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: Find moles of each solute.
Urea: 12 g, molar mass 60 g/mol, so \( n_{urea} = 12/60 = 0.2 \) mol. Glucose: 36 g, molar mass 180 g/mol, so \( n_{glucose} = 36/180 = 0.2 \) mol.
Step 2: Find moles of water.
Mass of water = 100 g, molar mass = 18 g/mol. So \( n_{water} = 100/18 \approx 5.556 \) mol.
Step 3: Find total moles of solute.
Both solutes are non-electrolytes (no dissociation). Total \( n_{solute} = 0.2 + 0.2 = 0.4 \) mol.
Step 4: Calculate mole fraction of water.
\[ X_{water} = \frac{5.556}{5.556 + 0.4} = \frac{5.556}{5.956} \approx 0.9329 \]
Step 5: Apply Raoult law.
For a solution with non-volatile solutes: \[ P = P^0 \times X_{water} = 25 \times 0.9329 \approx 23.32 \text{ torr} \]
Step 6: Match with options.
The vapour pressure is approximately 23.32 torr, matching option 4. \[ \boxed{23.32 \text{ torr}} \]