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Kitty [74]
3 years ago
10

The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What

is the heat of vaporization at 75oC?
Chemistry
1 answer:
pentagon [3]3 years ago
8 0

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

=  Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

or

The heat of vaporization at 75° C = 900 J/mol

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