Answer to Question #2657 in Physical Chemistry for Lina
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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