A solution containing 6 g of benzoic acid in 50 g ether `(C_(2)H_(5)OC_(2)H_(5))` has a vapour pressure of 410 mm of mercury at 293 K. Given that the vapour pressure of ether at the same temperature is 442 mm of mercury, calculate the molecular mass of benzoic acid. (Assume that the solution is dilute).

Explanation

To solve the problem step by step, we will use Raoult's Law and the concept of mole fractions. Here's how to approach it:

### Step 1: Write down the given data
- Weight of benzoic acid (solute), \( W_2 = 6 \, \text{g} \)
- Weight of ether (solvent), \( W_1 = 50 \, \text{g} \)
- Vapor pressure of pure ether, \( P_0 = 442 \, \text{mmHg} \)
- Vapor pressure of the solution, \( P_s = 410 \, \text{mmHg} \)

### Step 2: Calculate the change in vapor pressure
Using Raoult's Law, we can find the change in vapor pressure:
\[ \Delta P = P_0 - P_s = 442 \, \text{mmHg} - 410 \, \text{mmHg} = 32 \, \text{mmHg} \]

### Step 3: Use Raoult's Law to find the mole fraction of the solute
According to Raoult's Law, the mole fraction of the solute can be expressed as:
\[ \frac{\Delta P}{P_0} = \frac{n_2}{n_1 + n_2} \]
Where:
- \( n_2 \) = moles of solute (benzoic acid)
- \( n_1 \) = moles of solvent (ether)

Since the solution is dilute, we can approximate \( n_1 + n_2 \approx n_1 \) (neglecting \( n_2 \) in the denominator):
\[ \frac{\Delta P}{P_0} \approx \frac{n_2}{n_1} \]

### Step 4: Calculate the moles of solvent (ether)
First, we need to calculate the number of moles of ether:
- Molar mass of ether \( (C_2H_5OC_2H_5) \) = 74 g/mol
\[ n_1 = \frac{W_1}{M_1} = \frac{50 \, \text{g}}{74 \, \text{g/mol}} \approx 0.676 \, \text{mol} \]

### Step 5: Set up the equation for moles of solute
Now we can express the moles of solute in terms of the change in vapor pressure:
\[ \frac{32 \, \text{mmHg}}{442 \, \text{mmHg}} = \frac{n_2}{0.676} \]
Calculating the left side:
\[ \frac{32}{442} \approx 0.0723 \]
Now we can solve for \( n_2 \):
\[ n_2 = 0.0723 \times 0.676 \approx 0.0489 \, \text{mol} \]

### Step 6: Calculate the molecular mass of benzoic acid
Using the number of moles of benzoic acid, we can find its molecular mass:
\[ M_2 = \frac{W_2}{n_2} = \frac{6 \, \text{g}}{0.0489 \, \text{mol}} \approx 122.65 \, \text{g/mol} \]

### Final Answer
The molecular mass of benzoic acid is approximately **122.65 g/mol**.

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