Question

Given that the latent heat of fusion of naphthalene is `19.0 KJ mol^(-1)` and its melting point is `80.2^(@)C`.Estimate the solubility of naphthalene in benzene at `76.2^(@)C`.

Answer 1

Given
`T^(@) = 80.2^(@)C ,T_(S)=76.2^(@)C`
`Delta_(fus)H=90.0 kJ mol^(-1)`
For an ideal solution
`log chi_(2) =(DeltaH)/(2.303R(T^(@))^(2)) (T_(S)-T^(@))`..(i)
`DeltaH=19000 J,T_(S)-T^(@) =-4^(@)C`,
`R=8.314`
Substituting all values in Eq.(i),
`log chi_(2)=(19000)/(8.314xx2.303xx(353.20)^(2))xx(-4)`
or `chi_(2)=0.929~~0.9`