The density $\left(\text{ in }{\mathrm{gmL}}^{-1}\text{ )}\right.$ of a $3.60\mathrm{M}$ sulphuric acid solution that is $29\mathrm{\%}{\mathrm{H}}_2{\mathrm{SO}}_4$ $(\mathrm{molar} \mathrm{mass}= 98 \mathrm{g} {\mathrm{mol}}^{-1})$ by mass will be 

In $100 \mathrm{g}$ solution, $29 \mathrm{g} {\mathrm{H}}_2\mathrm{S}{0}_4$is present. The volume of this solution as determined from the given molarity is

$V=\frac{n}{M}=\frac{\left(29\mathrm{g}/98\mathrm{g}{\mathrm{mol}}^{-1}\right)}{3.60{\mathrm{molL}}^{-1}}=0.0822\mathrm{L}\equiv 82.2\mathrm{mL}$

Hence, density of the solution is $\rho =\frac{m}{V}=\frac{100\mathrm{g}}{82.2\mathrm{mL}}=1.22\mathrm{g}{\mathrm{mL}}^{-1}$