In ideal solutions of non volatile solute B in solvent A in `2:5` molar ratio has vapour pressure 250 mm, if a another solution in ratio `3:4` prepared then vapour pressure above this solution?

Options

A. 200mm
B. 250 mm
C. 350 mm
D. 400mm

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent in the solution.

### Step-by-step Solution:

1. **Identify the Molar Ratios and Initial Conditions**:
- The first solution has a molar ratio of solute B to solvent A as 2:5.
- The vapor pressure of this solution is given as 250 mm.

2. **Calculate the Mole Fraction of Solute B in the First Solution**:
- Let the number of moles of solute B = 2n and the number of moles of solvent A = 5n.
- Total moles = 2n + 5n = 7n.
- Mole fraction of solute B (X_B) = moles of B / total moles = \( \frac{2n}{7n} = \frac{2}{7} \).

3. **Use Raoult's Law to Find the Vapor Pressure of Pure Solvent A (P°_A)**:
- According to Raoult's Law:
\[ \frac{P°_A - P_s}{P°_A} = X_B \]
- Here, \( P_s \) is the vapor pressure of the solution, which is 250 mm.
- Rearranging gives:
\[ P°_A - 250 = X_B \cdot P°_A \]
- Substituting \( X_B = \frac{2}{7} \):
\[ P°_A - 250 = \frac{2}{7} P°_A \]
- Multiplying through by 7 to eliminate the fraction:
\[ 7P°_A - 1750 = 2P°_A \]
- Rearranging gives:
\[ 5P°_A = 1750 \implies P°_A = \frac{1750}{5} = 350 \text{ mm} \]

4. **Calculate the Mole Fraction of Solute B in the Second Solution**:
- The second solution has a molar ratio of solute B to solvent A as 3:4.
- Let the number of moles of solute B = 3m and the number of moles of solvent A = 4m.
- Total moles = 3m + 4m = 7m.
- Mole fraction of solute B (X_B) = moles of B / total moles = \( \frac{3m}{7m} = \frac{3}{7} \).

5. **Use Raoult's Law Again to Find the Vapor Pressure of the Second Solution (P_s)**:
- Using the same formula:
\[ \frac{P°_A - P_s}{P°_A} = X_B \]
- Substituting \( P°_A = 350 \) mm and \( X_B = \frac{3}{7} \):
\[ \frac{350 - P_s}{350} = \frac{3}{7} \]
- Rearranging gives:
\[ 350 - P_s = \frac{3}{7} \cdot 350 \]
- Calculating \( \frac{3}{7} \cdot 350 = 150 \):
\[ 350 - P_s = 150 \]
- Thus, solving for \( P_s \):
\[ P_s = 350 - 150 = 200 \text{ mm} \]

### Final Answer:
The vapor pressure above the second solution is **200 mm**.