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:

Sgas=kH·Pgas

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