Atmospheric pressure is 1.00 atm, but air is only composed of 21.0% (by moles) oxygen. thus, the partial pressure of oxygen in air is less than 1.00 atm. what is the millimolar concentration of dissolved oxygen for a lake or stream under the partial pressure of oxygen and a temperature of 20 ∘c?

8.74 millimolar.

Explanation:

The partial pressure of oxygen = (21.0 / 100) (1.0 atm) = 0.21 atm.Using the ideal gas law: PV = nRT, Where, P is the pressure in atm (P = 0.21 atm), V is the volume in L, n is the number of moles, R is the general gas constant (R = 0.082 l.atm/mo.K), and T is the temperature in K (T = 20 °C + 273 = 293 K). We can get the concentration in molar (n/V), The number of moles per 1.0 L of the solvent.The concentration of oxygen in molar (n/V) = P/RT = (0.21 atm) / (0.082 L.atm/mol.K) (293 K) = 8.74 x 10³⁻ Molar = 8.74 millimolar.

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