Standard enthalpy of vapourisation for $$CCl_4$$ is $$30.5$$ kJ mol$$^{-1}$$. Heat required for vapourisation of $$284$$ g of $$CCl_4$$ at constant temperature is ______ kJ. (Given molar mass in g mol$$^{-1}$$; C = 12, Cl = 35.5)

We need to find the heat required for vapourisation of $$284$$ g of $$CCl_4$$ at constant temperature.

• Standard enthalpy of vapourisation of $$CCl_4$$, $$\Delta H_{vap} = 30.5$$ kJ/mol
• Mass of $$CCl_4 = 284$$ g
• Molar mass of $$CCl_4 = 12 + 4 \times 35.5 = 12 + 142 = 154$$ g/mol

Using the formula: $$n = \frac{\text{mass}}{\text{molar mass}}$$

$$n = \frac{284}{154} = \frac{284}{154} \approx 1.844 \text{ mol}$$

Using the formula: $$q = n \times \Delta H_{vap}$$

$$q = 1.844 \times 30.5 = 56.24 \text{ kJ}$$

Rounding to the nearest integer: $$q \approx 56$$ kJ.

Therefore, the heat required for vapourisation is 56 kJ.