143. Calculate the molarity of 1 litre solution of 93% H₂SO₄ (weight/volume). The density of solution is 1.84 gm L^–1.

(i) Volume of solution = 1 L = 1000 mL
Density of solution = 1.84 g mL^–1
Mass of solution = V x d
= (1000 mL) x 1.84 mL^–1
= 18540 g
Mass of H₂SO₄ = 93g/100mL
 93 x 10 = 930 g
Mass of solvent (water)
= 1840 – 930 = 910 g.
(ii) Molarity (m)
   
   $= \frac{\mathrm{Mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4/\mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4}{\mathrm{Mass} \mathrm{of} \mathrm{water} \mathrm{in} \mathrm{kg}}$
  $$\begin{gathered} \mathrm{Mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4 \\ = 930 \mathrm{g}, \mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4 \\ = 98 \mathrm{g} {\mathrm{mol}}^{-1} \\ \mathrm{Mass} \mathrm{of} \mathrm{water} = \frac{910 \mathrm{g}}{1000} = 0.91 \mathrm{kg} \end{gathered}$$


$$\begin{gathered} \mathrm{Molarity} \left(\mathrm{m}\right) = \frac{930\mathrm{g}/98 {\mathrm{mol}}^{-1}}{(0.91 \mathrm{kg})} \\ = 10.43 \mathrm{mol} {\mathrm{kg}}^{-1} = 10.43 \mathrm{m}. \end{gathered}$$