The vapour pressure of pure liquid 'A' is 70 torr, at `27^(@)C`. It forms an ideal solution with another liquid 'B'. The mole fraction of 'B' is 0.2 and total pressure of the solution is 84 torr at `27^(@)C`. Find the vapour pressure of pure liquid B at `27^(@)C`.

Explanation

To find the vapor pressure of pure liquid B at 27°C, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial vapor pressures of each component in the solution.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure liquid A, \( P^0_A = 70 \, \text{torr} \)
- Mole fraction of liquid B, \( x_B = 0.2 \)
- Total pressure of the solution, \( P_{\text{total}} = 84 \, \text{torr} \)

2. **Calculate the Mole Fraction of Liquid A:**
- The mole fraction of liquid A, \( x_A \), can be calculated using the formula:
\[ x_A = 1 - x_B = 1 - 0.2 = 0.8 \]

3. **Apply Raoult's Law:**
- According to Raoult's Law, the total vapor pressure of the solution can be expressed as:
\[ P_{\text{total}} = P^0_A \cdot x_A + P^0_B \cdot x_B \]
- Substituting the known values:
\[ 84 = (70 \cdot 0.8) + (P^0_B \cdot 0.2) \]

4. **Calculate the Contribution of Liquid A:**
- Calculate \( 70 \cdot 0.8 \):
\[ 70 \cdot 0.8 = 56 \, \text{torr} \]

5. **Set Up the Equation:**
- Substitute this value back into the equation:
\[ 84 = 56 + (P^0_B \cdot 0.2) \]

6. **Solve for \( P^0_B \):**
- Rearranging the equation gives:
\[ 84 - 56 = P^0_B \cdot 0.2 \]
\[ 28 = P^0_B \cdot 0.2 \]
- Now, divide both sides by 0.2:
\[ P^0_B = \frac{28}{0.2} = 140 \, \text{torr} \]

### Final Answer:
The vapor pressure of pure liquid B at 27°C is \( P^0_B = 140 \, \text{torr} \).

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