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A student determines the molar mass of a liquid unknown by the method used in this experiment. She found that the equilibrium temperature of a mixture of ice and water was 1.0 degree C on her thermometer. When she added 11.1 g of her unknown sample to the mixture, the temperature, after thorough stirring, fell to -3.0 degree C. She then poured off the solution through a screen into a beaker. The mass of the solution was 90.4 g. K = 1.86 degree C/m What was the freezing point depression, Delta T? _____ degree C What was the molality, m, of the unknown solution? ______ mol/kg (Delta T = Km) How much unknown liquid was in the decanted solution? _______ g How much water was in the decanted solution? ______ kg What did she find to be the molar mass of the unknown liquid, assuming she made the calculation properly? _____ g/mol
Expert Answer
1)
freezing point depression =; dTf = freezing point of pure water – freezing point of solution
freezing point of pure water = 0.0C
freezing point of solution = -3.0 C
therefore;
dTf = 0 – (-3)
dTf = 3.0C
Thus the freezing point depression is 3 C
2)
freezing point depression;dTf = Kf x m
here ;
Kf for water = 1.86 C kg /mol
And freezing point depression= 3.0C
m = molaity
Then;
3 = 1.86 x m
m = 1.613
The molality of the unknown solution is 1.613 m
3)
11.1 g of unknown liquid was in the decanted solution
4)
Given that the mass of solution = 90.4 g
Mass of solution = mass of solute + mass of solvent
Here solvent is water. Thus
amount of water = 90.4g – 11.1g
amount of water = 79.3 g
so there is 79.3 g of water in the decanted solution .
5)
We know that molality is the number of moles in per kg of solvent then;
Here mass of water = 79.3 g
79.3 g *1.0 kg /1000 g = 0.0793 kg
molality = moles of liquid / mass of water (kg)
1.613 = moles of liquid / 0.0793
moles of liquid = 0.128 moles
and number of moles = amount in g / molar mass
then;
molar mass = amount in g / number of moles
= 11.1 g / 0.128 moles
=86.72 g / moles
