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

Since molarity of solution is 3.60 M. It means 3.6 moles  
of ${\mathrm{H}}_2{\mathrm{SO}}_4$is present in its 1 litre solution. 
$$\begin{gathered} \text{ Mass of }3.6\text{ moles of }{\mathrm{H}}_2{\mathrm{SO}}_4 \\ =\text{ Moles }\times \text{ Molecular mass } \\ =3.6\times 98\mathrm{g}=352.8\mathrm{g} \\ \therefore 1000\mathrm{ml}\text{ solution has }352.8\mathrm{g}\text{ of }{\mathrm{H}}_2{\mathrm{SO}}_4 \\ \text{ Given that }29\mathrm{g}\text{ of }{\mathrm{H}}_2{\mathrm{SO}}_4\text{ is present in }=100\mathrm{g}\text{ of solution } \\ \therefore 352.8\mathrm{g}\text{ of }{\mathrm{H}}_2{\mathrm{SO}}_4\text{ is present in } \\ =\frac{100}{29}\times 352.8\mathrm{g}\text{ of solution } \\ = 1216 \mathrm{g} \mathrm{of} \mathrm{solution} \\ \text{ Density }=\frac{\text{ Mass }}{\text{ Volume }}=\frac{1216}{1000}=1.216\mathrm{g}/\mathrm{ml}=1.22\mathrm{g}/\mathrm{ml} \end{gathered}$$