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)

Delta H_ 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=(284 / 154)=1.844 mol \ Delta H_ vap ^° for 1 mole=30.5 kJ / mol \ Delta H_ vap ^° for 1.844 mol=30.5 x 1.844 \ =56.242 kJ \

Standard enthalpy of vapourisation for CCl4 is 305 kJ m... | null (Chemistry)

\begin{aligned} & \Delta \mathrm{H}_{\text {vap }}^0 \mathrm{CCl}_4=30.5 \mathrm{~kJ} / \mathrm{mol} \\ & \text { Mass of } \mathrm{CCl}_4=284 \mathrm{~gm} \\ & \text { Molar mass of } \mathrm{CCl}_4=154 \mathrm{~g} / \mathrm{mol} \\ & \text { Moles of } \mathrm{CCl}_4=\frac{284}{154}=1.844 \mathrm{~mol} \\ & \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1 \mathrm{~mole}=30.5 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1.844 \mathrm{~mol}=30.5 \times 1.844 \\ & \quad=56.242 \mathrm{~kJ} \end{aligned}

Explanation

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