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.

We need to find the enthalpy change for vapourisation of 85 g of NH$$_3$$. The molar mass of NH$$_3$$ is calculated by adding the atomic mass of nitrogen and three hydrogens, giving 14 + 3(1) = 17 g mol$$^{-1}$$. Dividing 85 g by this molar mass yields $$n = \frac{85}{17} = 5$$ mol.

Given that the enthalpy of vapourisation of NH$$_3$$ is 23.4 kJ mol$$^{-1}$$, the total enthalpy change for vapourising 5 moles can be calculated as $$\Delta H = 5 \times 23.4 = 117$$ kJ. The enthalpy change for the vapourisation of 85 g of NH$$_3$$ is 117 kJ.