1. Use Raoult’s law for a non-volatile solute

For a dilute solution containing a non-volatile solute, $P_{\text{solution}} = X_{\text{water}} \, P^0_{\text{water}}$ where:

• $P_{\text{solution}}$ = vapour pressure of solution
• $P^0_{\text{water}}$ = vapour pressure of pure water
• $X_{\text{water}}$ = mole fraction of water

2. Given reduction in vapour pressure

The vapour pressure is reduced to one-half of pure water, so $P_{\text{solution}} = \frac{1}{2}P^0_{\text{water}}$

Thus, $X_{\text{water}} = \frac{P_{\text{solution}}}{P^0_{\text{water}}} = \frac{1}{2}$

3. Calculate moles of water

Mass of water = $100\,\text{g}$

Moles of water: $n_{\text{water}} = \frac{100}{18} = 5.56 \text{ mol (approx)}$

4. Use mole fraction expression

Let moles of solute $A$ be $n_A$.

Then, $X_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{water}} + n_A}$

Since $X_{\text{water}} = \frac{1}{2}$, $\frac{5.56}{5.56 + n_A} = \frac{1}{2}$

Cross-multiplying, $2(5.56) = 5.56 + n_A$ $n_A = 5.56$

5. Nearest integer

$n_A \approx 6$

Final Answer

The number of moles of solute added is $\boxed{6}$