What is the millimolar solubility of oxygen gas,O2, in water at 20 degree celsius, if the pressure of oxygen is 1.00atm?
15 years ago
15 years ago
answer is 1.4mM
part B answer is 0.29mM
13 years ago
What is the millimolar solubility of oxygen gas, , in water at 12 degrees celcius , if the pressure of oxygen is 1.00 ?
8 years ago
I am assuming that you are using mastering chemistry. If so you should have a graph that displays the answer. You just have to look at the graph follow 12 degrees C on the graph to where it intersects with the line and your answer would be the y value. In this case the value is 1.6.
Answer: 1.6mM
1 year ago
To calculate the millimolar solubility of oxygen gas in water at a specific temperature and pressure, we need to use the Henry's Law equation, which relates the solubility of a gas in a liquid to its partial pressure.
The Henry's Law equation is given by:
C = k * P
Where:
C is the concentration of the gas in the liquid (in this case, oxygen solubility in water)
k is the Henry's Law constant for the specific gas and liquid
P is the partial pressure of the gas
In this case, you are given the pressure of oxygen as 1.00 atm. The Henry's Law constant for oxygen in water at 20 degrees Celsius is approximately 769.23 mL/atm.
To convert from mL/atm to millimolar solubility, we can use the fact that 1 mole of ideal gas at standard temperature and pressure (STP) occupies 22.414 L (or 22,414 mL). Therefore, the conversion factor is:
1 millimole = 22.414 mL
Now, let's calculate the millimolar solubility:
C = k * P
C = 769.23 mL/atm * 1.00 atm
C = 769.23 mL
Now, convert mL to millimoles:
C = 769.23 mL * (1 millimole / 22.414 mL)
C ≈ 34.34 millimolar solubility
Therefore, the millimolar solubility of oxygen gas in water at 20 degrees Celsius, with an oxygen pressure of 1.00 atm, is approximately 34.34 millimolar.