The molar enthalpy of vaporization of benzene at its boiling point `(353 K)` is `29.7 KJ//"mole"`. For how long minute) would a `11.4` Volt source need to supply a `0. 5A` current in order to vaporise `7.8 g` of the sample at its boiling point ?

Answer

To solve the problem, we need to determine how long a 11.4 Volt source supplying a 0.5 A current would take to vaporize 7.8 g of benzene, given that the molar enthalpy of vaporization of benzene at its boiling point (353 K) is 29.7 kJ/mol.

### Step-by-Step Solution:

1. **Calculate the number of moles of benzene:**
The molar mass of benzene (C6H6) is approximately 78 g/mol.
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{7.8 \, \text{g}}{78 \, \text{g/mol}} = 0.1 \, \text{mol} \]

2. **Calculate the total energy required to vaporize the benzene:**
The energy required (in kJ) can be calculated using the molar enthalpy of vaporization:
\[ \text{Energy required} = \text{moles} \times \text{molar enthalpy} = 0.1 \, \text{mol} \times 29.7 \, \text{kJ/mol} = 2.97 \, \text{kJ} \]

3. **Convert the energy required from kJ to Joules:**
Since 1 kJ = 1000 J,
\[ 2.97 \, \text{kJ} = 2.97 \times 1000 \, \text{J} = 2970 \, \text{J} \]

4. **Calculate the power supplied by the electrical source:**
Power (P) can be calculated using the formula:
\[ P = V \times I \]
where \( V = 11.4 \, \text{V} \) and \( I = 0.5 \, \text{A} \).
\[ P = 11.4 \, \text{V} \times 0.5 \, \text{A} = 5.7 \, \text{W} \]

5. **Calculate the time required to supply the necessary energy:**
The time (t) can be calculated using the formula:
\[ t = \frac{\text{Energy}}{\text{Power}} = \frac{2970 \, \text{J}}{5.7 \, \text{W}} \approx 521.05 \, \text{s} \]

6. **Convert time from seconds to minutes:**
\[ t \approx \frac{521.05 \, \text{s}}{60} \approx 8.68 \, \text{minutes} \]

### Final Answer:
The time required to vaporize 7.8 g of benzene is approximately **8.68 minutes**.