1.

The freezing point depression of 0.1 m NaCI solution is 0.372^(@)C. What conlusion would you draw about the state in aqueous soluion ? (K_(f) "for water"=1.86 "K kg mol"^(-1)).

Answer»


Solution :Step. Calculation of Van't HAFF factor.
`"FREEZING point DEPRESSION"(DeltaT_(f))=ixxK_(f)xxm`
`DeltaT_(f)=0.372^(@)C or 0.372 K, K_(f)=1.86" K kg MOL"^(-1), m=0.1" m"or 0.1" mol kg"^(-1)`
`i=(DeltaT_(f))/(K_(f)xxm)=((0.372 K))/((1.86" K kg mol"^(-1))xx(0.1" mol"^(-1)kg))=2.`
Step II.`"Calculation of degree of DISSOCIATION"(alpah).`
`NaCI to Na^(+)+CI^(-)`
`therefore""alpha=(i-1)/(n-1)=(2-1)/(2-1)1.`
`"Thus, degree of dissociation of "alpha=1 "or it is 100% dissociated in dqueous soltion.`


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