Question

Calculate the weight of non-volatile solute (in grams) having molecular weight $40\text{u}$, which should be dissolved in $57\text{g}$ octane to reduce its vapour pressure to $80\%$:

Answer 1

$\text{Here, molecular weight of the solute is}=40\text{u}$
Again, solvent is octane $\left(C_8{H}_{18}\right)$
$\text{So, molar mass of the solvent is}=114{\text{g mol}}^{-1}$
Let, the vapour pressure of pure solvent be $P^o$
Hence according to the relative lowering of vapour pressure,
$\frac{P^0-0.8P^0}{P^0}=\frac{\frac{w}{40}}{\frac{w}{40}+\frac{57}{114}}$
Where, $w$ is the mass of solute present in the solution.
$\therefore \frac{P^0-0.8P^0}{P^0}=\frac{w}{w+20}$
$\Rightarrow 0.2=\frac{w}{w+20}$
$\Rightarrow 0.2w+4=w$
$\Rightarrow w=5\text{g}$