The enthalpy of vaporisation of ` C CI_(4)` is `30.5 kJ "mol"^(-1)`.Calculate the heat required for the vapourisation of 284 g of `C CI_(4)` at constant pressure. (Molar mass of ` C CI_(4) = 154u`). (Keep your answer in KJ and two digits only)

Explanation

To calculate the heat required for the vaporization of 284 g of carbon tetrachloride (CCl₄) at constant pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Molar Mass of CCl₄**:
The molar mass of carbon tetrachloride (CCl₄) is given as 154 g/mol.

2. **Calculate the Number of Moles of CCl₄ in 284 g**:
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{284 \, \text{g}}{154 \, \text{g/mol}} \approx 1.8435 \, \text{mol} \]

3. **Use the Enthalpy of Vaporization**:
The enthalpy of vaporization of CCl₄ is given as 30.5 kJ/mol. This means that for every mole of CCl₄, 30.5 kJ of heat is required for vaporization.

4. **Calculate the Total Heat Required for Vaporization**:
To find the total heat required for the vaporization of 1.8435 moles, we multiply the number of moles by the enthalpy of vaporization:
\[ \text{Heat required} = \text{Number of moles} \times \text{Enthalpy of vaporization} \]
Substituting the values:
\[ \text{Heat required} = 1.8435 \, \text{mol} \times 30.5 \, \text{kJ/mol} \approx 56.25 \, \text{kJ} \]

5. **Final Answer**:
The heat required for the vaporization of 284 g of CCl₄ at constant pressure is approximately **56.25 kJ**.