Mole fraction of solute is 0.2 in solution then lowering in V.P `DeltaP=10`. If lowering in V.P. `DeltaP=20` then mole fraction of solvent will be in solution :

Options

A. 0.2
B. 0.4
C. 0.6
D. 0.8

Answer

To solve the problem step by step, we will use Raoult's Law and the relationship between the mole fractions of solute and solvent.

### Step 1: Understand the Given Information
- Mole fraction of solute (Xb) = 0.2
- Lowering in vapor pressure (ΔP) = 10
- We need to find the mole fraction of solvent (Xa) when ΔP = 20.

### Step 2: Apply Raoult's Law
According to Raoult's Law, the lowering of vapor pressure (ΔP) can be expressed as:
\[ \Delta P = P_{A0} \cdot X_b \]
where:
- \( P_{A0} \) = vapor pressure of the pure solvent
- \( X_b \) = mole fraction of the solute

### Step 3: Calculate \( P_{A0} \) Using the First ΔP
From the first condition (ΔP = 10):
\[ 10 = P_{A0} \cdot 0.2 \]
Rearranging gives:
\[ P_{A0} = \frac{10}{0.2} = 50 \]

### Step 4: Calculate the Mole Fraction of Solute for ΔP = 20
Now, using the second condition (ΔP = 20):
\[ 20 = P_{A0} \cdot X_b \]
Substituting \( P_{A0} = 50 \):
\[ 20 = 50 \cdot X_b \]
Rearranging gives:
\[ X_b = \frac{20}{50} = \frac{2}{5} \]

### Step 5: Calculate the Mole Fraction of Solvent
Using the relationship between mole fractions:
\[ X_a + X_b = 1 \]
Substituting \( X_b = \frac{2}{5} \):
\[ X_a + \frac{2}{5} = 1 \]
Rearranging gives:
\[ X_a = 1 - \frac{2}{5} = \frac{3}{5} = 0.6 \]

### Final Answer
The mole fraction of the solvent (Xa) when ΔP = 20 is:
\[ \boxed{0.6} \]