The density of 95.2 mass $\mathrm{\%}{\mathrm{H}}_2{\mathrm{SO}}_4$ is $1.53\mathrm{g}{\mathrm{cm}}^{-3}$ The molarity of this solution is

Let there be 100 g of solution. Its volume will be $V=\frac{m}{\rho }=\frac{100\mathrm{g}}{1.53\mathrm{g}{\mathrm{cm}}^{-3}}=\left(\frac{100}{1.53}\right){\mathrm{cm}}^3=\frac{0.10}{1.53}{\mathrm{dm}}^3$

This solution contains 95.2 g of ${\mathrm{H}}_2{\mathrm{SO}}_4$ Its amount is $n=\frac{m}{M_{\mathrm{m}}}=\frac{95.2\mathrm{g}}{98\mathrm{g}{\mathrm{mol}}^{-1}}$

Hence, the molarity of solution is $M=\frac{n}{V}=\frac{(95.2/98)}{(0.1/1.53){\mathrm{dm}}^3}=14.86\mathrm{mol}{\mathrm{dm}}^{-3}$