Question

Knowing that at 0 ^oC and an oxygen pressure of 1.00 atm, the aqueous solubility of O₂ (g) is 48.9 mL O2 per liter. The approximate values for the molarity of O₂ (g) in solution and the partial pressure of O₂ above a 8.23 x 10^-4 M O2 (g) saturated aqueous solution at 0 ^oC are respectively:

A. 2.18 M and 3.78 x 10^-4 atm

B. 2.18 x 10^-3 M and 0.378 atm

C. 0.0446 M and 0.0179 atm

D. 44.6 M and 1.84 x 10^-5 atm

Answer 1

Given:

Solubility of O₂ at 0°C and 25°C = 48.9 mL per L of water

Assume O₂ to be an ideal gas, we can use the ideal gas law to calculate the number of moles of O₂ in 49 mL .

49 mL is 0.049 L or 0.049 x 10^-3 m³ as 1000 L is equal to 1 m³

P = 1 atm or 10⁵ Pa

n= ?

R= 8.314 J / K mol

T = 273.15 K

Number of moles of O₂ present = 0.00217 moles

Therefore molarity = Number of moles / 1 Litre of solution

Therefore, Molarity = 0.00217 / 1 = 2.17 x 10^-3 M

The henry's law of solubility states that the solubility of a gas in a liquid is directly proportional to the pressure applied.

Let the solubility be S and the pressure be P

Let the proportional constant be K and is called the henry's constant

When P = 1 atm

As Solubility is 0.0021 moles per liter at 0°C,

Therefore, the question asks to find the value of P at S = 8.23 x 10^-4

Therefore we get:

Therefore, The correct option is option (B)