When 25g of a non-volatile solute is dissolved in 100g of water, the vapour pressure is lowered by 0.225 mm. If the vapour pressure of water at `25^(@)C` is 17.5 mm, what is the molecular mass of solute ?

Options

A. 206
B. 302
C. 318
D. 276

Explanation

To find the molecular mass of the non-volatile solute, we can use the concept of relative lowering of vapor pressure. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Mass of solute (W_solute) = 25 g
- Mass of solvent (W_solvent) = 100 g (water)
- Vapor pressure lowering (ΔP) = 0.225 mm
- Vapor pressure of pure water (P₀) = 17.5 mm

### Step 2: Calculate the Vapor Pressure of the Solution
The vapor pressure of the solution (P_s) can be calculated using the formula:
\[ P_s = P_0 - \Delta P \]
Substituting the values:
\[ P_s = 17.5 \, \text{mm} - 0.225 \, \text{mm} = 17.275 \, \text{mm} \]

### Step 3: Calculate the Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure (X_B) can be expressed as:
\[ \frac{P_0 - P_s}{P_0} = \frac{0.225 \, \text{mm}}{17.5 \, \text{mm}} \]
Calculating this gives:
\[ X_B = \frac{0.225}{17.5} \approx 0.012857 \]

### Step 4: Relate Mole Fraction to Moles of Solute and Solvent
The mole fraction of solute (X_B) is given by:
\[ X_B = \frac{n_B}{n_B + n_A} \]
Where:
- \( n_B \) = moles of solute = \( \frac{W_{solute}}{M} \)
- \( n_A \) = moles of solvent (water) = \( \frac{W_{solvent}}{M_{water}} \)

The molecular weight of water (M_water) is 18 g/mol. Thus:
\[ n_A = \frac{100 \, \text{g}}{18 \, \text{g/mol}} \approx 5.56 \, \text{mol} \]

### Step 5: Substitute Values into the Mole Fraction Equation
Substituting \( n_B \) and \( n_A \) into the mole fraction formula:
\[ 0.012857 = \frac{\frac{25}{M}}{\frac{25}{M} + 5.56} \]

### Step 6: Cross-Multiply and Rearrange
Cross-multiplying gives:
\[ 0.012857 \left( \frac{25}{M} + 5.56 \right) = \frac{25}{M} \]
Expanding and rearranging:
\[ 0.012857 \cdot 5.56 = \frac{25}{M} - 0.012857 \cdot \frac{25}{M} \]
\[ 0.0715 = \frac{25}{M} (1 - 0.012857) \]
\[ 0.0715 = \frac{25}{M} \cdot 0.987143 \]

### Step 7: Solve for M
Now, solving for M:
\[ M = \frac{25 \cdot 0.987143}{0.0715} \]
Calculating this gives:
\[ M \approx 345.5 \, \text{g/mol} \]

### Final Answer
The molecular mass of the solute is approximately **345.5 g/mol**.

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