Question

For ethyl alcohol, C2H5OH, the enthalpy of fusion is 108.9 J/g, and the entropy of fusion is 31.6 J/mol •K. The enthalpy of vaporization at the boiling point is 837 J/g, and the molar entropy of vaporization is 109.9 J/mol •K.

Answer 1

`T_m=157.22=-115.8^oC\\T_b=350.34K=77.34^oC`

Step-by-step explanation:

Hello,

In this case, given the enthalpies and entropies of boh fusion and boiling, we can compute both the fusion and boiling points as shown below, considering ethanol's molar mass is 46 g/mol:

`T_m=(\Delta _fH)/(\Delta _fS) =(108.9(J)/(g)*(46g)/(1mol) )/(31.6(J)/(mol*K) ) \\\\T_m=157.22=-115.8^oC\\\\T_b=(\Delta _bH)/(\Delta _bS) =(837(J)/(g)*(46g)/(1mol) )/(109.9(J)/(mol*K) ) \\\\T_b=350.34K=77.34^oC`

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