Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Answer 1

What's the equation for solving for a DeltaT?

Answer 2

Freezing Point Depression

Equation: ΔT = i Kf m

ΔT= freezing point depression (T_f_ (pure solvent) - T_f_ (solux)) [°C]

Kf = freezing point depression constant [cryogenic constant] (depends on the properties of the solvent) [(°C*kg/mol)]

m = molality (moles of solute per kilogram of solvent) [(mol/kg)]

i = Van't Hoff Factor (number of ion particles per individual molecule of solute, e.g. i = 2 for NaCl, 3 for BaCl2)

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To solve:

1.) Convert grams of glucose to moles of glucose:
25.5g gluc. * (1 mol/180g gluc.) = 1.41E-1 mol gluc.
Note. Molar mass gluc. = 180g/mol

2.) Convert grams of solvent to kilograms of solvent:
398 g H2O * (1kg H2O/1000g H2O) = 0.398 kg H2O

3.) Determine molality (moles solute/kilogram solvent) of solux:
1.41E-1 mol gluc./0.398 kg H2O = 3.5427E-1 mol/kg

4.) Employ Freezing Point Depression equation and determine the freezing point of the solux:

ΔT = i Kf m
ΔT= ? (°C)
Kf = 1.86 (°C kg mol¯1)
m = 3.5427E-1 (mol/kg)
i = (assumed negligible)

ΔΤ = (1.86)(3.5427E-1)
ΔT = 6.589Ε-1°C

5.) ΔT = 6.589Ε-1°C

Freezing Point Depression: http://en.wikipedia.org/wiki/Freezing-point_depression
Freezing Point Depression Calculator: http://www.calctool.org/CALC/chem/substance/fp_depression
Freezing Pt. Depression (2): http://www.chemteam.info/Solutions/fp-depression-problems.html

Answer 3

Wouldn't 25.5g gluc. * (1 mol/180g gluc.) = equal 0.14166, not 1.41E-1 mol gluc.?

Answer 4

0.141 IS equal to 1.41E-1