Question

When a certain amount of solid A is dissolved in 100 g of water at 25°C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is _________. (Nearest Integer).

Answer 1

Correct Answer: 6

To solve the problem, we use Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. The given problem states that the vapor pressure of the solution is half that of pure water. 
Therefore, \(P_{\text{solution}} = \frac{1}{2} \times P_{\text{pure water}}\).
Purer water vapor pressure: \(P_{\text{pure water}} = 23.76 \text{ mmHg}\).
Thus, \(P_{\text{solution}} = \frac{1}{2} \times 23.76 = 11.88 \text{ mmHg}\).
By Raoult's Law: \[ P_{\text{solution}} = X_{\text{water}} \times P_{\text{pure water}} \] where \(X_{\text{water}}\) is the mole fraction of water. So, \[ X_{\text{water}} \times 23.76 = 11.88 \] Solving for \(X_{\text{water}}\), \[ X_{\text{water}} = \frac{11.88}{23.76} = \frac{1}{2} \] Since \(X_{\text{water}} + X_{\text{solute}} = 1\), the mole fraction of solute \(X_{\text{solute}} = \frac{1}{2}\). Assume the number of moles of water \(n_{\text{water}}\) is given by: \[ n_{\text{water}} = \frac{100 \text{ g}}{18 \text{ g/mol}} = 5.56 \text{ mol} \] Since \(X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}}\), thus: \[ \frac{n_{\text{solute}}}{n_{\text{solute}} + 5.56} = \frac{1}{2} \] This implies: \[ 2n_{\text{solute}} = n_{\text{solute}} + 5.56\] \[ n_{\text{solute}} = 5.56 \text{ mol} \] Rounding 5.56 gives approximately 6 moles. Therefore, the number of moles of solute A added is 6, which is within the expected range (6,6).