If benzoic acid (mol. Mass = 122) is associated into double molecules when dissolved in benzene and the osmotic pressure of a solution of 5g of benzoic acid in 100 mL benzene is 5.73 atm at `10^(@)C`, what is the percentage association of benzoic acid ?

Explanation

To solve the problem of determining the percentage association of benzoic acid when dissolved in benzene, we can follow these steps:

### Step 1: Calculate the number of moles of benzoic acid
Given:
- Mass of benzoic acid (solute) = 5 g
- Molar mass of benzoic acid = 122 g/mol

Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \, \text{g}}{122 \, \text{g/mol}} \approx 0.0410 \, \text{mol} \]

### Step 2: Convert the volume of the solution to liters
Given:
- Volume of benzene = 100 mL = 0.1 L

### Step 3: Use the osmotic pressure formula
The osmotic pressure (\(\pi\)) is given by the formula:
\[ \pi = iCRT \]
Where:
- \(\pi = 5.73 \, \text{atm}\)
- \(R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
- \(T = 10^\circ C = 283 \, \text{K}\)
- \(C\) is the concentration in mol/L, which can be calculated as:
\[ C = \frac{\text{number of moles}}{\text{volume in liters}} = \frac{0.0410 \, \text{mol}}{0.1 \, \text{L}} = 0.410 \, \text{mol/L} \]

### Step 4: Rearranging the osmotic pressure formula to find \(i\)
Rearranging the formula to solve for \(i\):
\[ i = \frac{\pi}{CRT} \]
Substituting the known values:
\[ i = \frac{5.73 \, \text{atm}}{(0.410 \, \text{mol/L})(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(283 \, \text{K})} \]

Calculating \(i\):
\[ i \approx \frac{5.73}{(0.410)(0.0821)(283)} \approx 0.6017 \]

### Step 5: Determine the degree of association (\(\alpha\))
For benzoic acid associating into dimers, the relationship between \(i\) and \(\alpha\) is given by:
\[ i = \frac{1 - \alpha}{1 - \frac{\alpha}{2}} \]
Rearranging gives:
\[ 1 - \alpha = i \left(1 - \frac{\alpha}{2}\right) \]
Substituting \(i\):
\[ 1 - \alpha = 0.6017 \left(1 - \frac{\alpha}{2}\right) \]

### Step 6: Solve for \(\alpha\)
Expanding and rearranging:
\[ 1 - \alpha = 0.6017 - 0.30085\alpha \]
\[ 1 - 0.6017 = \alpha - 0.30085\alpha \]
\[ 0.3983 = 0.69915\alpha \]
\[ \alpha \approx \frac{0.3983}{0.69915} \approx 0.570 \]

### Step 7: Calculate the percentage association
The percentage association is given by:
\[ \text{Percentage association} = \alpha \times 100 \approx 0.570 \times 100 \approx 57.0\% \]

### Final Answer
The percentage association of benzoic acid in benzene is approximately **57.0%**.

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