The vapour pressure of benzene is `53.3 k P_(a)` at `60.6^(@)` but it falls to `51.5 kP_(a)` when 19 g of a non-volatile organic compound is dissolved in 500 g benzene . The molar mass of the non-volatile compound is close to :

Options

A. 82
B. 85
C. 88
D. 92

Answer

To solve the problem, we will apply Raoult's law, which relates the vapor pressure of a solvent to the concentration of a solute in a solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- Vapor pressure of pure benzene (P₀) = 53.3 kPa
- Vapor pressure of benzene with solute (P₁) = 51.5 kPa
- Mass of the non-volatile organic compound (solute) = 19 g
- Mass of benzene (solvent) = 500 g

### Step 2: Calculate the relative lowering of vapor pressure
Using the formula for relative lowering of vapor pressure:
\[ \Delta P = P₀ - P₁ \]
Substituting the values:
\[ \Delta P = 53.3 \, \text{kPa} - 51.5 \, \text{kPa} = 1.8 \, \text{kPa} \]

### Step 3: Apply Raoult's Law
According to Raoult's law:
\[ \frac{\Delta P}{P₀} = X_B \]
Where \(X_B\) is the mole fraction of the solute. Thus:
\[ X_B = \frac{1.8 \, \text{kPa}}{53.3 \, \text{kPa}} \]

### Step 4: Calculate the mole fraction of the solute
Calculating \(X_B\):
\[ X_B = \frac{1.8}{53.3} \approx 0.0338 \]

### Step 5: Calculate the number of moles of benzene (solvent)
The molar mass of benzene (C₆H₆) is:
\[ \text{Molar mass of benzene} = 12 \times 6 + 1 \times 6 = 78 \, \text{g/mol} \]
Now, calculate the number of moles of benzene:
\[ n_A = \frac{\text{mass}}{\text{molar mass}} = \frac{500 \, \text{g}}{78 \, \text{g/mol}} \approx 6.41 \, \text{mol} \]

### Step 6: Relate mole fraction to moles of solute
Using the mole fraction formula:
\[ X_B = \frac{n_B}{n_A + n_B} \]
Where \(n_B\) is the number of moles of the solute. Rearranging gives:
\[ X_B \approx \frac{n_B}{n_A} \quad \text{(since } n_B \text{ is small)} \]
Thus:
\[ n_B = X_B \times n_A \approx 0.0338 \times 6.41 \approx 0.216 \, \text{mol} \]

### Step 7: Calculate the molar mass of the solute
Now, we can find the molar mass (M) of the solute:
\[ \text{Molar mass} = \frac{\text{mass of solute}}{\text{number of moles of solute}} = \frac{19 \, \text{g}}{0.216 \, \text{mol}} \approx 87.96 \, \text{g/mol} \]

### Step 8: Conclusion
The molar mass of the non-volatile organic compound is approximately 88 g/mol.