90 g non - volatile, non - dissociative solution is added to 1746 g water to form a dilute, ideal solution. The vapour pressure of water has decreased from 300 mm of Hg to 291 mm of Hg. The molecular weight of solute is.

Options

A. 90
B. 60
C. 30
D. 15

Answer

To solve the problem step by step, we will use Raoult's Law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the mole fraction of the solvent in the solution.

### Step 1: Understand the given data
- Mass of solute (non-volatile, non-dissociative) = 90 g
- Mass of solvent (water) = 1746 g
- Vapor pressure of pure water (P₀) = 300 mm Hg
- Vapor pressure of the solution (P) = 291 mm Hg

### Step 2: Calculate the decrease in vapor pressure
The decrease in vapor pressure (ΔP) can be calculated as:
\[ \Delta P = P₀ - P = 300 \, \text{mm Hg} - 291 \, \text{mm Hg} = 9 \, \text{mm Hg} \]

### Step 3: Use Raoult's Law
According to Raoult's Law:
\[ \frac{P}{P₀} = X_A \]
where \(X_A\) is the mole fraction of the solvent (water).

### Step 4: Calculate the mole fraction of the solvent
Substituting the values:
\[ \frac{291}{300} = X_A \]
Calculating \(X_A\):
\[ X_A = \frac{291}{300} = 0.97 \]

### Step 5: Calculate the mole fraction of the solute
Since the total mole fraction must equal 1:
\[ X_B = 1 - X_A = 1 - 0.97 = 0.03 \]

### Step 6: Calculate the number of moles of the solvent (water)
The number of moles of water (solvent) can be calculated using its mass and molar mass:
\[ \text{Moles of water} = \frac{\text{mass}}{\text{molar mass}} = \frac{1746 \, \text{g}}{18 \, \text{g/mol}} = 97 \, \text{mol} \]

### Step 7: Relate the mole fraction to the number of moles
Let \(n_B\) be the number of moles of the solute. The mole fraction of the solute can be expressed as:
\[ X_B = \frac{n_B}{n_A + n_B} \]
Substituting the values:
\[ 0.03 = \frac{n_B}{97 + n_B} \]

### Step 8: Solve for the number of moles of solute
Cross-multiplying gives:
\[ 0.03(97 + n_B) = n_B \]
Expanding and rearranging:
\[ 2.91 + 0.03n_B = n_B \]
\[ 2.91 = n_B - 0.03n_B \]
\[ 2.91 = 0.97n_B \]
\[ n_B = \frac{2.91}{0.97} \approx 3 \, \text{mol} \]

### Step 9: Calculate the molecular weight of the solute
The molecular weight (M) of the solute can be calculated as:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{90 \, \text{g}}{3 \, \text{mol}} = 30 \, \text{g/mol} \]

### Final Answer
The molecular weight of the solute is **30 g/mol**.

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