Name R number TA Laboratory Date Section 2. A solution is prepared by dissolving

Question

Name R number TA Laboratory Date Section 2. A solution is prepared by dissolving 0.074 g of glucose, a molecular solid with the formula CH1 Os in in 927 ml of water at 20"C a. Use Raoult's law to determine the water vapor pressuring lowering at 20 C where the vapor pressure and density of pure water are 17.5424 torr and 998.2 g/em, respectively b. Calculate the freezing point decrease for this solution compared to pure water. c. Calculate the boiling point increase for this solution compared to pure water. Calculate the osmotic pressure of this solution assuming no volume change when the solution is prepared. d. e. Considering the results from parts a-d, which colligative property would you use to ine the formula weight of glucose (assuming it was unknown) if your decision was based only on the relative magnitude of the physical property change.

Explanation / Answer

a) Vapor pressure:

First we must calculate the moles of glucose that were added to the pure water and the moles of pure water present:

nGlucose = 0.074 g / 180.156 g / mol = 4.11 x10 ^ -4 moles

mass H2O = 927 mL * 0.9982 g / mL = 925.3314 g

n H2O = 925.3314 g / 18.02 g / mol = 51.35024 moles

The total moles are calculated:

Total n = n Glucose + n H2O = 51.35026 moles

The molar fraction of the water in solution is calculated:

X H2O = n H2O / n Totals = 51.35024 moles / 51.35026 moles = 0.9999

finally the vapor pressure is calculated:

Pv = XH2O * Pvº = 0.9999 * 17.5424 torr = 17.5406 torr.

b) Cryoscopic descent: the molality of the solution must be calculated:

m = n Glucose / Kg H2O = 4.11 x10 ^ -4 moles / 0.9253314 Kg = 4.4417x 10 ^ -4 mol / Kg

the decrease in the freezing point is calculated with the water solvent constant:

Tc = -1.86 ºC * Kg / mol * 4.4417x 10 ^ -4 mol / Kg = - 8.2616 x10 ^ -4ºC

c) Boiling temperature: the change in the boiling temperature is calculated with the water solvent constant:

Te = 0.52 ºC * Kg / mol * 4.4417x 10 ^ -4 mol / Kg = 2.3097 ^ -4 ºC

d) Osmotic pressure: the molarity of the solution is calculated:

M = 4.11x10 ^ -4 moles / 0.927 L = 4.43x10 ^ -4 M

then the osmotic pressure is calculated:

= M * R * T = 4.43x10 ^ -4 mol / L * 0.082 atm * L / mol * K * 293 K = 0.0106 atm.

e) You can choose change in freezing point and / or change in the boiling point to determine the molecular weight of the unknown solute, for its ease and simplicity to determine with the calculations.

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