Question

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1.) The partial molar volume of water and ethanol in a solution with Xx20=0.60 at 25°C are 17.0 and 57.0 cm mol-, respectively. Calculate the volume change upon mixing sufficient ethanol with 2.00 mol of water to give this concentration. The densities of water and ethanol are 0.997 and 0.7893 g cm , respectively, at this temperature.

Answer 1

Given before mixing water a. ethanol → 2 v = 17 cm Te = 57 cm? I mor mol 81 0.997 g/cm3. 12 = 0.1893 g/cm3 XH2O = 0.60 (mole fraction of water) before miting. Total volume. Vi - Vitve Ti = 74 com mot- nomo Now Xtra = Xto - nhorn & H2O = n = min2 Now 0.24 and ni= a mot Louilt thin n = 6.33 mo! molar this in solution. volume of - Mi Pi = = 18.05 cm3 18 g/mol 0.991 g/cm3 mol mol similarly, V2 = M2 = 46 g/mol = 58.28 cm3 0-1843910m3 Totat volume of solution, VF = V + VZ' : 7633 olar change in molas vot out

V = vini = 1 V2 = Tam = 1782 = 34 cm3 STX.6.33 = 360.81 cm3 18.05x 2 = 36:10 cm3 58.28 x 6.33 = 369 cm3 vil = U21 = v'ni = Jln2 = chanse in volume. Dv = (vi'+V2) - (VI+ V2) lov- 10.29 cm3