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 âˆ™ 300 K ) = 0.048 mol/l;

Freezing point of water is depressed:

âˆ†T = KF âˆ™ 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/lâˆ™0.06 kg/mol)=0.0538 mol/kg

KF for water is 1.853 Kâˆ™kg/mol.

âˆ†T=1.853 âˆ™ 0.0538 = 0.0997 K.

Therefore freezing point of solution is -0.0997ËšC

Therefore:

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

Freezing point of water is depressed:

âˆ†T = KF âˆ™ 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/lâˆ™0.06 kg/mol)=0.0538 mol/kg

KF for water is 1.853 Kâˆ™kg/mol.

âˆ†T=1.853 âˆ™ 0.0538 = 0.0997 K.

Therefore freezing point of solution is -0.0997ËšC

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