Question

Imagine an aquarium with a tight-fitting lid of total volume 1.0 m3 that is two thirds filled with water. A small amount of benzene (1.56 g) is added and allowed to equilibrate between these two phases at 25 ̊C. Determine the final masses (in g) of benzene in the air and in the water. Take Henry's law constant KH = 557 m3 mol-1 Pa.

Answer 1

volume of water = 2/3 m^3
density of water = 1000 Kg/m^3
mass of water = density * volume = 1000*2/3 = 666.7 Kg = 666700 g
molar mass of water = 18 g/mol
moles of water,n1 = 666700/18 = 37037 mol

mass of benzene = 1.56 g
molar mass of benzene = 78 g/mol
number of moles of benzene = mass/molar mass
= 1.56/78
= 0.02 mol

mole fraction, X = n2/(n1+n2)
= 0.02/(37037+0.02)
= 5.4*10^-7

pressure of benzene = KH*X
= 557*(5.4*10^-7)
=3.01*10^-4 pa

use:
p*V = n*R*T
here n is number of moles of benzene in gas phase
3.01*10^-4 * (1/3) = n*8.314*(25+273)
n=4.05*10^-8 mol

mass of benzene in air = number of moles * molar mass
= 4.05*10^-8 * 78
= 3.2*10^-6 g

mass of benzene in water = 1.56 g - 3.2*10^-6 = 1.559997 g