Q. 43

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Answer :

The enthalpy of vapourisation is given for 1 mole of CCl4 = 30.5 kJ mol–1


Hence, for 284 g, it will be = (mole no. × 30.5) kJ


Molar mass of CCl4 = 154 g mol–1 that means 154 g = 1 mole.


Therefore , 284 g = (284g/154gmol-1)= 1.84 mole.


Hence,


the heat required for the vapourisation of 284 g of CCl4 at constant pressure = (1.84 mol ×30.5 KJ mol-1) kJ


=56.12 KJ


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