The freezing point depression of 0.1 molal solution acid in benzene is 0.256 K. For benzene `K_(f)` is 5.12 K kg `mol^(-1)` Calculate the value of Van't Hoff factor for enzoic acid in benzene. What conclusion can you drow about the molecular state of benzonic acid in cenzene ?

Answer

Benzoic acid is a dimer in benzene

Explanation

Step I. Calculation of Van't Hoff factor.
`"Freezing point depression "(DeltaT_(f))=ixxK_(f)xxm`
`DeltaT_(f)=0.256 K, K_(f)=5.12" K kg mol"^(-1),m=0.1m=(0.1" mol kg"^(-1))`
`i=(DeltaT_(f))/(K_(f)xxm)=((0.256 K))/((5.12" K kg mol"^(-1))xx(0.1"mol"^(-1)kg))=0.5`
Step II. Predicting the molecular state of benzoic acid.
`i=("Normal molar mass")/(Observed molar mass")=0.=1/2.`
This shows that the observed molar mass of benzoni acid is double the normal molar mass or benzoic axists as a dimer in benzene.