Correct Answer - 3.59 K
`DeltaT_(f)=K_(f)xxm" and " DeltaT_(b)=K_(b)m `
`(DeltaT_(f))/(DeltaT_(b))=K_(f)/K_(f),DeltaT_(b)=((30"Kg mol"^(-1)))/((5.02"K kg mool"^(1)))xx(0.60K)=3.59 K.`
`DeltaT_(f)=K_(f)/K_(b)xxDeltaT_(b)=((30" K J mol"^(-1)))/((5.02" K kg mol"^(-1)))xx(0.60 K)=3.59 K.`