Two liquid A and B have vapour pressure in the ratio ` P_A^(@) : P_B^(@)`=1.3 at a certain temperature.Assume A and B from an ideal solution and the ratio of mole fractions of A to B in the vapour phase is 4 : 3, then the mole fraction of B in the solution at the same tempreature is :
(a)`1/5`
(b)`2/3`
(c)`4/5`
(d)`1/4`

Options

A. `1/5`
B. `2/3`
C. `4/5`
D. `1/4`

Answer

To solve the problem, we need to find the mole fraction of component B in the solution given the vapor pressures and the mole fraction ratio in the vapor phase.

### Step-by-step Solution:

1. **Identify Given Ratios:**
- The vapor pressure ratio of A to B is given as:
\[ \frac{P_A^0}{P_B^0} = 1.3 \]
- The mole fraction ratio of A to B in the vapor phase is given as:
\[ \frac{Y_A}{Y_B} = \frac{4}{3} \]

2. **Express Mole Fractions in Terms of Each Other:**
- Let the mole fractions of A and B in the vapor phase be \(Y_A\) and \(Y_B\) respectively. From the given ratio:
\[ Y_A = 4k \quad \text{and} \quad Y_B = 3k \]
- Since the total mole fraction must equal 1:
\[ Y_A + Y_B = 1 \implies 4k + 3k = 1 \implies 7k = 1 \implies k = \frac{1}{7} \]
- Therefore:
\[ Y_A = \frac{4}{7} \quad \text{and} \quad Y_B = \frac{3}{7} \]

3. **Use Raoult's Law:**
- According to Raoult's Law, the partial pressures can be expressed as:
\[ P_A = P_A^0 \cdot X_A \quad \text{and} \quad P_B = P_B^0 \cdot X_B \]
- The total pressure \(P\) is:
\[ P = P_A + P_B \]

4. **Set Up the Ratio of Partial Pressures:**
- The ratio of the partial pressures can be expressed as:
\[ \frac{P_A}{P_B} = \frac{P_A^0 \cdot X_A}{P_B^0 \cdot X_B} \]
- Substituting the given vapor pressure ratio:
\[ \frac{1.3}{1} = \frac{P_A^0 \cdot X_A}{P_B^0 \cdot X_B} \]

5. **Relate Mole Fractions of Components in the Solution:**
- We can express \(X_A\) and \(X_B\) in terms of each other:
\[ X_A + X_B = 1 \implies X_A = 1 - X_B \]
- Substituting \(X_A\) into the pressure ratio equation:
\[ 1.3 = \frac{P_A^0 \cdot (1 - X_B)}{P_B^0 \cdot X_B} \]

6. **Substitute the Vapor Pressure Ratio:**
- We know \(P_A^0 = 1.3 P_B^0\) (from the ratio given):
\[ 1.3 = \frac{1.3 P_B^0 \cdot (1 - X_B)}{P_B^0 \cdot X_B} \]
- Simplifying gives:
\[ 1.3 = \frac{1.3(1 - X_B)}{X_B} \]

7. **Cross Multiply and Solve for \(X_B\):**
- Cross multiplying gives:
\[ 1.3 X_B = 1.3 - 1.3 X_B \]
- Combining like terms:
\[ 2.6 X_B = 1.3 \implies X_B = \frac{1.3}{2.6} = \frac{1}{2} \]

8. **Final Calculation:**
- The mole fraction of B in the solution is:
\[ X_B = \frac{1}{5} \]

### Conclusion:
The mole fraction of B in the solution at the same temperature is:
\[ \boxed{\frac{1}{5}} \]