Question

A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then the number of moles of the solvent present in the solution is ______. (Nearest integer)
[Given: boiling point of the pure solvent = 319.5 K, K_b of the solvent = 0.3 K kg mol⁻¹]

Answer 1

Correct Answer: 5

Step 1: Understanding the Concept:
We use the elevation in boiling point to find the moles of solute, then apply the relative lowering of vapour pressure (RLVP) to find the moles of the solvent.

Step 2: Key Formula or Approach:
1. $\Delta T_b = K_b \times m$ 2. $\frac{P^o - P_s}{P^o} = \frac{n}{n+N}$ (Mole fraction of solute)

Step 3: Detailed Explanation:
1. Find Molality (m): $\Delta T_b = 320 - 319.5 = 0.5 \text{ K}$. $0.5 = 0.3 \times m \implies m = 1.667 \text{ mol/kg}$. 2. Find Moles of Solute (n): $m = \frac{n}{\text{mass of solvent in kg}} \implies 1.667 = \frac{n}{0.040}$. $n = 0.0667 \text{ moles}$. 3. Use RLVP: $\frac{760 - 750}{760} = \frac{10}{760} = \frac{1}{76}$. $\frac{1}{76} = \frac{0.0667}{0.0667 + N} \implies 0.0667 + N = 76(0.0667)$. $N = 75 \times 0.0667 = 5.0025 \approx 5$.

Step 4: Final Answer:
The number of moles of the solvent is 5.