Onserved mol. Mass
`= (1000xxK_(f)xxw)/(WxxDeltaT)`
`=(1000xx5.13xx20xx10^(-3))/(1xx0.69)=1.48.4`
Normal mol. Mass of phenol `(C_(6)H_(5)OH)=94`
So, `("Normal mol. mass")/("Observed mol. mass")=(94)/(148.4)`
`=1+((1)/(n)-1)alpha=1+((1)/(2)-1)alpha`
`(94)/(148.4)=1-(alpha)/(2)`
or `alpha=0.733 or73.3%`.