The freezing point depression of 0.1 molal NaCl solution is 0.372 K. What conclusion can you draw about the molecular state of NaCl in water. `K_f` of water = 1.86 k/m.

Answer

NaCl is dissociated completely

Explanation

`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." But i"=("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^(-)`