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»


Solution :`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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