1.

Depression in freezing of 0.10 molal solution of HF is -0.201^(@)C. Calculate the percentage degree of dissociation of HF. (K_(f)="1.86 K kg mol"^(-1)).

Answer»


Solution :Observed `DeltaT_(f)=0.201^(@)C`. Calculated `DeltaT_(f)-K_(f)xxm=1.86xx0.1=0.186^(@)C""therefore""i=(0.201)/(0.186)=1.0806`
`{:(HF,hArr,H^(+),+,F^(-),),("1 mol",,"0",,"0",),(1-alpha,,alpha,,alpha",", "TOTAL "=1+alpha.." HENCE, i"=1+alpha"or"alpha=i-1=0.0806=8.06%):}`


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