A 600.0 ml sample of pure water is allowed to come to equilibrium with pure oxyg

Question

A 600.0 ml sample of pure water is allowed to come to equilibrium with pure oxygen gas at a pressure of 920 mmHg.

            What mass of oxygen gas dissolves in the water? (The Henry's law constant for oxygen gas is 1.3 A 600.0 ml sample of pure water is allowed to come to equilibrium with pure oxygen gas at a pressure of 920 mmHg.

A 600.0 ml sample of pure water is allowed to come to equilibrium with pure oxygen gas at a pressure of 920 mmHg.

A 600.0 ml sample of pure water is allowed to come to equilibrium with pure oxygen gas at a pressure of 920 mmHg.

            What mass of oxygen gas dissolves in the water? (The Henry's law constant for oxygen gas is 1.3

Explanation / Answer

First you have to find Molarity (M).

(1.3*10^-3M/atm)(920mmhg/760mmhg)=.00157M

Molarity=amount solute (in mol)/volume solution (in L)

.00157M = mol/.6L

.6L*.00157M = 9.42*10^-4 mol O2

to find grams of O2 multiply by the molar mass of O2

(9.42*10^-4 mol O2)(32g O2/1molO2)= .030gO2

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