Two liquids having vapour pressure `P_1^0 and P_2^0` in pure state in the ratio of `2:1` are mixed in the molar ratio `1:2` The ratio of their moles in the vapour state would be

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. `3:2`

Answer

To solve the problem, we need to find the ratio of the moles of two liquids in the vapor state after mixing them in a given molar ratio. We will use Raoult's law to help us with this.

### Step-by-step Solution:

1. **Identify the Given Information:**
- The vapor pressures of the two liquids in pure state are in the ratio \( P_1^0 : P_2^0 = 2 : 1 \).
- The liquids are mixed in the molar ratio \( 1 : 2 \).

2. **Assign Variables:**
- Let \( P_1^0 = 2y \) and \( P_2^0 = y \), where \( y \) is a common factor.
- Let the moles of liquid 1 be \( n_1 = 1 \) and the moles of liquid 2 be \( n_2 = 2 \).

3. **Calculate the Total Moles:**
- Total moles \( n_{total} = n_1 + n_2 = 1 + 2 = 3 \).

4. **Calculate the Mole Fractions:**
- Mole fraction of liquid 1, \( X_1 = \frac{n_1}{n_{total}} = \frac{1}{3} \).
- Mole fraction of liquid 2, \( X_2 = \frac{n_2}{n_{total}} = \frac{2}{3} \).

5. **Apply Raoult's Law:**
- According to Raoult's law, the partial vapor pressure of each liquid in the mixture is given by:
- \( P_1 = X_1 \cdot P_1^0 \)
- \( P_2 = X_2 \cdot P_2^0 \)

6. **Calculate Partial Vapor Pressures:**
- For liquid 1:
\[ P_1 = X_1 \cdot P_1^0 = \frac{1}{3} \cdot (2y) = \frac{2y}{3} \]
- For liquid 2:
\[ P_2 = X_2 \cdot P_2^0 = \frac{2}{3} \cdot y = \frac{2y}{3} \]

7. **Calculate the Ratio of Partial Vapor Pressures:**
- Now, we find the ratio of the partial vapor pressures:
\[ \frac{P_1}{P_2} = \frac{\frac{2y}{3}}{\frac{2y}{3}} = 1 \]

8. **Determine the Ratio of Moles in Vapor State:**
- Since the ratio of the partial pressures is equal to the ratio of the moles in the vapor state, we have:
\[ \text{Ratio of moles in vapor state} = 1 : 1 \]

### Final Answer:
The ratio of their moles in the vapor state is \( 1 : 1 \).

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