# Answer to Question #2657 in Physical Chemistry for Lina

Question #2657

The osmotic pressure of an aqueous solution of urea at 300 K is 120 kPa. Calculate the freezing point of the same solution.

Expert's answer

π =CRT ; where π-osmotic pressure, C& - molar concentration, T – temperature , R – gas constant;

Therefore:

C = 120 kPa/(8.314 J/mol K ∙ 300 K ) = 0.048 mol/l;

Freezing point of water is depressed:

∆T = KF ∙ m, where KF – cryoscopic constant, m – molality. To get molality from molar concentration use:

& m=C/(ρ-CM)=(0.048 mol/l)/(1 kg/l-0.048 mol/l∙0.06 kg/mol)=0.0538 mol/kg

KF for water is 1.853 K∙kg/mol.

∆T=1.853 ∙ 0.0538 = 0.0997 K.

Therefore freezing point of solution is -0.0997˚C

Therefore:

C = 120 kPa/(8.314 J/mol K ∙ 300 K ) = 0.048 mol/l;

Freezing point of water is depressed:

∆T = KF ∙ m, where KF – cryoscopic constant, m – molality. To get molality from molar concentration use:

& m=C/(ρ-CM)=(0.048 mol/l)/(1 kg/l-0.048 mol/l∙0.06 kg/mol)=0.0538 mol/kg

KF for water is 1.853 K∙kg/mol.

∆T=1.853 ∙ 0.0538 = 0.0997 K.

Therefore freezing point of solution is -0.0997˚C

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