathrm{HNO}_3=1+14+3 \times 16=63 \mathrm{~g} \mathrm{~mol}^{-1}$
$\because$ Molarity
$\begin{aligned}
& =\frac{\text { weight of } \mathrm{HNO}_3}{\text { mass of } \left.\mathrm{HNO}_3 \times \text { volume of solution ( } L\right)} \\
& \therefore \text { Weight of } \mathrm{HNO}_3=\text { molarity } \times \text { mol. mass } \\
& \times \text { volume }(\mathrm{L}) \\
& =2 \times 63 \times \frac{1}{4} \mathrm{~g}=31.5 \mathrm{~g} \\
&
\end{aligned}$
It is the weight of $100 \% \mathrm{HNO}_3$.
But the given acid is $70 \% \mathrm{HNO}_3$.
$\therefore \text { Its weight }=31.5 \times \frac{100}{70} \mathrm{~g}=45 \mathrm{~g}$