An aqueous solution of K_2 CO_3 is prepared by dissolving 15.6 g K_2 CO_3 in 44.

Question

An aqueous solution of K_2 CO_3 is prepared by dissolving 15.6 g K_2 CO_3 in 44.3 g H_2 O. The density of the solution is 1.20 g/mL at 25 degree C What is the molarity of the solution? What is the molality of the solution? What is the mole fraction of solvent in the solution Estimate the normal boiling point of the solution(K_b of H_2 O = 0.512 degree C/m; normal boiling point of H_2O = 100.00 degree C Estimate the freezing point of the solution(K_r of H_2 O = 1.86^degree C m; freezing point of H_2O = 0.00^degree C Estimate the osmotic pressure of this solution at 25^degree C Estimate the vapor pressure of this solution at 25^degree C. The vapor pressure of pure H_2O at this temperature is 23.8 torr

Explanation / Answer

a] Molarity = Molesof solute / Volume of solution in L

Moles = Mass / MW [MW of K2CO3 = 138 gms ]

Mass of solution = 15.6+44.3 = 59.9 gms

density = Mass / Volume

1.2 = 59.9 / Volume

Volume = 49.9167 ml

Molarity = 15.6*1000 / 49.9167*138 = 2.264 M

b]

Molality = moles of solute / Solvent weight in Kg

Molality = 15.6 *1000 / 44.3*138 = 2.55 m

c]

Mole fraction of solvent = moles of solvent / TOtal moles

Molesof solvent = 44.3/18 = 2.46

Moles of solute = 15.6/138 = 0.113

Mole fraction of solvent = 2.46 / 2.46+0.113 = 0.956

d]

K2CO3 ---> 2K+ + Co3-2 [ i=3 no of particles]

delta T = i*Kb*m [m = molality]

delta T = 3.9168

Boiling point of solution = 103.9168 C

e]

delta Tf = i*Kf*m = 14.229

Freezing point = 0 - 14.229 = -14.229 C

f]

Osmotic pressure = MRT

M = molarity ; R - 0.0821 ; T in K

Osmotic pressure = 55.39 atm

g]

Vapour pressure of solution = Mole fraction*Pure vapour pressure = 0.956*23.8 = 22.75 torr

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