Question

Calculate the molality of a sulphuric acid solution of specific gravity $1.2$ containing $27\%$ $H_{2}S{O}_4$ by weight.

Answer 1

1000 g of solution contain 270 g of sulphuric acid which corresponds to $\frac{270}{98}=2.755$ moles.
The mass of the solvent is $1000-270=730g$
The molality of the solution is $\frac{\text{number of moles of sulphuric acid}}{\text{mass of water (in kg)}}=\frac{2.755}{0.730}=3.8m\approx 4m.$