The vapour pressure of ethanol and methanol ate `44.5 mm Hg` and `88.7 mm Hg`, respectively. An ideal solution is formed at the same temperature by mixing `60g` of ethanol and `40g` of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of ethanol and methanol.

1. **Molar mass of ethanol (C₂H₅OH)**:
- Molar mass = 46 g/mol
2. **Molar mass of methanol (CH₃OH)**:
- Molar mass = 32 g/mol

Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- For ethanol:
\[ \text{Moles of ethanol} = \frac{60 \text{ g}}{46 \text{ g/mol}} \approx 1.3043 \text{ moles} \]

- For methanol:
\[ \text{Moles of methanol} = \frac{40 \text{ g}}{32 \text{ g/mol}} = 1.25 \text{ moles} \]

### Step 2: Calculate the total number of moles in the solution.

\[ \text{Total moles} = \text{Moles of ethanol} + \text{Moles of methanol} = 1.3043 + 1.25 \approx 2.5543 \text{ moles} \]

### Step 3: Calculate the mole fraction of ethanol and methanol.

- Mole fraction of ethanol (X₁):
\[ X_{\text{ethanol}} = \frac{\text{Moles of ethanol}}{\text{Total moles}} = \frac{1.3043}{2.5543} \approx 0.5105 \]

- Mole fraction of methanol (X₂):
\[ X_{\text{methanol}} = \frac{\text{Moles of methanol}}{\text{Total moles}} = \frac{1.25}{2.5543} \approx 0.4895 \]

### Step 4: Calculate the partial vapor pressures of ethanol and methanol.

Using Raoult's Law:
\[ P_{\text{ethanol}} = X_{\text{ethanol}} \cdot P^0_{\text{ethanol}} = 0.5105 \cdot 44.5 \text{ mm Hg} \approx 22.7 \text{ mm Hg} \]
\[ P_{\text{methanol}} = X_{\text{methanol}} \cdot P^0_{\text{methanol}} = 0.4895 \cdot 88.7 \text{ mm Hg} \approx 43.4 \text{ mm Hg} \]

### Step 5: Calculate the total vapor pressure of the solution.

\[ P_{\text{total}} = P_{\text{ethanol}} + P_{\text{methanol}} = 22.7 \text{ mm Hg} + 43.4 \text{ mm Hg} \approx 66.1 \text{ mm Hg} \]

### Step 6: Calculate the mole fraction of methanol in the vapor.

Using the formula:
\[ Y_{\text{methanol}} = \frac{P_{\text{methanol}}}{P_{\text{total}}} = \frac{43.4 \text{ mm Hg}}{66.1 \text{ mm Hg}} \approx 0.656 \]

### Final Answers:
- Total vapor pressure of the solution: **66.1 mm Hg**
- Mole fraction of methanol in the vapor: **0.656**