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Under an O2(g) pressure of 1.00 atm, 28.31 mL of O2(g) dissolves in 1.00 L H2O at 25 °C. What will be the molarity of O2 in the saturated solution at 25 °C when the O2 pressure is 4.47 atm? Assume that the solution volume remains at 1.00 L.

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In the first case, the molarity of oxygen gas is 0.0012 mol/L. In the second case, the molarity of oxygen gas is 0.0052 mol/L .

We have to obtain the number of moles of oxygen as follows;

P = 1.00 atm

V = 28.31 mL or 0.02831 L

n = ?

R = 0.082 atm LK-1mol-1

T = 25 °C + 273 = 298 K

n = PV/RT

n=  1.00 atm × 0.02831 L/ 0.082 atm LK-1mol-1 ×  298 K

n = 0.0012 moles

Molarity = Number of moles/Volume = 0.0012 moles/1.00 L =  0.0012 mol/L

For the second case;

P = 4.47 atm

V = 28.31 mL or 0.02831 L

n = ?

R =  0.082 atm LK-1mol-1

T = 25 °C + 273 = 298 K

n = PV/RT

n = 4.47 atm ×  0.02831 L/0.082 atm LK-1mol-1 × 298 K

n = 0.1265/24.436

n = 0.0052 moles

Molarity of oxygen = 0.0052 moles/1.00 L = 0.0052 mol/L

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