The following experiment is conducted to measure the Henry’s law constant of chl

Question

The following experiment is conducted to measure the Henry’s law constant of chloroform (CHCl3). A vessel with a volume of 2 L is half filled with water and then sealed, so it contains 1 L of water and 1 L of air at 1 atm and 293 K. One gram of liquid chloroform is injected into the water through a rubber septum. The contents are thoroughly mixed and then allowed to equilibrate. A sample of gas is extracted from the air space above the water. Using a gas chromatograph, it is determined that the mass concentration of chloroform in air is 89 mg/L. Determine the value of the Henry’s law constant in units of M/atm.

Explanation / Answer

CHCl3 concentration = 89 mg /L

CHCl3 molar mass = 119.3 g /mol

CHCl3 molarity = (89 x 10^-3 g / 119 .3 g/mol)   / L

                          = 7.46 x 10^-4 mol / L

                          = 7.46 x 10^-4 M

pressure = 1 atm

concentration = KH x pressure

7.46 x 10^-4 M = KH x 1 atm

KH = 7.46 x 10^-4 M / atm

Henry’s law constant =KH = 7.46 x 10^-4 M / atm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.