The Henry’s law constant for O2 dissolved in water is 4.34 × 104 atm at certain temperature. If the partial pressure of O2 in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of O2 in the solution?

Question

The Henry’s law constant for O2 dissolved in water is 4.34 × 104 atm at certain temperature. If the partial pressure of O2 in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of O2 in the solution? (A) 1 × 10-5 (B) 1 × 10-4 (C) 2 × 10-5 (D) 1 × 10-6

Tardigrade Admin · Accepted Answer

Given, KH for O 2 dissolved in water =4.34 × 104 atm PO 2 =0.434 atm x(. mole fraction of O 2 in solution )=? According to Henry's law, P =KH × x (0.434/4.34 × 104) =x O 2 1 × 10-5 =x O 2 ∴ Mole fraction of O 2 in water =1 × 10-5.

Q. The Henry’s law constant for $O_{2}$ dissolved in water is $4.34 \times 1 0^{4}$ atm at certain temperature. If the partial pressure of $O_{2}$ in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of $O_{2}$ in the solution?

$1 \times 1 0^{- 5}$

$1 \times 1 0^{- 4}$

$2 \times 1 0^{- 5}$

$1 \times 1 0^{- 6}$

$\times 1 0^{- 6}$

Q. The Henry’s law constant for O2​ dissolved in water is 4.34×104 atm at certain temperature. If the partial pressure of O2​ in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of O2​ in the solution?

1×10−5

1×10−4

2×10−5

1×10−6

×10−6