The vapour pressures of ethanol and methanol are `42.0mm`and`88.0mmHg`respectively .An ideal solution is formed at the same temperature by mixing `46.0g`of ethanol with `16.0g`of methanol. The mole fraction of methanol in the vapour is:

Options

A. `0.467`
B. `0.502`
C. `0.513`
D. `0.556`

Explanation

To solve the problem of finding the mole fraction of methanol in the vapor phase, we will follow these steps:

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

**Given:**
- Mass of ethanol (C2H5OH) = 46.0 g
- Mass of methanol (CH3OH) = 16.0 g

**Molar Masses:**
- Molar mass of ethanol (C2H5OH) = 2(12) + 6(1) + 16 = 46 g/mol
- Molar mass of methanol (CH3OH) = 12 + 4 + 16 = 32 g/mol

**Calculating moles:**
- Moles of ethanol = Mass / Molar mass = 46.0 g / 46 g/mol = 1.0 mol
- Moles of methanol = Mass / Molar mass = 16.0 g / 32 g/mol = 0.5 mol

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

**Total moles = Moles of ethanol + Moles of methanol**
- Total moles = 1.0 mol + 0.5 mol = 1.5 mol

### Step 3: Calculate the mole fractions of ethanol and methanol in the liquid phase.

**Mole fraction of methanol (X_Methanol) = Moles of methanol / Total moles**
- X_Methanol = 0.5 mol / 1.5 mol = 1/3 ≈ 0.333

**Mole fraction of ethanol (X_Ethanol) = Moles of ethanol / Total moles**
- X_Ethanol = 1.0 mol / 1.5 mol = 2/3 ≈ 0.667

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

**Given vapor pressures:**
- Vapor pressure of ethanol (P°_Ethanol) = 42.0 mmHg
- Vapor pressure of methanol (P°_Methanol) = 88.0 mmHg

**Using Raoult's Law:**
- Partial pressure of ethanol (P_E) = X_Ethanol * P°_Ethanol
- P_E = 0.667 * 42.0 mmHg = 28.014 mmHg

- Partial pressure of methanol (P_M) = X_Methanol * P°_Methanol
- P_M = 0.333 * 88.0 mmHg = 29.304 mmHg

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

**Total vapor pressure (P_total) = P_E + P_M**
- P_total = 28.014 mmHg + 29.304 mmHg = 57.318 mmHg

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

**Using the formula:**
- Mole fraction of methanol in vapor (Y_Methanol) = P_M / P_total
- Y_Methanol = 29.304 mmHg / 57.318 mmHg ≈ 0.512

### Final Answer:
The mole fraction of methanol in the vapor phase is approximately **0.512**.

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