A solution of glyceraldehyde, C3H6O3, has a density of 1.15 g/mL, and a mass per

Question

A solution of glyceraldehyde, C3H6O3, has a density of 1.15 g/mL, and a mass percentage of 28.1% glyceraldhyde at 25 deg C. (molar mass = 90.078g/mol)

For this solution, calculate the following:

A. Moles of Glyceraldhyde in 100g of solution.

B. Mole fraction of glyceraldhyde

C. Vapor pressure of the solution ( given Pvapor= 23.8 mmHg at 25 deg C)

D. Molal concentration of Glyceraldhyde

E. Freezing Point of the solution ( given Kf= 1.86 deg C/m)

F. Molar concentration of glyceraldehyde

(Please give the equation used)

Explanation / Answer

(A) Mass of glyceraldehyde in 100 g of solution = 28.1% of mass of solution

= 28.1/100 x 100 = 28.1 g


Moles of glyceraldehyde in 100 g of solution = mass/molar mass of glyceraldehyde

= 28.1/90.078 = 0.312 mol


(B) Mass of water 100 g of solution = 100 - 28.1 = 71.9 g


Moles of water in 100 g of solution = mass/molar mass of water

= 71.9/18.02 = 3.99 mol


Mole fraction of glyceraldehyde = 0.312/(0.312 + 3.99) = 0.0725


(C) Mole fraction of water = 1 - 0.0725 = 0.9275


Vapor pressure of solution = mole fraction of water x vapor pressure of water

= 0.9275 x 23.8 = 22.1 mmHg


(D) Molality = moles of glyceraldehyde/mass of water in kg

= 0.312/0.0719 = 4.34m


(E) Freezing point depression DTf = Kf x molality

= 1.86 x 4.34 = 8.07 deg C


Freezing point of solution = freezing point of water - DTf

= 0.0 - 8.07 = -8.07 deg C


(F) Volume of 100 g of solution = mass/density

= 100/1.15 = 86.96 mL = 0.08696 L


Molarity = moles of glyceraldehyde/volume of solution

= 0.312/0.08696 = 3.59 M

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