1.

The freezing point depression of 0.1 m NaCl solution is 0.372^(@)C. What conclusion would you draw about its molecular state ? K_(f) for water is "1.86 K kg mol"^(-1).

Answer»


Solution :`DeltaT_(f)"(obs.)"=0.372^(@)C, DeltaT_(f)"(cal.)"=1.86xx0.1=0.186^(@)C`
As `DeltaT_(f)"(obs.)" gt DeltaT_(f)"(cal.)"` this MEANS NaCl is dissociated in solution.
`"Further,i"=(DeltaT_(f)"(obs.)")/(DeltaT_(f)"(cal.)")=(0.372)/(0.186)=2."Buti"=("No. of particles after dissociation")/("No. of molecules taken")`
This means that each NaCl MOLECULE ionizes to give two particles, i.e., `Na^(+) and Cl^(-)` ions as follows:
`NaCl overset(+AQ)rarrNa^(+)+Cl^(-)`


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