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When a liquid converts into a vapor state it absorbs heat. The amount of heat absorbed when one mole of a substance converts into a vapor state is known as molar heat of vaporization.

The molar heat of vaporization of ammonia is \( = 4.8\;{\rm{kJ}}/{\rm{mol}}\)

The amount of ammonia is \( = 100.0\;{\rm{g}}\)

The molar mass of ammonia is \( = 17.031\;{\rm{g}}/{\rm{mol}}\)

Hence, we can say that:

when 1 mole of ammonia converts into vapor the amount of heat absorbed is \( = 4.8\;{\rm{kJ}}\)

when \(17.031\;{\rm{g}}(1\) mole) of ammonia converts into vapor the amount of heat absorbed is \( = 4.8\;{\rm{kJ}}\)

when \(1\;{\rm{g}}\) of ammonia converts into vapor the amount of heat absorbed is \( = \frac{{4.8\;{\rm{kJ}}}}{{17.031}}\)

when \(100.0\;{\rm{g}}\) of ammonia converts into vapor the amount of heat absorbed is \( = \frac{{4.8\;{\rm{kJ}}}}{{17.031}} \times 100.0\)

\( = 28.2\;{\rm{kJ}}\)