Question

What are the boiling point and freezing point of a 0.625 m aqueous solution of any nonvolatile, nonelectrolyte solute?

Answer 1

The boiling water is 69.87 and cold water is-8.69 degree Celsius

Answer 2

BOILING POINT:

ΔT = Kb*m

where,

ΔT=> elevation in boiling point of the solution

Kb=> Molal elevation constant of water, which is equal to 0.515 K kg/mol

m=> Molality of the solution

So,

ΔT = 0.515 * 0.539 = 0.278 K or 0.278 degree C

Please note that ΔT can have the unit degree Celsius or Kelvin because it is the “difference” in the boiling points of solution and solvent.

The boiling point of solution (solvent + non-volatile solute) = Boiling point of solvent (water) + ΔT

= (100 + 0.278) degree C = 100.278 degree C = 373.278 K

FREEZING POINT:

ΔT = Kf*m

where,

ΔT=> depression in freezing point of the solution

Kf=> Molal depression constant of water, which is equal to 1.853 K kg/mol

m=> Molality of the solution

So,

ΔT = 1.853 * 0.539 = 0.999 K or 0.999 degree C

Please note that ΔT can have the unit degree Celsius or Kelvin because it is the “difference” in the freezing points of solution and solvent.

The freezing point of solution (solvent + non-volatile solute) = Freezing point of pure solvent (water) - ΔT

= (0 - 0.999) degree C = -0.999 degree C = 272.001 K

Charu