Question

At \(298K\), the vapour pressure of pure water is \(25\ torr\). The vapour pressure of water, when \(12\ g\) of urea \((\text{molar mass }=60\ g\ mol^{-1})\) and \(36\ g\) of glucose \((\text{molar mass }=180\ g\ mol^{-1})\) is dissolved in \(100\ g\) of water at the same temperature \((in\ torr)\) is

• A. \(25.02\)
• B. \(24.12\)
• C. \(23.92\)
• D. \(23.32\)

Hint:

For non-volatile solutes, vapour pressure of solution is calculated by: \[ P=P^0X_{\text{solvent}} \] where \(X_{\text{solvent}}\) is the mole fraction of solvent.

Answer 1

The Correct Option is D

Step 1: Use Raoult's law.
For a solution containing non-volatile solutes, \[ P=P^0X_{\text{water}} \] where, \[ P^0=25\ torr \] and \[ X_{\text{water}}=\frac{n_{\text{water}}}{n_{\text{water}}+n_{\text{solute}}} \]

Step 2: Calculate moles of water.
Mass of water: \[ 100\ g \] Molar mass of water: \[ 18\ g\ mol^{-1} \] Therefore, \[ n_{\text{water}}=\frac{100}{18} \] \[ n_{\text{water}}=5.56\ mol \]

Step 3: Calculate moles of solutes.
Moles of urea: \[ n_{\text{urea}}=\frac{12}{60} \] \[ n_{\text{urea}}=0.2\ mol \] Moles of glucose: \[ n_{\text{glucose}}=\frac{36}{180} \] \[ n_{\text{glucose}}=0.2\ mol \] Total moles of solute: \[ n_{\text{solute}}=0.2+0.2 \] \[ n_{\text{solute}}=0.4\ mol \]

Step 4: Calculate mole fraction of water.
\[ X_{\text{water}}=\frac{5.56}{5.56+0.4} \] \[ X_{\text{water}}=\frac{5.56}{5.96} \] \[ X_{\text{water}}\approx0.933 \]

Step 5: Calculate vapour pressure of solution.
\[ P=P^0X_{\text{water}} \] \[ P=25\times0.933 \] \[ P=23.32\ torr \]

Step 6: Final conclusion.
Hence, the vapour pressure of water in the solution is \[ \boxed{23.32\ torr} \]