Standard enthalpy of vapourisation for is 3035 kJ . Heat required for vapourisation of 284 g of at constant temperature is _____ KJ.

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Solution

Q.
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, CI = 35.5) [2024]

Ans.
(56)

$Molar mass of {CCl}_{4} (M) = (12 + 4 \times 35.5) {gmol}^{- 1}$

$= 154 g m o l^{- 1}$

$Given mass of {CCl}_{4} (W) = 284 g$

$Number of moles of {CCl}_{4} (n) = \frac{W}{M} = \frac{284 g}{154 g {mol}^{- 1}} = 1.844 mol$

$Heat required for vapourization :$

$= Molar enthalpy of vapourization \times n$

$= 30.5 kJ {mol}^{- 1} \times 1.844 mol = 56.242 kJ$

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