Thermodynamics C Question 6
Question 6 - 2024 (29 Jan Shift 2)
Standard enthalpy of vapourisation for $\mathrm{CCl}{4}$ is $30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Heat required for vapourisation of $284 \mathrm{~g}$ of $\mathrm{CCl}{4}$ at constant temperature is $\mathrm{kJ}$.
(Given molar mass in $\mathrm{gmol}^{-1} ; \mathrm{C}=12, \mathrm{Cl}=35.5$ )
Show Answer
Answer (56)
Solution
$\Delta \mathrm{H}{\text {vap }}^{0} \mathrm{CCl}{4}=30.5 \mathrm{~kJ} / \mathrm{mol}$
Mass of $\mathrm{CCl}_{4}=284 \mathrm{gm}$
Molar mass of $\mathrm{CCl}_{4}=154 \mathrm{~g} / \mathrm{mol}$
Moles of $\mathrm{CCl}_{4}=\frac{284}{154}=1.844 \mathrm{~mol}$
$\Delta \mathrm{H}_{\text {vap }}{ }^{\circ}$ for $1 \mathrm{~mole}=30.5 \mathrm{~kJ} / \mathrm{mol}$
$\Delta \mathrm{H}_{\mathrm{vap}}{ }^{\circ}$ for $1.844 \mathrm{~mol}=30.5 \times 1.844$
$=56.242 \mathrm{~kJ}$
