Question

Calculate the molality of a 1 L solution of 93% $H_{2}S{O}_4$ (w/v). The density of the solution is 1.84 g/mol.

• A. 10.4 m
• B. 5.1 m
• C. 12.7 m
• D. 9.3 m

Answer 1

The correct option is A 10.4 m

93% (w/v) means 93 g of the solute is present in 100 mL of the solution.
So, 1000 mL of the solution will contain 930 g of the solute.
$\text{Density of solution}=\frac{\text{weight of solution}}{\text{voulme of the solution}}$
Weight of the solution = $volume\times density=1000\times 1.84=1840g$
Weight of the solvent
= weight of solution - weight of solute
Weight of the solvent
= 1840 - 930 = 910 g
Number of moles of $H_{2}S{O}_4=\frac{\text{given mass}}{\text{molar mass}}=\frac{930}{98}=9.49mol$
Molality = $\frac{\text{moles of solute}}{\text{mass of solvent in g}}\times 1000$
Molality = $\frac{9.49}{910}\times 1000$ = 10.4 m