Question

The density of $2.45$ M aqueous methanol $\left(C{H}_{3}OH\right)$ is $0.976$ g/mL. What is the molality of the solution (Molecular weight of $C{H}_{3}OH=32)$?

• A. $27.3$ m
• B. $0.273$ m
• C. $7.23$ m
• D. $2.73$ m

Answer 1

Answer: (D)

$2.73$ m

$1$ L of solution contains $2.45$ moles of methanol.
$1$ L of solution weighs $0.976\times 1000=976$ g.
$2.45$ moles of methanol weighs $2.45\times 32=78.4$ g.
Hence, the mass of water is $976-78.4=897.6=0.8976$ kg.
The molality of the solution is $m=\frac{\text{Number of moles of methanol}}{\text{Mass of water (in kg)}}=\frac{2.45mol}{0.8976kg}=2.73m$.