Mole fraction of component `A` in vapour phase is `chi_(1)` and that of component `A` in liquid mixture is `chi_2`, then (`p_(A)^@`)= vapour pressure of pure A, `p_(B)^@` = vapour pressure of pure B), the total vapour pressure of liquid mixture is

Options

A. `(P_(A)^(@)chi_(2))/chi_(1)`
B. `(P_(A)^(@)chi_(1))/chi_(2)`
C. `(P_(A)^(@)chi_(1))/chi_(2)`
D. `(P_(B)^(@)chi_(2))/chi_(1)`

Explanation

To find the total vapor pressure of the liquid mixture, we can use the relationship between the mole fractions in both the vapor and liquid phases and the vapor pressures of the pure components.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( \chi_1 \) be the mole fraction of component A in the vapor phase (often denoted as \( y_1 \)).
- Let \( \chi_2 \) be the mole fraction of component A in the liquid phase (often denoted as \( x_2 \)).
- Let \( P_A^0 \) be the vapor pressure of pure A.
- Let \( P_B^0 \) be the vapor pressure of pure B.
- Let \( P_{total} \) be the total vapor pressure of the liquid mixture.

2. **Use Raoult's Law**:
According to Raoult's Law, the partial vapor pressure of component A in the mixture is given by:
\[ P_A = P_A^0 \cdot \chi_2 \]
where \( \chi_2 \) is the mole fraction of A in the liquid phase.

3. **Relate Mole Fractions to Total Pressure**:
The mole fraction of component A in the vapor phase can be expressed as:
\[ \chi_1 = \frac{P_A}{P_{total}} \]
Substituting for \( P_A \) from Raoult's Law:
\[ \chi_1 = \frac{P_A^0 \cdot \chi_2}{P_{total}} \]

4. **Rearranging the Equation**:
To find the total vapor pressure \( P_{total} \), we can rearrange the equation:
\[ P_{total} = \frac{P_A^0 \cdot \chi_2}{\chi_1} \]

5. **Final Expression**:
Therefore, the total vapor pressure of the liquid mixture can be expressed as:
\[ P_{total} = \frac{\chi_2}{\chi_1} \cdot P_A^0 \]

### Final Answer:
The total vapor pressure of the liquid mixture is given by:
\[ P_{total} = \frac{\chi_2}{\chi_1} \cdot P_A^0 \]