A 2L container has 1 L water and 1 L air headspace. The headspace has a total pr

Question

A 2L container has 1 L water and 1 L air headspace. The headspace has a total pressure of 1 atm. The temperature of both water and air is 298K. 1 mg of formaldehyde (HCHO) was added into the water. Since formaldehyde is a volatile compound, some formaldehyde will escape into the air. The Henry’s constant (KH) for formaldehyde is 6300 M?atm-1. What is the mass concentration of formaldehyde in the water after the equilibrium between the air and the water has been reached? What is the molar fraction of formaldehyde (YHCHO) in the air headspace at equilibrium?

Explanation / Answer

P= 1 atm

Concentration of Formaldehyde in solution = kH * PA

PA = (1*10^-3)/30.026/6300 = 5.2864*10^-9 atm

Mole fraction in water = 5.2864*10^-9/1 = 5.2864*10^-9 mole *30.0260 = 1.5873*10^-7 gm

Mass concentration = 1.5873*10^-7 gm

Molar fraction in air = 5.2864*10^-9

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