The volume required to prepare $2.0\text{L}$ of $3.0\text{M}$ $H_{2}S{O}_4$ solution from $96\%\text{( w/w)}$ $H_{2}S{O}_4$ solution (density $=2g{\text{mL}}^{-1}$) is

The correct option is C $306.25\text{mL}$
Moles of $HCl$ in $2.0\text{L}$ of $3.0\text{M}$ solution
$=\left(2.0\text{L}\right)\times \left(3.0{\text{mol L}}^{-1}\right)=6.0\text{mol}$
Mass of $H_{2}S{O}_4$ in this solution $=\left(6.0\text{mol}\right)\times \left(98{\text{g mol}}^{-1}\right)=588\text{g}$
Mass of $96\%$ $H_{2}S{O}_4$ solution containing $588\text{g}$ of $H_{2}S{O}_4$
$=\frac{588\times 100\text{g}}{96}=612.5\text{g}$
Volume of $96\%$ $H_{2}S{O}_4$ solution weighing $612.5\text{g}$
$=\frac{612.5\text{g}}{2{\text{g~mL}}^{-1}}=306.25\text{mL}$