A solution containing 0.460 g in 100 mol solution has an osmotic pressure of 38

Question

A solution containing 0.460 g in 100 mol solution has an osmotic pressure of 38 torr at 25 degree C. What are the molarity of the solution and the molar mass of solute? The vapor pressure of pure water at 25 degree C is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by dissolving 18.0 g of glucose (a nonelectrolyte, MW = 180.0 g/mol) in 95 0 g of water? The freezing point of ethanol (C_2H_5O_H) is -114.6 degree C. The molal freezing point depression constant for ethanol is 2.00 degree C/m. What is the freezing point (degree C) of a solution prepared by dissolving 50.0 g of glycerin (C_3H_8O_3, a nonelectrolyte) in 200.0 g of ethanol?

Explanation / Answer

ANSWER

5(1)

The question is based on colligative property relative lowering of vapour pressur

(Po-P)/Po = x2

Po = vapour pressure in pure state, P = vapour pressure in solution

x2 = mole fraction of solute = n2 / n2+n1

n2 = no. of moles of solute = mass/molar mass = 18 / 180 = 0.1moles

n1 = moles of solvnt = 95.0 / 18 = 5.27moles

x2 = 0.1/5.27 + 0.1 = 0.0186

(Po-P)/Po = 0.0186

Po = 23.8torr

(23.8 - P)/23.8 = 0.0186

P = 23.35 = 23.4

Hence OPTION (B) IS CORRECT ANSWER.

(2)

THE QUESTION IS BASED ON DEPRESSION IN FREEZING POINT

T = Tsolvent - Tsolution = Kf X m

Tsolvent = freezing point of solvent = -114oC

Tsolution =  freezing point of solution

Kf = 2.00C/m

m = molality of solutr = no. of moles of solute / ,mass of solvent in kg

no. of moles of solute = 50.0 / 92.0 = 0.54 moles

mass of solvent in kg = 200/1000 = 0.2kg

m = 0.54 / 0.2 = 2.7

-114 - Tsolution = 2 X 2.7

Tsolution = 108.6oC

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