Given data The equilibrium partial pressure of A in the equimolar mixture of A a
ID: 699530 • Letter: G
Question
Given data
The equilibrium partial pressure of A in the equimolar mixture of A and B = 7.84 kPa
The density of A = 0.8 g/cm3 = 0.8 g/mL (since 1 cm3 = 1 mL) and its molecular weight = 160 g/mol
Therefore, the liquid molar voluem of A in the solution = 160/0.8 = 200 mL/mol (at 25 oC)
The saturation pressure of A in the solution = 13.3 kPa
The liquid molar volume of B = 107 mL/mol (At 25 oC)
The solubility parameter of B = 18.2 (J/mL)1/2 (At 25 oC)
The saturation pressure of B = 25.5 kPa (at 10 oC)
The solubility parameter of the species similar to A = 10-20 (J/mL)1/2
Here, the molefraction of A in the solution (xA) = 200/(200+107) = 0.65 and xB = 107/(200+107) = 0.35
The liquid B with 0.35 molefraction has the solubility parameter of 18.2 (J/mL)1/2
Therefore, the solubility parameter of liquid A having the molefraction 0.65 in the solution = (0.65/0.35) * 18.2 = 33.8 (J/mL)1/2
Explanation / Answer
te: 12/15/2017 Question 4 (25 points) An equimolar mixture of A and B at 10°C gives an equilibrium partial pressure of A equal to 7.84 kPa. Assuming regular-solution theory describes the mixture well, estimate the solubility parameter of A at 25°C The following data are available: I) A has a density of 0.8 g cm3 at 25°C and its molecular weight is 160 2) A has a saturation pressure of 13.3 kPa at 10°C 3) At 25°C, liquid molar volume of B is 107 cm mol and solubility parameter of B is 18.2 (J cm2) 4) B has a saturation pressure of 25.5 kPa at 10°C 5) Species similar to A have solubility parameters in the range of 10-20 (J cm3)12
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