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 step by step, we will use Raoult's law and the information provided in the question.

### Step 1: Understand the given data
We have an ideal solution of a non-volatile solute B in solvent A with a molar ratio of 2:5, and the vapor pressure of this solution is given as 250 mm. We need to find the vapor pressure of another solution with a molar ratio of 3:4.

### Step 2: Calculate the mole fraction of the solvent in the first solution
The molar ratio of solute (B) to solvent (A) is 2:5.

- Total moles = 2 (solute) + 5 (solvent) = 7 moles
- Mole fraction of solvent (A), \( X_A \) = \( \frac{moles \ of \ solvent}{total \ moles} = \frac{5}{7} \)

Calculating \( X_A \):
\[ X_A = \frac{5}{7} \approx 0.714 \]

### Step 3: Calculate the partial pressure of the pure solvent
According to Raoult's law:
\[ P_{solution} = P^0_A \cdot X_A \]
Where:
- \( P_{solution} \) = vapor pressure of the solution = 250 mm
- \( P^0_A \) = vapor pressure of the pure solvent
- \( X_A \) = mole fraction of the solvent

Rearranging the equation to find \( P^0_A \):
\[ P^0_A = \frac{P_{solution}}{X_A} = \frac{250 \ mm}{0.714} \approx 350.35 \ mm \]

### Step 4: Calculate the mole fraction of the solvent in the second solution
For the second solution with a molar ratio of 3:4:
- Total moles = 3 (solute) + 4 (solvent) = 7 moles
- Mole fraction of solvent (A), \( X_A' \) = \( \frac{4}{7} \)

Calculating \( X_A' \):
\[ X_A' = \frac{4}{7} \approx 0.571 \]

### Step 5: Calculate the vapor pressure of the second solution
Using Raoult's law again:
\[ P_{solution}' = P^0_A \cdot X_A' \]
Substituting the values:
\[ P_{solution}' = 350.35 \ mm \cdot 0.571 \approx 200.7 \ mm \]

### Step 6: Round the answer
Rounding 200.7 mm gives us approximately 200 mm.

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

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