At 293 K , ethyl acetate has vapour pressure of 72.8 torr of Hg and ethyl propionate has vapour pressure of 27.7 torr of Hg. Assuming their mixtures to obey Raoult's law determine the vapour pressure of a mixture containing 25 g of ethyl acetate and 50 g of ethyl propionate .

Explanation

To solve the problem of determining the vapor pressure of a mixture containing 25 g of ethyl acetate and 50 g of ethyl propionate at 293 K, we will follow these steps:

### Step 1: Calculate the number of moles of ethyl acetate
1. **Find the molecular weight of ethyl acetate (C4H8O2)**:
- Carbon (C): 12.01 g/mol × 4 = 48.04 g/mol
- Hydrogen (H): 1.008 g/mol × 8 = 8.064 g/mol
- Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol
- Total = 48.04 + 8.064 + 32.00 = **88.104 g/mol** (approximately 88 g/mol)

2. **Calculate the number of moles of ethyl acetate**:
\[ \text{Moles of ethyl acetate} = \frac{\text{mass}}{\text{molecular weight}} = \frac{25 \text{ g}}{88 \text{ g/mol}} \approx 0.284 \text{ moles} \]

### Step 2: Calculate the number of moles of ethyl propionate
1. **Find the molecular weight of ethyl propionate (C5H10O2)**:
- Carbon (C): 12.01 g/mol × 5 = 60.05 g/mol
- Hydrogen (H): 1.008 g/mol × 10 = 10.08 g/mol
- Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol
- Total = 60.05 + 10.08 + 32.00 = **102.13 g/mol** (approximately 102 g/mol)

2. **Calculate the number of moles of ethyl propionate**:
\[ \text{Moles of ethyl propionate} = \frac{50 \text{ g}}{102 \text{ g/mol}} \approx 0.490 \text{ moles} \]

### Step 3: Calculate the mole fractions of both components
1. **Total moles in the mixture**:
\[ \text{Total moles} = 0.284 + 0.490 = 0.774 \text{ moles} \]

2. **Mole fraction of ethyl acetate (\(X_{EA}\))**:
\[ X_{EA} = \frac{\text{Moles of ethyl acetate}}{\text{Total moles}} = \frac{0.284}{0.774} \approx 0.367 \]

3. **Mole fraction of ethyl propionate (\(X_{EP}\))**:
\[ X_{EP} = \frac{\text{Moles of ethyl propionate}}{\text{Total moles}} = \frac{0.490}{0.774} \approx 0.633 \]

### Step 4: Calculate the vapor pressures of the components in the mixture
1. **Vapor pressure of ethyl acetate**:
\[ P_{EA} = X_{EA} \times P^0_{EA} = 0.367 \times 72.8 \text{ torr} \approx 26.72 \text{ torr} \]

2. **Vapor pressure of ethyl propionate**:
\[ P_{EP} = X_{EP} \times P^0_{EP} = 0.633 \times 27.7 \text{ torr} \approx 17.53 \text{ torr} \]

### Step 5: Calculate the total vapor pressure of the mixture
\[ P_{total} = P_{EA} + P_{EP} = 26.72 \text{ torr} + 17.53 \text{ torr} \approx 44.25 \text{ torr} \]

### Final Answer
The total vapor pressure of the mixture is approximately **44.25 torr**.

---