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2 years ago

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Substances a and b are both volatile liquids with pa = 300 torr, pb = 250 torr, and kb = 200 torr (concentration expressed in

mole fraction). when xa = 0.9, bb = 2.22 mol kg−1, pa = 250 torr, and pb = 25 torr. calculate the activities and activity coefficients of a and
b. use the mole fraction, raoult's law basis system for a and the henry's law basis system (both mole fractions and molalities) for

b.

1 answer:

castortr0y [4]2 years ago

3 0

Calculating activity and activity coefficient:

I. using mole fraction and Raoult's law for a:

Activity α $\alpha a = \frac{Pa}{P*a}$

$\alpha a = \frac{250 torr}{300 torr} = 0.83$

and activity of a = activity coefficient of a x mole fraction of a

αa = γa . xa

γa = αa/xa = 0.83/0.90 = 0.926

II. using the henry's law basis system and:

<em>1. Mole fraction: </em>

αb = Pb/Kb = 25 torr/ 200 torr = 0.13

αb = γb . xb

γb = αb/xb = 0.13/ 0.1 = 1.3

<em>2. Molality:</em>

<em>K*b = Xa.Ma.b0.Kb</em>

<em>(Xa, b0 and Kb are known)</em>

<em>Ma = Xb/Xa.bb</em>

<em>K*b = Xa.Ma.b0.Kb = Xa.Xb.b0.Kb/Xa.bb = Xb.b0.Kb/bb</em>

<em>K*b = (0.1)(1.0 mol/Kg)(200 torr)/ (2.22 mol/Kg) = 9.0 torr</em>

αb = Pb/K*b = 25 torr / 9 torr = 2.8

αb = γb.bb/bo

γb = αb.bo/bb = (1.0 <em>mol/Kg)(2.8) / (2.22 mol/Kg) = 1.3</em>

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