Question

Calculate the weight of non - volatile solute having molecular weight $40$, which should be dissoved in $57\text{g}$ octane to reduce its vapour pressure to $80\%$.

• A. $2.5\text{g}$
• B. $5\text{g}$
• C. $0.66\text{g}$
• D. $10g$

Answer 1

The correct option is B $5\text{g}$

Let, $P_s\text{be the vapour pressure of solution}$
$P^0\text{be the vapour pressure of pure octane}$
$P_s=\left(\frac{80}{100}\right)P^{\circ },w=?M_{solute}=40{\text{g mol}}^{-1},W_{solvent}=57\text{g},M_{solvent}=114{\text{g mol}}^{-1}\because \frac{P^{\circ }-P_s}{P_s}=\frac{w\times M_{solvent}}{M_{solute}\times W}\Rightarrow \frac{P^{\circ }-\left(\frac{80}{100}\right)P^{\circ }}{\left(\frac{80}{100}\right)P^{\circ }}=\frac{w\times 114}{40\times 57};\therefore w=5\text{g}$