Problem Set 11 Solutions

6. The Henry's Law constant for carbon dioxide in water is 1.25×10 6 torr at 25˚C. Calculate the solubilityof carbon dioxide in water at 25˚C when its partial pressure in air is (a) 4.0 kPa and (b) 100 kPa.9Henry's Law isp i = K H x i . Solving for the mole fraction yields€x i =p iK H.(a) For a partial pressure of 4.0 kPa or 30.0 torr (1 torr = 133 Pa), the mole fraction of carbon dioxide dissolvedin water is€30.0 torrx i =1.25×10 6 torrx i = 2.4 ×10 −5 .(b) For a partial pressure of 100 kPa € or 750 torr, the mole fraction of carbon dioxide dissolved in water is750 torrx i =1.25×10 6 torrx i = 6.0 ×10 −4 .€7. At 25˚C, the mole fraction of air dissolved in water is 1.388×10 –5 .(a) Determine the molarity of the solution.The molarity is defined as moles of air divided by liters of solution,[ air] =n airL solution.Since the mole fraction is so small, we can assume that the liters of solution are equal to liters of water,€[ air] =n airL H 2 O.In order to calculate the molarity, we need to determine the moles of air. The definition of mole fraction is€x air =n air.n air + n H 2 O€