The vapour pressure of a solution of a non-volatile electrolyte `B` in a solvent `A` is `95%` of the vapour pressure of the solvent at the same temperature. If the molecular weight of the solvent is `0.3` times, the molecular weight of solute, the weight ratio of the solvent and solute are:

Options

A. (a) `0.15`
B. (b) `5.7`
C. (c ) `0.2`
D. (d) `4.0`

Explanation

To solve the problem, we will use Raoult's Law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the mole fraction of the solvent in the solution. Let's break down the steps:

### Step 1: Understand the Given Information
- The vapor pressure of the solution (PS) is 95% of the vapor pressure of the pure solvent (P0).
- The molecular weight of the solvent (M) is 0.3 times the molecular weight of the solute (m).

### Step 2: Apply Raoult's Law
According to Raoult's Law:
\[ P_S = X_A \cdot P_0 \]
Where:
- \( P_S \) = vapor pressure of the solution
- \( X_A \) = mole fraction of the solvent
- \( P_0 \) = vapor pressure of the pure solvent

Since \( P_S = 0.95 P_0 \), we can write:
\[ 0.95 P_0 = X_A \cdot P_0 \]
Dividing both sides by \( P_0 \):
\[ X_A = 0.95 \]

### Step 3: Relate Mole Fraction to Weights
The mole fraction \( X_A \) can also be expressed in terms of the number of moles of solvent and solute:
\[ X_A = \frac{n_A}{n_A + n_B} \]
Where:
- \( n_A \) = moles of solvent
- \( n_B \) = moles of solute

Using the molecular weights, we can express the number of moles:
\[ n_A = \frac{W_A}{M} \]
\[ n_B = \frac{W_B}{m} \]
Where:
- \( W_A \) = weight of solvent
- \( W_B \) = weight of solute

Substituting these into the mole fraction equation:
\[ X_A = \frac{\frac{W_A}{M}}{\frac{W_A}{M} + \frac{W_B}{m}} \]

### Step 4: Substitute Known Values
We know that \( M = 0.3m \). Thus:
\[ X_A = \frac{\frac{W_A}{M}}{\frac{W_A}{M} + \frac{W_B}{0.3M}} \]

This simplifies to:
\[ X_A = \frac{W_A}{W_A + \frac{W_B}{0.3}} \]

### Step 5: Set Up the Equation
Since \( X_A = 0.95 \):
\[ 0.95 = \frac{W_A}{W_A + \frac{W_B}{0.3}} \]

### Step 6: Cross Multiply and Solve for Weight Ratio
Cross-multiplying gives:
\[ 0.95 \left(W_A + \frac{W_B}{0.3}\right) = W_A \]
Expanding this:
\[ 0.95 W_A + \frac{0.95 W_B}{0.3} = W_A \]
Rearranging gives:
\[ W_A - 0.95 W_A = \frac{0.95 W_B}{0.3} \]
\[ 0.05 W_A = \frac{0.95 W_B}{0.3} \]

### Step 7: Solve for the Weight Ratio
Rearranging gives:
\[ \frac{W_A}{W_B} = \frac{0.95}{0.3 \times 0.05} \]
Calculating the right side:
\[ \frac{W_A}{W_B} = \frac{0.95}{0.015} = 63.33 \]

### Step 8: Final Weight Ratio
Thus, the weight ratio of the solvent to the solute is:
\[ \frac{W_A}{W_B} = 63.33:1 \]