Question

Calculate the molality of $1$ litre solution containing $93$% $H_2{SO}_4$ (W/V) if the density of the solution is $1.84gm{L}^{-1}$.

Answer 1

93 % $H_{2}S{O}_4$ w/v means 93 g of $H_{2}S{O}_4$ is present in 100cc of solution.

Since volume is 1000ml, amount of $H_{2}S{O}_4$ is 930 g/litre. Also density is given 1.84g/ml

M= density x volume = 1.84 x 1000 = 1840 g

wt of solvent = 1840 - 930 = 910 g

$\because$ molality$=\frac{\text{moles of solute}}{\text{Kg of solvent}}$

$\therefore$ molality $=\frac{930}{98}\times \frac{1000}{910}=10.43mol/Kg$