The density $\left(\mathrm{in}\mathrm{g}{\mathrm{mL}}^{-1}\right)$ of a 3.6 M sulphuric acid solution that is 29% ${\mathrm{H}}_2{\mathrm{SO}}_4$ (Molar mass = 98 g ${\mathrm{mol}}^{-1})$ by mass will be 

Let the density of solution be ‘d’

Molarity of solution given = 3.6 M

i.e., 1 liter of solution contains 3.6 moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ or 1 liter of solution contains $3.6\times 98\mathrm{g}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4$ 

Since, the solution contains 29 % by mass.

100 g solution contains 29 g of ${\mathrm{H}}_2{\mathrm{SO}}_4$

$\begin{array}{l}\frac{100}{\mathrm{d}}\mathrm{mL}\mathrm{solution}\mathrm{contains}29\mathrm{g}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4\\ 1000\mathrm{mL}\mathrm{solution}\mathrm{contains}3.6\times 98\mathrm{g}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4\\ \therefore 3.6\times 98=\frac{29\times \mathrm{d}}{100}\times 1000\\ \mathrm{d}=1.22\end{array}$

Another method:

                                 $$\begin{gathered} \mathrm{M}=\frac{\%\mathrm{by} \mathrm{mass}\times 10\times \mathrm{d}}{\mathrm{Molecular} \mathrm{weight}} \\ 3.6=\frac{29\times 10\times \mathrm{d}}{98} \\ \mathrm{d}=\frac{3.6\times 98}{29\times 10}=1.22 \mathrm{g}/\mathrm{ml} \end{gathered}$$