Thermodynamics Question 6
Question 6 - 2024 (29 Jan Shift 2)
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 $gmol^{-1} ; C=12, Cl=35.5$ )
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Answer (56)
Solution
$\Delta H _{\text {vap }}^{0} CCl _4=30.5 kJ / mol$
Mass of $CCl _4=284 gm$
Molar mass of $CCl _4=154 g / mol$
Moles of $CCl _4=\frac{284}{154}=1.844 mol$
$\Delta H _{\text {vap }}{ }^{\circ}$ for $1 mole=30.5 kJ / mol$
$\Delta H _{vap}{ }^{\circ}$ for $1.844 mol=30.5 \times 1.844$
$=56.242 kJ$
