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

Options

A. 43.46 mm and 0.51
B. 66.15 mm and 0.657
C. 66.15 mm and 0.791
D. 70.59 mm and 0.657

Explanation

To solve the problem, we will follow these steps:

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

1. **Molar Mass Calculation**:
- Molar mass of ethanol (C₂H₅OH) = 46.07 g/mol
- Molar mass of methanol (CH₃OH) = 32.04 g/mol

2. **Number of Moles**:
- Moles of ethanol = mass of ethanol / molar mass of ethanol
- Moles of methanol = mass of methanol / molar mass of methanol

\[ \text{Moles of ethanol} = \frac{60 \text{ g}}{46.07 \text{ g/mol}} \approx 1.30 \text{ moles} \]

\[ \text{Moles of methanol} = \frac{40 \text{ g}}{32.04 \text{ g/mol}} \approx 1.25 \text{ moles} \]

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

\[ \text{Total moles} = \text{Moles of ethanol} + \text{Moles of methanol} = 1.30 + 1.25 = 2.55 \text{ moles} \]

### Step 3: Calculate the mole fractions of ethanol and methanol

1. **Mole Fraction of Ethanol (Xₑ)**:
\[ Xₑ = \frac{\text{Moles of ethanol}}{\text{Total moles}} = \frac{1.30}{2.55} \approx 0.51 \]

2. **Mole Fraction of Methanol (Xₕ)**:
\[ Xₕ = \frac{\text{Moles of methanol}}{\text{Total moles}} = \frac{1.25}{2.55} \approx 0.49 \]

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

Using Raoult's Law:
\[ Pₑ = Xₑ \cdot Pₑ^0 \]
\[ Pₕ = Xₕ \cdot Pₕ^0 \]

Where:
- \(Pₑ^0 = 44.5 \text{ mm Hg}\) (vapor pressure of ethanol)
- \(Pₕ^0 = 88.7 \text{ mm Hg}\) (vapor pressure of methanol)

Calculating partial pressures:
\[ Pₑ = 0.51 \cdot 44.5 \approx 22.66 \text{ mm Hg} \]
\[ Pₕ = 0.49 \cdot 88.7 \approx 43.44 \text{ mm Hg} \]

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

\[ P_{\text{total}} = Pₑ + Pₕ = 22.66 + 43.44 \approx 66.10 \text{ mm Hg} \]

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

Using the formula:
\[ Xₕ^{\text{vapor}} = \frac{Pₕ}{P_{\text{total}}} \]

Calculating:
\[ Xₕ^{\text{vapor}} = \frac{43.44}{66.10} \approx 0.66 \]

### Final Answers:
- Total vapor pressure of the solution = **66.10 mm Hg**
- Mole fraction of methanol in the vapor = **0.66**

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