A stream of air is bubbled slowly through liquid benzene in a flask at `20.0^(@)C` against an ambient pressure of `100.56kPa`. After the passage of `4.80L` of air, measured at `20.0^(@)C` and `100.56kPa` before it contains benzonb vapor, it is found that `1.705g` of benzene have been evaporated. Assuming that the air saturated benzene vapor when it leaves the flask. Calculate the equilibrium vapor pressure of the benzene at `20.0^(@)C`.

Explanation

To calculate the equilibrium vapor pressure of benzene at \(20.0^\circ C\), we will follow these steps:

### Step 1: Calculate the moles of benzene evaporated
We are given the mass of benzene evaporated as \(1.705 \, \text{g}\). The molar mass of benzene (C₆H₆) is calculated as follows:

\[ \text{Molar mass of benzene} = (6 \times 12) + (6 \times 1) = 72 + 6 = 78 \, \text{g/mol} \]

Now, we can calculate the number of moles of benzene using the formula:

\[ \text{Moles of benzene} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.705 \, \text{g}}{78 \, \text{g/mol}} \approx 0.02185 \, \text{mol} \]

### Step 2: Use the Ideal Gas Law to find the pressure of benzene vapor
We will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \(P\) = pressure (in atm)
- \(V\) = volume (in liters)
- \(n\) = number of moles (in mol)
- \(R\) = universal gas constant \(0.0821 \, \text{atm} \cdot \text{L} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}\)
- \(T\) = temperature (in Kelvin)

Given:
- Volume \(V = 4.80 \, \text{L}\)
- Temperature \(T = 20.0^\circ C = 20 + 273 = 293 \, \text{K}\)

Now, we can rearrange the Ideal Gas Law to solve for pressure \(P\):

\[ P = \frac{nRT}{V} \]

Substituting the values we have:

\[ P = \frac{(0.02185 \, \text{mol}) \times (0.0821 \, \text{atm} \cdot \text{L} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}) \times (293 \, \text{K})}{4.80 \, \text{L}} \]

Calculating this gives:

\[ P \approx \frac{0.529 \, \text{atm} \cdot \text{L}}{4.80 \, \text{L}} \approx 0.109 \, \text{atm} \]

### Step 3: Convert the pressure to kPa
Since the problem asks for the vapor pressure in kPa, we convert \(0.109 \, \text{atm}\) to kPa using the conversion factor \(1 \, \text{atm} = 101.325 \, \text{kPa}\):

\[ P \approx 0.109 \, \text{atm} \times 101.325 \, \text{kPa/atm} \approx 11.04 \, \text{kPa} \]

### Final Answer
The equilibrium vapor pressure of benzene at \(20.0^\circ C\) is approximately **11.04 kPa**.

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