The freezing point of pure water is 0.00 degree. An 8.00 g quantity of glucose (

Question

The freezing point of pure water is 0.00 degree. An 8.00 g quantity of glucose (sugar) is added to 100 g of water. The molecular weight of glucose is 160 g /mol. The freezing point depression constant for water is -1.86 degree C/kg. What is the freezing point for this glucose solution 0.00 degree C -0.93 degree C +1.86 degree C -1.86 degree C| If you repeated the experiment mentioned in question 1 above with an equal amount of sodium chloride (NaCl) in terms of moles of solute, which one of the following situations would you expect to occur The freezing point of the sodium chloride solution would be lower than that observed for the glucose solution. The freezing point of the sodium chloride solution would be higher than that observed for the glucose solution. The freezing point of sodium chloride solution is exactly the same as that observed for the glucose solution. It is impossible to determine whether the sodium chloride chloride solution would have a higher freezing point, lower freezing point or the same freezing point as the glucose solution with the information given. In the freezing point depression experiment you did with sugar and water (the first experiment), the greatest source of error was due to which one of the following factors The error in measuring the amount of water used as a solvent. The error in measuring the amount of sugar used as a solute. The error in measuring the change in temperature for the freezing point of water. Some other sort of error not mentioned above.

Explanation / Answer

moles of glucose = mass / molar mass

given

molar mass of glucose = 160 g /mol

so

moles of glucose = 8 / 160 = 0.05

now

molality = moles of glucose / mass of water (kg)

molaltiy = 0.05 / 0.1

molality = 0.5

now

dTf = kf x m

dTf = 1.86 x 0.5

dTf = 0.93

0 - Tf = 0.93

Tf = -0.93

so

the freezing point for this solution is   C) -0.93

2)

for NaCl

NaCl---> Na+ + Cl-

two particles are formed

now

dTf = i x kb x m

dTf = 2 x 0.93

dTf = 1.86

TF = -1.86

so

A) the freezing point of the NaCl solution would be lower than that observed for glucose

3)

the greastest error will be measuring the change in the temperature

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