Question

Calculate the mass of a non-volatile solute (molar mass $40 \, \text{gm mol}^{-1}$) which should be dissolved in $114 \, \text{gm$ octane to reduce its vapour pressure to 80%.}

Hint:

In Raoult’s law problems, start with mole fraction formula and use the relation $\frac{P}{P^\circ} = X_{solvent}$.

Answer 1

Step 1: Apply Raoult’s Law.
\[ \frac{P}{P^\circ} = X_{solvent} \] Given: $\frac{P}{P^\circ} = 0.8$
Thus, $X_{solvent} = 0.8$ Step 2: Relation between mole fractions.
\[ X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}} = 0.8 \] Step 3: Calculate moles of solvent.
Molar mass of octane = $114 \, g mol^{-1}$
Mass of octane = $114 \, g$
\[ n_{solvent} = \frac{114}{114} = 1 \, mol \] Step 4: Substitute values.
\[ \frac{1}{1 + n_{solute}} = 0.8 \] \[ 1 = 0.8 (1 + n_{solute}) \] \[ 1 = 0.8 + 0.8n_{solute} \] \[ 0.2 = 0.8n_{solute} \] \[ n_{solute} = 0.25 \, mol \] Step 5: Calculate mass of solute.
Molar mass = $40 \, g mol^{-1}$
\[ \text{Mass} = n \times M = 0.25 \times 40 = 10 \, g \] Conclusion:
The required mass of solute is: \[ \boxed{10 \, g} \]