The density of a solution containing 13% by mass of sulphuric acid is 1.09 g/mL. Calculate the molarity and normality of the solution

Volume of 100 g of the solution $=\frac{100}{\mathrm{d}}$

$=\frac{100}{1.09}\mathrm{mL}=\frac{100}{1.09\times 1000}\mathrm{litre}$

$=\frac{1}{1.09\times 10} \mathrm{litre}$

Number of moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ in 100 g of the solution

                  $=\frac{13}{98}$

Molarity $=\frac{\text{No. of moles of Solute}}{\text{Volume in litre}}$

$=\frac{13}{98}\times \frac{1.09\times 10}{1}=1.445\mathrm{M}$

                    $\left[OR\right]$

M = $$\begin{gathered} \mathrm{M} =\frac{\%\times \mathrm{d}\times 10}{\mathrm{GMW}} \\ \mathrm{M} =\frac{ 13\times 1.09 \times 10}{98} \\ \mathrm{M} =1.445\mathrm{M} \end{gathered}$$

N = M $\times$basicity of acid

N = 1.445 $\times$2 = 2.89