Question 2 (15 pts): A 2L container has 1 L water and 1 L air headspace. The hea

Question

Question 2 (15 pts): 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 atm1. 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 fractiorn of formaldehyde (YHcHo) in the air headspace at equilibrium?

Explanation / Answer

Henry's law formula

P = K? H C

C = P/K H = 1atm/6300M atm-1 = 0.000157M

C is the concentration of formaldehyde in water.

Therefore moles of formaldehyde in 1L of water = 0.000157mol

Therefore mass of formaldehyde

= 0.000157mol / 30.031gmo-1 = 5.23e-6 g

Mass concentration of formaldehyde = 5.23e-6g L-1

grams of formaldehyde in airhow space = 0.001g - 5.23e-6g

=0.000994 g

Moles of formaldehyde in air= 0.000994g / 30.031gmol-1

=3.31e-5 mol

PV = n? total RT

n total = PV / RT = 1atm* 1L / 0.082057L atm mol-1K-1*298K

= 0.0409 mol

Therefore,

Molar fraction of formaldehyde in air

= 3.31e-5mol / 0.0409 mol = 0.00081

     

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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