The question is incomplete. The complete question is:
At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the equation:
ΔV = x1x2(45x1 + 25x2)
Where ΔV is in cm3-mol-1. At these conditions, the molar volumes of pure liquid 1 and 2 are V1= 110 and V2= 90 cm3-mol-1. Determine the partial molar volumes 1VE and 2VE in a mixture containing 40 mole percent of specie 1.
Answer:
1VE = 117.92 cm³.mol⁻¹, 2VE = 97.92 cm³.mol⁻¹
Explanation:
In the equation given, x represents the molar fraction of each substance, thus x1 = 0.4 and x2 = 0.6. Because of the mixture, the molar partial volume of each substance will change by a same amount, which will be:
ΔV = 0.4*0.6(45*0.4+ 25*0.6)
ΔV = 7.92 cm³.mol⁻¹
1VE - V1 = 7.92
1VE = 7.92 + 110
1VE = 117.92 cm³.mol⁻¹
2VE - V2 = 7.92
2VE = 7.92 + 90
2VE = 97.92 cm³.mol⁻¹
