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17 . 0   g of NH 3 completely vapourises at - 33 . 42   ° C and 1 bar pressure and the enthalpy change in the process is 23 . 4   kJ   mol - 1 . The enthalpy change for the vapourisation of 85   g of NH 3 under the same conditions is kJ .

Correct answer

117

Step-by-step solution

17 . 0   g of NH 3   =   17 17   =   1   mole and the enthalpy change = 23 . 4   kJ   mol - 1 . 85   g of NH 3   =   85 17   =   5   moles enthalpy change under the same conditions will be = 23 . 4 × 5 = 117   kJ .

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