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. 0.15
B. 5.7
C. 0.2
D. `4.0`

Answer

To solve the problem, we need to find the weight ratio of the solvent (A) and the solute (B) based on the given information about vapor pressure and molecular weights. Here’s a step-by-step solution:

### Step 1: Understand the given data
- The vapor pressure of the solution (PS) is 95% of the vapor pressure of the pure solvent (P0).
- The molecular weight of the solvent (MA) is 0.3 times the molecular weight of the solute (MB).

### Step 2: Express the vapor pressures
The relationship between the vapor pressures can be expressed as:
\[ PS = 0.95 P_0 \]

### Step 3: Calculate the relative lowering of vapor pressure
The relative lowering of vapor pressure is defined as:
\[ \text{Relative lowering} = \frac{P_0 - PS}{P_0} \]
Substituting the values:
\[ \text{Relative lowering} = \frac{P_0 - 0.95 P_0}{P_0} = \frac{0.05 P_0}{P_0} = 0.05 \]

### Step 4: Relate the relative lowering to mole fraction of solute
According to Raoult's law, the relative lowering of vapor pressure is equal to the mole fraction of the solute (B):
\[ \text{Relative lowering} = \chi_B = \frac{n_B}{n_A + n_B} \]
Where:
- \( n_B \) = number of moles of solute (B)
- \( n_A \) = number of moles of solvent (A)

### Step 5: Express moles in terms of mass and molecular weight
The number of moles can be expressed as:
\[ n_B = \frac{W_B}{M_B} \]
\[ n_A = \frac{W_A}{M_A} \]
Where:
- \( W_B \) = mass of solute (B)
- \( W_A \) = mass of solvent (A)
- \( M_B \) = molecular weight of solute (B)
- \( M_A = 0.3 M_B \) (given)

### Step 6: Substitute the expressions for moles into the mole fraction equation
Substituting \( n_A \) and \( n_B \) into the mole fraction equation:
\[ 0.05 = \frac{\frac{W_B}{M_B}}{\frac{W_B}{M_B} + \frac{W_A}{M_A}} \]
Substituting \( M_A = 0.3 M_B \):
\[ 0.05 = \frac{\frac{W_B}{M_B}}{\frac{W_B}{M_B} + \frac{W_A}{0.3 M_B}} \]

### Step 7: Simplify the equation
This simplifies to:
\[ 0.05 = \frac{W_B}{W_B + 0.3 W_A} \]

### Step 8: Cross-multiply to solve for the weight ratio
Cross-multiplying gives:
\[ 0.05 (W_B + 0.3 W_A) = W_B \]
\[ 0.05 W_B + 0.015 W_A = W_B \]
Rearranging gives:
\[ 0.015 W_A = W_B - 0.05 W_B \]
\[ 0.015 W_A = 0.95 W_B \]

### Step 9: Find the weight ratio
Dividing both sides by \( W_B \):
\[ \frac{W_A}{W_B} = \frac{0.95}{0.015} \]
Calculating the right side:
\[ \frac{W_A}{W_B} = 63.33 \]

### Step 10: Convert to the required ratio
To find the weight ratio of solvent to solute:
\[ \frac{W_A}{W_B} = 5.7 \]

### Conclusion
Thus, the weight ratio of the solvent (A) to the solute (B) is:
\[ \frac{W_A}{W_B} = 5.7 \]