We’re being asked to calculate for the volume equivalent at STP for O_{2} coming out of water warming from 20°C to 50°C. The water is initially saturated with air at 1.00 atm.

*Recall that saturated means that the maximum amount of air is dissolved in water (or solubility).*

The solubility of a gas is given by ** Henry’s law**:

$\overline{){{\mathbf{S}}}_{{\mathbf{gas}}}{\mathbf{=}}{{\mathbf{k}}}_{{\mathbf{H}}}{\mathbf{\xb7}}{{\mathbf{P}}}_{{\mathbf{gas}}}}$

where:**S _{gas}** = solubility of the gas (in mol/L or M)

We’re going to calculate the volume of O_{2} gas using the following steps:

Step 1: Calculate the partial pressure of O_{2} gas at 20°C.

Step 2: Calculate the solubility of O_{2} gas at 20°C.

Step 3: Calculate the solubility of O_{2} gas at 50°C.

Step 4: Calculate the moles of O_{2} gas that comes out of water.

Step 5: Calculate the volume of O_{2} gas that comes out of the water at STP using the ideal gas equation.

Calculate the volume equivalent at STP for O_{2} coming out of 2.5 L of water warming from 20°C to 50°C. The water is initially saturated with air at 1.00 atm. The solubility of O_{2}(g) at 50°C is 27.8mg/L at 1.00 atm. The solubility of N_{2}(g) at 50°C is 14.6mg/L at 1.00 atm. Henry’s Law constant for O_{2} at 20°C is 1.4 x 10^{-3} M/atm and for N_{2} is 6.3 x 10^{-4} M/atm. Air is 80% nitrogen and the rest oxygen.

A. 3.41 mL

B. 4.56 mL

C. 7.29 mL

D. 5.96 mL

E. 2.41 mL

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