A sample of O2 gas is collected over water at 23C at a barometric pressure of 751 mmHg (vapor pressure of water at 23C is 21 mmHg). The partial pressure of O2 gas in the sample collected is:
(a) 21 mmHg
(b) 751 mm Hg
(c) 0.96 atm; or
(d) 1.02 atm
The answer is (c). Not sure how the problem is worked out. thank you.
I figured it out. P total=751 mmHg. Partial pressure O2 + partial pressure H2O=751 mmHg. partial pressure O2=730 mmHg and 1atm/760 mmHg=x atm/730 mmHg. answer=0.96 atm! It was so easy I missed it....
To solve this problem, we need to use Dalton's law of partial pressures, which states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each individual gas.
In this case, we have O2 gas collected over water, so we need to consider the partial pressure of the O2 gas and the vapor pressure of water.
Given:
Barometric pressure (total pressure) = 751 mmHg
Vapor pressure of water at 23°C = 21 mmHg
To find the partial pressure of O2 gas, we subtract the vapor pressure of water from the total pressure:
Partial pressure of O2 gas = Total pressure - Vapor pressure of water
Partial pressure of O2 gas = 751 mmHg - 21 mmHg
Partial pressure of O2 gas = 730 mmHg
To convert mmHg to atm, we divide by 760 mmHg (which is equal to 1 atm):
Partial pressure of O2 gas = 730 mmHg / 760 mmHg/atm
Partial pressure of O2 gas = 0.96 atm
Therefore, the partial pressure of O2 gas in the collected sample is approximately 0.96 atm. Hence, the correct answer is option (c).
To find the partial pressure of O2 gas in the sample collected, you need to subtract the vapor pressure of water from the barometric pressure.
Given:
Barometric pressure = 751 mmHg
Vapor pressure of water at 23°C = 21 mmHg
First, convert the vapor pressure of water to atm by dividing it by 760 (since 1 atm = 760 mmHg):
Vapor pressure of water = 21 mmHg / 760 mmHg/atm = 0.0276 atm
Next, subtract the vapor pressure of water from the barometric pressure to find the partial pressure of O2 gas:
Partial pressure of O2 gas = Barometric pressure - Vapor pressure of water
= 751 mmHg - 21 mmHg
= 730 mmHg
Finally, convert the partial pressure of O2 gas to atm by dividing it by 760:
Partial pressure of O2 gas = 730 mmHg / 760 mmHg/atm = 0.961 atm
So, the partial pressure of O2 gas in the sample collected is approximately 0.96 atm, which corresponds to option (c).