Question

Calculate the molality of a 1 L solution of 80% $H_{2}S{O}_4$$\left(\frac{w}{v}\right)$, given that the density of the solution is 1.80 g $m{L}^{-1}$.

• A. 8.16 m
• B. 8.60 m
• C. 1.02 m
• D. 10.8 m

Answer 1

The correct option is A 8.16 m

Let the volume of the solution be 1000 mL.
The sample of $H_{2}S{O}_4$ is 80% by volume.
Hence, weight of $H_{2}S{O}_4$ = $w\left(H_{2}S{O}_4\right)$ = 800 g in 1000 mL
Moles of $H_{2}S{O}_4=\frac{\text{given mass}}{\text{molar mass}}=\frac{800}{98}=8.16$
Weight of solution $=1000\times 1.80=1800$ g
Weight of solvent = $1800–800=1000g$
Molality $=\frac{\text{Moles of solute}}{\text{weight of solvent (in g)}}\times 1000$
Molality $=\frac{8.16}{1000}\times 1000=8.16$ m