Question

An aqueous sucrose (C12H22O11) solution of unknown concentration is found to have a freezing point of –0.912°C. What is the normal boiling point and the partial pressure (in torr) of water at 25°C of this solution? Kb (H2O) = 0.51°C/m; P°(H2O) = 23.800 torr at 25°C.

Answer 1

∆T = (0-(-0.912) = 0.912°C

∆T = Kf* molality

Kf = 1.86°C/molality

= 0.912 = 1.86 * molality

Molality = 0.49

∆Tb = kB * molality

= 0.512 * 0.49

= 0.251 °C

T = 100 + ∆T

= 100 + 0.251

= 100.251 °C

Moles / mass of solvent (in Kg) = 0.49

Suppose we have 1 kg of water

Moles of sucrose = 0.49

Moles of water = mass / molar mass

= 1000 / 18

= 55.55

Mole fraction of water = moles of water / total moles

= 55.55 / ( 55.55 + 0.49)

= 0.99126

P(solution) = mole fraction of water * P°(water)

= 0.99126 * 23.8

= 23.592 torr

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