Vapour pressure of pure A is 70 mm of Hg at `25^(@)C`. If it forms an ideal solution with B in which mole fraction of A is 0.8 and vapour pressure of the solution is 84 mm of Hg at `25^(@)C`, then the vapour pressure of pure B at `25^(@)C` is

Options

A. 56mm
B. 70mm
C. 140mm
D. 28mm

Answer

To find the vapor pressure of pure B at 25°C, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the vapor pressures of the individual components multiplied by their respective mole fractions.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure A (\(P^0_A\)) = 70 mm Hg
- Mole fraction of A (\(X_A\)) = 0.8
- Vapor pressure of the solution (\(P_{solution}\)) = 84 mm Hg

2. **Calculate Mole Fraction of B:**
- Since the sum of the mole fractions in a solution is equal to 1:
\[ X_B = 1 - X_A = 1 - 0.8 = 0.2 \]

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

4. **Calculate the Contribution of A:**
- Calculate \(70 \cdot 0.8\):
\[ 70 \cdot 0.8 = 56 \]

5. **Substitute and Solve for \(P^0_B\):**
- Now substitute back into the equation:
\[ 84 = 56 + P^0_B \cdot 0.2 \]
- Rearranging gives:
\[ P^0_B \cdot 0.2 = 84 - 56 \]
\[ P^0_B \cdot 0.2 = 28 \]
- Now, divide both sides by 0.2 to find \(P^0_B\):
\[ P^0_B = \frac{28}{0.2} = 140 \text{ mm Hg} \]

6. **Final Answer:**
- The vapor pressure of pure B at 25°C is **140 mm Hg**.