What volume of water, in mL, is needed to dissolve 1.3*103 mg of carbon dioxide gas when the partial pressure of carbon dioxide above the water is 3.7 atm at 25 °C? The Henry's constant for carbon dioxide gas in water at 25 °C is 0.034 M/atm.

Question

Chemistry

Pyscho

11 August, 22:02

What volume of water, in mL, is needed to dissolve 1.3*103 mg of carbon dioxide gas when the partial pressure of carbon dioxide above the water is 3.7 atm at 25 °C? The Henry's constant for carbon dioxide gas in water at 25 °C is 0.034 M/atm.

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Answers (1)

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Aimee · Answer 1

V = 0.2714 mL

Explanation:

Henry's Law:

Pi = Ci*Kh

∴ Kh CO2 = 0.034 M/atm ... The Henry's constant

∴ Partial pressure: P CO2 = 3.7 atm

⇒ C CO2 = P CO2 / Kh

⇒ C CO2 = 3.7 atm / 0.034 M/atm = 108.823 M (mol/L)

∴ molar mass CO2 = 44.01 g/mol

⇒ mol CO2 = (1.3 E3 mg) (g/1000 mg) (mol/44.01 g) = 0.03 mol CO2

volumen solution:

⇒ V sln = (0.03 mol) (L/108.823 mol) = 2.714 E-4 L

⇒ V sln = 0.2714 mL