Q. 135.0( 1 Vote )

# A solution of glu

Answer :

Given, boiling point of solution, Tb = 100.20°C = 373.35 K

Kf and Kb = 1.86 K kg mol–1 and 0.512 K kg mol–1 respectively.

We know, boiling point of pure water, Tb° = 100°C = 373.15 K

And freezing point of pure water, Tb°= 0°C = 273.15 K

Δ Tb = Kbm

Tb - Tb°= 0.512 m

0.20 = 0.512 m

m = 0.390 mol/kg

Now, putting the above value in Δ Tf = Kfm

Δ Tf = 1.86 0.390

= 0.725 K

And, Δ Tf = Tf° - Tf

Tf = 273.15 – 0.725

= 272.425 K

Therefore, the freezing point of the same solution is 272.425K.

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