# The volume of 96% $H_2SO_4$ (density $1.83gmL^{-1}$) required to prepare 2.0L of 3.0M $H_2SO_4$ solution is

$\begin{array}{1 1}33.47mL\\3.347mL\\334.7mL\\343.7mL\end{array}$

Volume of 96% of $H_2SO_4=(2.0L)(3.0molL^{-1})(98gmol^{-1})\big(\large\frac{1}{1.83gmL^{-1}}\big)$$=334.7mL$