1.

Depression in freezing point of 1.10-molal solution of HF is `0.201^(@)C`. Calculate percentage degree of dissoviation of HF (`K_(f)`=1.856 K kg `mol^(-1)`).

Answer» Correct Answer - 8.06%18.67 atm
`DeltaT_(f)=ixxK_(f)xxmori=(DeltaT_(f))/(K_(f)xxm)`
`i=((0.201K))/((1.86" K kg mol"^(-1))(0.10"mol kg"^(-1)))=1.0806.`
`"Degree of dissociation of HF "(alpha)=(i-1)/(n-1)=(1.0806-1)/(2-1)`
=0.0806 = 8.06 %.


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