Question

\text{Standard enthalpy of vaporisation for CCl}_4 \text{ is 30.5 kJ mol}^{-1}. \text{ Heat required for vaporisation of } 284 \, \text{g of CCl}_4 \text{ at constant temperature is } \\_\\_\\_\\_ \, \text{kJ.} \\\ \text{(Given molar mass in g mol}^{-1}\text{; C = 12, Cl = 35.5)}

Answer 1

$\Delta H_{\text{vap}}^\circ \, \text{for } \, \text{CCl}_4 = 30.5 \, \text{kJ/mol}$

Mass of $\text{CCl}_4 = 284 \, \text{g}$
Molar mass of $\text{CCl}_4$:
$= 154 \, \text{g/mol}$

Moles of $\text{CCl}_4$:
$= \frac{284}{154} = 1.844 \, \text{mol}$

$\Delta H_{\text{vap}}^\circ \, \text{for 1 mole} = 30.5 \, \text{kJ/mol}$

$\Delta H_{\text{vap}}^\circ \, \text{for 1.844 moles} = 30.5 \times 1.844$

$= 56.242 \, \text{kJ}$

The Correct Answer is:= $56.242 \, \text{kJ}$