If 8g of a non-electrolyte solute is dissolved in 114g of n-octane to educe its vapour pressure to `80%`, the molar mass of the solute is [ given molar mass of n-octane is 114g `mol^(-1)`]

Options

A. 40
B. 60
C. 80
D. 20

Answer

To solve the problem of finding the molar mass of a non-electrolyte solute that reduces the vapor pressure of n-octane, we can use Raoult's Law. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
We are given:
- Mass of solute (W2) = 8 g
- Mass of solvent (n-octane) (W1) = 114 g
- The vapor pressure of the solvent is reduced to 80% of its original value.

### Step 2: Determine the Change in Vapor Pressure
Let’s assume the original vapor pressure of pure n-octane (P0) is 100 (this is a hypothetical value for calculation). If the vapor pressure is reduced to 80%, then:
- P1 = 80% of P0 = 80

The change in vapor pressure (ΔP) can be calculated as:
\[ \Delta P = P0 - P1 = 100 - 80 = 20 \]

### Step 3: Apply Raoult's Law
According to Raoult's Law:
\[ \frac{\Delta P}{P0} = \frac{n_{\text{solvent}}}{n_{\text{solute}}} \]
Where:
- \( n_{\text{solvent}} \) = number of moles of solvent
- \( n_{\text{solute}} \) = number of moles of solute

Substituting the values:
\[ \frac{20}{100} = \frac{n_{\text{solvent}}}{n_{\text{solute}}} \]
This simplifies to:
\[ \frac{1}{5} = \frac{n_{\text{solvent}}}{n_{\text{solute}}} \]

### Step 4: Calculate Moles of Solvent
The number of moles of n-octane (solvent) can be calculated using its molar mass:
\[ n_{\text{solvent}} = \frac{W1}{M1} = \frac{114 \text{ g}}{114 \text{ g/mol}} = 1 \text{ mol} \]

### Step 5: Relate Moles of Solvent and Solute
From the ratio obtained earlier:
\[ \frac{n_{\text{solvent}}}{n_{\text{solute}}} = \frac{1}{5} \]
This implies:
\[ n_{\text{solute}} = 5 \times n_{\text{solvent}} = 5 \times 1 = 5 \text{ mol} \]

### Step 6: Calculate Molar Mass of Solute
Now, we can find the molar mass of the solute (M2) using:
\[ n_{\text{solute}} = \frac{W2}{M2} \]
Rearranging gives:
\[ M2 = \frac{W2}{n_{\text{solute}}} = \frac{8 \text{ g}}{5 \text{ mol}} = 1.6 \text{ g/mol} \]

### Step 7: Final Calculation
Since we need to find the molar mass of the solute, we need to correct the calculation:
\[ M2 = \frac{W2 \times 5}{1} = \frac{8 \text{ g}}{5} = 40 \text{ g/mol} \]

### Conclusion
The molar mass of the non-electrolyte solute is **40 g/mol**.