Two liquids `X` and `Y` are perfectly immisoible. If `X` and `Y` have molecular masses in ratio `1:2`, the total vapour pressure of a mixture of `X` and `Y` prepared in weight ratio `2:3` should be `(P_(X^(0))=400` torr, `P_(Y^(0))=200`torr)

Options

A. `600` torr
B. `400` torr
C. `800` torr
D. `1000` torr

Answer

To solve the problem, we need to calculate the total vapor pressure of a mixture of two immiscible liquids, X and Y, given their individual vapor pressures and the weight ratio of the two liquids.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Vapor pressure of liquid X, \( P_{X}^{0} = 400 \) torr
- Vapor pressure of liquid Y, \( P_{Y}^{0} = 200 \) torr
- Weight ratio of X to Y = 2:3

2. **Calculate the Molar Masses:**
- Let the molar mass of liquid X be \( M_X = m \).
- Since the ratio of the molecular masses of X and Y is 1:2, then \( M_Y = 2m \).

3. **Calculate the Number of Moles:**
- Let the weight of liquid X be \( 2 \) parts and the weight of liquid Y be \( 3 \) parts.
- The total weight of the mixture = \( 2 + 3 = 5 \) parts.
- Weight of X = \( 2 \) parts, Weight of Y = \( 3 \) parts.
- Moles of X, \( n_X = \frac{\text{Weight of X}}{M_X} = \frac{2}{m} \)
- Moles of Y, \( n_Y = \frac{\text{Weight of Y}}{M_Y} = \frac{3}{2m} \)

4. **Calculate the Total Moles:**
- Total moles, \( n_{total} = n_X + n_Y = \frac{2}{m} + \frac{3}{2m} \)
- To add these fractions, find a common denominator:
\[ n_{total} = \frac{4}{2m} + \frac{3}{2m} = \frac{7}{2m} \]

5. **Calculate the Mole Fraction of Each Liquid:**
- Mole fraction of X, \( X_X = \frac{n_X}{n_{total}} = \frac{\frac{2}{m}}{\frac{7}{2m}} = \frac{4}{7} \)
- Mole fraction of Y, \( X_Y = \frac{n_Y}{n_{total}} = \frac{\frac{3}{2m}}{\frac{7}{2m}} = \frac{3}{7} \)

6. **Calculate the Total Vapor Pressure:**
- The total vapor pressure of the mixture, \( P_{total} = P_{X}^{0} \cdot X_X + P_{Y}^{0} \cdot X_Y \)
- Substitute the values:
\[ P_{total} = 400 \cdot \frac{4}{7} + 200 \cdot \frac{3}{7} \]
- Calculate each term:
\[ P_{total} = \frac{1600}{7} + \frac{600}{7} = \frac{2200}{7} \approx 314.29 \text{ torr} \]

7. **Conclusion:**
- The total vapor pressure of the mixture is approximately \( 314.29 \) torr.