1. Using Henry\'s Law constants in the table 6.1 given below, calculate the solu

Question

1.

Using Henry's Law constants in the table 6.1 given below, calculate the solubility in M of each gas in water at 25C if

PO2=0.2bar where K is 42*103 bar

PN2=0.75bar where K is 84*103 bar

PO2=0.05bar where K is 42*103 bar

Express your answer to one significant figure and include the appropriate units

2. what will be the vapor pressure at 25oC of water in this solution if Raoult's law holds? The vapour pressure of pure water at 250C is 3167 Pa.

Exercise 6.8 Using Henry's Law constants in the table given below, calculate the solubility in M of each gas in water at 25°C if TABLE 6.1 Henry's Law coefficients k pvB/xB, n bar, for aqueous solutions Gas 370C 31 x 103 139x 103 138 x 103 He 84 x 103 50 x 103 98 x 10 35 x 103 57 x 103 67 x 103 CO 26 x 103 42 x 103 50 x 10 23 x 10 38 x 10 46 x 10 CH 45 x 103 x 10 38 x 103 1.59 x 103 0.71 x 10 2.13 x 103 1.32 x 103 0.71 x 10 1.69 x 103 2 2 Part A 0.2 bar Express your answer to one significant figure and include the appropriate units. Value Units Subm My Answers Give U Part B 0.75 bar Express your answer to two significant figures and include the appropriate units. Value Units Subm My Answers Give U Part C CO 0.05 bar Express your answer to one significant figure and include the appropriate units.

Explanation / Answer

according to Henry's law

solubility of gas and its partial pressure are related as (at a given temperature)

Solubility= K X parital pressure

Where K = Henry's constant

a) PO2 = 0.2 bar

Henry's constant = 42*103 bar

so we will obtain mole fraction here.

mole fraction = partial pressure / K = 0.2 bar / 42*103 bar = 4.76 X 10^-6

so moles of water = 1-4.76 X 10^-6 = 0.99moles = 0.99 X 18 = 17.82 grams = 17.82 mL

so concentration = 4.76 X 10^-6 / 17.82 X 10^-3 = 2.67 X 10^-4 moles / L

2) PN2=0.75bar where K is 84*103 bar

mole fraction = partial pressure / K = 0.75 bar / 84*103 bar = 8.93 X 10^-6

so moles of water = 1-8.93 X 10^-6 = 0.99moles = 0.99 X 18 = 17.82 grams = 17.82 mL

so concentration =8.93 X 10^-6 / 17.82 X 10^-3 = 5.01 X 10^-4 moles / L

c) PO2=0.05bar

Henry's constant = 42*103 bar

so we will obtain mole fraction here.

mole fraction = partial pressure / K = 0.05 bar / 42*103 bar = 1.19 X 10^-6

so moles of water = 1-4.76 X 10^-6 = 0.99moles = 0.99 X 18 = 17.82 grams = 17.82 mL

so concentration = 1.19 X 10^-6 / 17.82 X 10^-3 = 6.67 X 10^-5 moles / L

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