If a 0.690 m aqueous solution freezes at-2.40 ?, what is the van\'t Hoff factor,

Question

If a 0.690 m aqueous solution freezes at-2.40 ?, what is the van't Hoff factor, i, of the solute? Kr values can be found here. Number Colligative Constants Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Solvent Formula Kf value Normal freezing Kb value Normal boiling point (°c) 0.00 5.49 6.59 C/m) point C°C) 0.512 2.53 2.92 1.22 5.03 C/m 1.86 H20 C6H6 5.12 water benzene cyclohexane C6H12 20.8 ethanol carbon tetrachloride camphor 100.00 80.1 C2H60 1.99 CCI4 80.7 78.4 76.8 29.8 -22.9 C10H160 37.8 176 ewhen using positive Kr values, assume that ? is the absolute value of the change in temperature. If you would prefer to define ? as "final minus initial" temperature, then ? will be negative and so you must use negative Kf values. Either way, the freezing point of the solution should be lower than that of the pure solvent.

Explanation / Answer

Aqueous solution means solute in water (solvent)

Freezing point depression

Freezing point of water - freezing point of solution = van't Hoff factor x freezing point depression constant x Molality

Tf - Ts = i x Kf x m

0 - (-2.40) °C = i x 1.86 °C/m x 0.690 m

2.40 = 1.2834 x i

i = 1.87

i = 2 ( in whole number)

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