Benzene and toluene form nearly ideal solutions. Consider an equimolar solution

Question

Benzene and toluene form nearly ideal solutions. Consider an equimolar solution benzene and toluene. At 20 degree C the the of vapour pressures of pure benzene and toluene are 9.9 kPa and 2.9 kPa, respectively.The solution is boiled by reducing external pressure below the vapour pressure. Calculate (a) the pressure when boiling begins, (b) the of each component in the vapour, and (c) the vapour pressure when only a few drops of liquid remain. Assume that the rate of vaporization is low enough for the temperature to remain constant at 20 degree C.

Explanation / Answer

(a) the Pressure when boiling begins

since equimolar solution

P = C b Pob + Ct Pot

= 0.5 (9.9 + 2.9 )

= 6.4 kPa

(b) the Composition of each component in the vapour

Pb = Yb P = C b Pob

Therefore Y b = C b Pob / P = 0.5 x 9.9 kPa / 6.4 kPa = 0.77

(c) the vapour pressure when only few drops of liquid remains

Y b = Y t = 0.5

Y b / Y t = Pob x Cb /   Pot Ct = 1

Therefore P = Pob x Cb - (1 - Cb ) Pot    Where Cb = 0.226

P = 9.9 x 0.226 - (1 - 0.226) 2.9

P = 4.48 k Pa

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