# Problem: Calculate the volume equivalent at STP for O2 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 O2(g) at 50°C is 27.8mg/L at 1.00 atm. The solubility of N2(g) at 50°C is 14.6mg/L at 1.00 atm. Henry’s Law constant for O2 at 20°C is 1.4 x 10-3 M/atm and for N2 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

###### FREE Expert Solution

We’re being asked to calculate for the volume equivalent at STP for O2 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:

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

where:
Sgas = solubility of the gas (in mol/L or M)
KH = Henry’s law constant for the gas
Pgas = partial pressure of the gas

We’re going to calculate the volume of O2 gas using the following steps:

Step 1: Calculate the partial pressure of O2 gas at 20°C.
Step 2: Calculate the solubility of O gas at 20°C.
Step 3: Calculate the solubility of O2 gas at 50°C.
Step 4: Calculate the moles of O2 gas that comes out of water.
Step 5: Calculate the volume of O2 gas that comes out of the water at STP using the ideal gas equation.

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###### Problem Details

Calculate the volume equivalent at STP for O2 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 O2(g) at 50°C is 27.8mg/L at 1.00 atm. The solubility of N2(g) at 50°C is 14.6mg/L at 1.00 atm. Henry’s Law constant for O2 at 20°C is 1.4 x 10-3 M/atm and for N2 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