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, we need to find the molar mass of a non-electrolyte solute that is dissolved in n-octane, which reduces its vapor pressure to 80% of the original vapor pressure. Here’s the step-by-step solution:

### Step 1: Understand the Given Data
- Mass of solute (non-electrolyte) = 8 g
- Mass of n-octane = 114 g
- Molar mass of n-octane = 114 g/mol
- Vapor pressure of the solution = 80% of the vapor pressure of pure n-octane

### Step 2: Calculate the Vapor Pressure Reduction
The reduction in vapor pressure can be expressed as:
\[ \frac{P_0 - P}{P_0} = \frac{n_b}{n_a + n_b} \]
Where:
- \( P_0 \) = vapor pressure of pure n-octane
- \( P \) = vapor pressure of the solution
- \( n_a \) = number of moles of n-octane
- \( n_b \) = number of moles of solute

Given that the vapor pressure of the solution is 80% of the pure vapor pressure:
\[ \frac{P_0 - 0.8P_0}{P_0} = \frac{0.2P_0}{P_0} = 0.2 \]
This means:
\[ \frac{n_b}{n_a + n_b} = 0.2 \]

### Step 3: Express the Mole Fraction
From the equation above, we can rearrange it to find the relationship between \( n_b \) and \( n_a \):
\[ 0.2(n_a + n_b) = n_b \]
\[ 0.2n_a + 0.2n_b = n_b \]
\[ 0.2n_a = n_b - 0.2n_b \]
\[ 0.2n_a = 0.8n_b \]
\[ n_a = 4n_b \]

### Step 4: Calculate the Number of Moles of n-Octane
Using the molar mass of n-octane, we can calculate the number of moles of n-octane:
\[ n_a = \frac{\text{mass of n-octane}}{\text{molar mass of n-octane}} = \frac{114 \text{ g}}{114 \text{ g/mol}} = 1 \text{ mol} \]

### Step 5: Find the Number of Moles of Solute
Using the relationship we derived:
\[ n_a = 4n_b \implies 1 = 4n_b \implies n_b = \frac{1}{4} = 0.25 \text{ mol} \]

### Step 6: Calculate the Molar Mass of the Solute
The molar mass of the solute can be calculated using the formula:
\[ \text{Molar Mass} = \frac{\text{mass of solute}}{n_b} = \frac{8 \text{ g}}{0.25 \text{ mol}} = 32 \text{ g/mol} \]

### Final Answer
The molar mass of the solute is **32 g/mol**.

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