physical chemistry A solution is prepared by dissolving 122 g of a nervolaite vo

Question

physical chemistry

A solution is prepared by dissolving 122 g of a nervolaite volume whose molecular mass is M = 241.g/mol in 920 g of water (H_2O M = 13.02 g/mol). The vapor pressure of water above the solution is pass = 17.02 mol. The vapor pressure of pure water of the same temperature is pass = 17.54 torr. a) What are a(H_2O) and T(H_2O), the activity and activity coefficient for water in the above solution. b) The freezing point depression constant for water is K_f = 1.85 kg.K/mol. Assuming ideal behavior, find normal freezing point for the above solution Give your final answer in degree C. In a recent experiment (C. W. Wu, H Matsui, N.S Wag M C) Ln Chem A 115, 8086 (2011) the thermal decomposition of ethanol at high temperature was investigated

Explanation / Answer

(A)

Moles of compound taken = 122/241 = 0.506

Moles of water = 920/18.02 = 51.05

Mole fraction of water, Xw = 51.05/(51.05+0.506) = 0.99

Vapor pressure of water according to Raoult's law = P0*Xw = 17.54*0.99 = 17.36 torr

Actual vapor pressure = 17.02 torr

Thus activity coefficient for water, fH2O = Actual vapor pressure / raoult's law v.p. = 17.02 / 17.36 = 0.98

Activity , aH2O = Xw*fH2O = 0.99*0.98 = 0.97

(B)

Since we are assuming ideal behaviour, we will be using the following eqn :

dT = Kf * m

Here m is molality of solution = moles of solute/kg of solvent =0.506/0.920 = 0.55

Thus, dT = 1.86*0.55 = 1.023

Thus normal freezing point for this solution is -1.023 0C

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