Question

What volume of $96\%$ $H_{2}S{O}_4$ solution $\left(\text{density}1.83\text{g/mL}\right)$ is required to prepare $4\text{liter}$ of $3.0\text{M}$ $H_{2}S{O}_4$ solution?

• A. $669.39\text{mL}$
• B. $700.00\text{mL}$
• C. $500.50\text{mL}$
• D. $600.45\text{mL}$

Answer 1

The correct option is A $669.39\text{mL}$

$\text{Molarity}=\frac{\text{moles of solute}}{\text{volume of solution in L}}$
$\Rightarrow \text{Moles = Molarity}\times \text{Volume}$
$\Rightarrow \text{Moles}=3\times 4=12$
$\text{Mass of}{H}_{2}S{O}_4=98\text{g/mol}\times 12\text{mol}=1176\text{g}$
$\Rightarrow \frac{96}{100}\times x=1176$
$\Rightarrow x=1225\text{g}$
Density given, $d=1.83\text{g/mL}$
$\therefore$ Volume $=\frac{1225}{1.83}=669.39\text{mL}$