When 45 grams of a solute is added to 900 gm of water, its vapour pressure decreased from 30 mm to 24 mm. The mole fraction of the solvent in the solution is

Options

A. `0.2`
B. `0.8`
C. `0.1`
D. `0.9`

Explanation

To find the mole fraction of the solvent in the solution, we can follow these steps:

### Step 1: Understand the problem
We are given:
- Mass of solute = 45 grams
- Mass of solvent (water) = 900 grams
- Initial vapor pressure of pure solvent (P0) = 30 mm
- Vapor pressure of the solution (P) = 24 mm

### Step 2: Apply Raoult's Law
According to Raoult's Law, the relative lowering of vapor pressure is given by the formula:
\[ \frac{P_0 - P}{P_0} = x_2 \]
Where:
- \(P_0\) = vapor pressure of pure solvent
- \(P\) = vapor pressure of the solution
- \(x_2\) = mole fraction of the solute

### Step 3: Calculate the relative lowering of vapor pressure
Substituting the values into the formula:
\[ \frac{30 \, \text{mm} - 24 \, \text{mm}}{30 \, \text{mm}} = x_2 \]
Calculating the difference:
\[ \frac{6 \, \text{mm}}{30 \, \text{mm}} = x_2 \]
This simplifies to:
\[ x_2 = \frac{1}{5} = 0.2 \]

### Step 4: Calculate the mole fraction of the solvent
Since the sum of the mole fractions of solute and solvent must equal 1:
\[ x_1 + x_2 = 1 \]
Where \(x_1\) is the mole fraction of the solvent. Therefore:
\[ x_1 = 1 - x_2 = 1 - 0.2 = 0.8 \]

### Final Answer
The mole fraction of the solvent in the solution is:
\[ x_1 = 0.8 \]

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