The enthalpy of vaporisation of $CCl_4$ is $30.5 \ kJ \ mol^{-1}$. Calculate the heat required for the vaporisation of $284 \ g$ of $CCl_4$ at constant pressure. (Molar mass of $CCl_4 = 154 \ g \ mol^{-1}$).
(N/A) The molar mass of $CCl_4$ is $154 \ g \ mol^{-1}$.
The number of moles of $CCl_4$ in $284 \ g$ is given by $n = \frac{\text{mass}}{\text{molar mass}} = \frac{284 \ g}{154 \ g \ mol^{-1}} \approx 1.844 \ mol$.
The heat required for vaporisation is given by $q = n \times \Delta_{vap}H$.
$q = 1.844 \ mol \times 30.5 \ kJ \ mol^{-1} = 56.242 \ kJ$.
Rounding to two decimal places,the heat required is $56.24 \ kJ$.
The number of moles of $CCl_4$ in $284 \ g$ is given by $n = \frac{\text{mass}}{\text{molar mass}} = \frac{284 \ g}{154 \ g \ mol^{-1}} \approx 1.844 \ mol$.
The heat required for vaporisation is given by $q = n \times \Delta_{vap}H$.
$q = 1.844 \ mol \times 30.5 \ kJ \ mol^{-1} = 56.242 \ kJ$.
Rounding to two decimal places,the heat required is $56.24 \ kJ$.
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