The vapour pressure of ether at `20^(@)C` is 442 mm. When 7.2 g of a solute is dissolved in 60 g ether, vapour pressure is lowered by 32 units. If molecular mass of ether is 74 then molecular mass of solute is:

Options

A. 113.77 g/mole
B. 150.35 g/mole
C. 116.23 g/mole
D. 190.26 g/mole

Answer

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

### Step 1: Understand the Given Information
- Vapor pressure of pure ether (P₀) = 442 mm
- Lowering of vapor pressure (ΔP) = 32 mm
- Mass of solute (W) = 7.2 g
- Mass of ether (m₁) = 60 g
- Molecular mass of ether (M₁) = 74 g/mol

### Step 2: Calculate the Vapor Pressure of the Solution
The vapor pressure of the solution (P) can be calculated as:
\[ P = P₀ - ΔP = 442 \, \text{mm} - 32 \, \text{mm} = 410 \, \text{mm} \]

### Step 3: Calculate the Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure is given by:
\[ \text{Relative lowering} = \frac{ΔP}{P₀} = \frac{32}{442} \]

### Step 4: Use Raoult's Law
According to Raoult's Law, the relative lowering of vapor pressure is equal to the mole fraction of the solute (X₂):
\[ \frac{ΔP}{P₀} = X₂ \]

### Step 5: Calculate the Mole Fraction of the Solute
The mole fraction of the solute (X₂) can also be expressed in terms of the number of moles of solute and solvent:
\[ X₂ = \frac{n₂}{n₁ + n₂} \]
Where:
- \( n₁ \) = number of moles of ether
- \( n₂ \) = number of moles of solute

### Step 6: Calculate the Number of Moles of Ether
The number of moles of ether (n₁) can be calculated as:
\[ n₁ = \frac{m₁}{M₁} = \frac{60 \, \text{g}}{74 \, \text{g/mol}} \approx 0.8108 \, \text{mol} \]

### Step 7: Set Up the Equation for Mole Fraction
Let the molecular mass of the solute be M. The number of moles of solute (n₂) is:
\[ n₂ = \frac{W}{M} = \frac{7.2 \, \text{g}}{M} \]

Now, substituting into the mole fraction equation:
\[ X₂ = \frac{n₂}{n₁ + n₂} = \frac{\frac{7.2}{M}}{0.8108 + \frac{7.2}{M}} \]

### Step 8: Substitute and Solve for M
Using the relative lowering calculated earlier:
\[ \frac{32}{442} = \frac{\frac{7.2}{M}}{0.8108 + \frac{7.2}{M}} \]

Cross-multiplying gives:
\[ 32(0.8108 + \frac{7.2}{M}) = 442 \cdot \frac{7.2}{M} \]

### Step 9: Simplify the Equation
Expanding and rearranging:
\[ 25.9136 + \frac{230.4}{M} = \frac{3195.6}{M} \]

Combining terms:
\[ 25.9136 = \frac{3195.6 - 230.4}{M} \]
\[ 25.9136 = \frac{2965.2}{M} \]

### Step 10: Solve for M
Rearranging gives:
\[ M = \frac{2965.2}{25.9136} \approx 114.0 \, \text{g/mol} \]

### Conclusion
The molecular mass of the solute is approximately **114.0 g/mol**.

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