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Question: A sample of \({O_2}\) gas is collected over water at \({23^ \circ }C\) at a barometric pressure of \...

A sample of O2{O_2} gas is collected over water at 23∘C{23^ \circ }C at a barometric pressure of 751 mm Hg751\,mm\,Hg (vapor pressure of water at 23∘C{23^ \circ }C is 21 mm Hg21\,mm\,Hg). The partial pressure of O2{O_2} gas in the sample collected is:
A. 21 mm Hg21\,mm\,Hg
B. 751 mm Hg751\,mm\,Hg
C. 0.96 atm0.96\,atm
D. 1.02 atm1.02\,atm

Explanation

Solution

Dalton’s law of partial pressure states that total pressure of a mixture of gases is equal to the sum of the partial pressure of the individual gases in the mixture. Ideally the ratio of partial pressures is equal to the ratio of numbers of molecules.

Complete answer:
Dalton’s law expresses the fact that the total pressure of the mixture of gases is equal to the sum of the partial pressure of gases present in that mixture. This law is applicable to ideal gas, as in ideal gas conditions gas molecules are far apart to even interact with each other.
P=PN2+PH2+PNH3P = {P_{{N_2}}} + {P_{{H_2}}} + {P_{N{H_3}}}.
Here P is total pressure, and PN2{P_{{N_2}}} is partial pressure of nitrogen gas, PH2{P_{{H_2}}} is partial pressure of hydrogen gas and PNH3{P_{N{H_3}}} is partial pressure of ammonia gas in a container where ammonia gas is formed from nitrogen and hydrogen gas.
Here in the given question,
Total pressure is 751 mm Hg751\,mm\,Hg at 23∘C{23^ \circ }C
xi=pip=nin{x_i} = \dfrac{{{p_i}}}{p} = \dfrac{{{n_i}}}{n}
Partial pressure of water given as 21 mm Hg21\,mm\,Hg at 23∘C{23^ \circ }C and we have to calculate the partial pressure of oxygen gas. Applying Dalton’s law of partial pressure, we have
Ptotal=Poxygen+Pwater{P_{total}} = {P_{oxygen}} + {P_{water}}
By putting the values, we get 751 mm Hg=Poxygen+21 mm Hg751\,mm\,Hg = {P_{oxygen}} + 21\,mm\,Hg
After calculation we get
Poxygen=751 mm Hg−21 mm Hg Poxygen=730 mm Hg {P_{oxygen}} = 751\,mm\,Hg - 21\,mm\,Hg \\\ {P_{oxygen}} = 730\,mm\,Hg
By converting partial pressure of oxygen to atmospheric pressure, 730 mm Hg760 mm Hg×1 atm=0.96 atm\dfrac{{730\,mm\,Hg}}{{760\,mm\,Hg}} \times 1\,atm = 0.96\,atm

**Hence the correct option will be C.

Note:**
The ratio of partial pressure equals the ratio of the number of molecules. That means, the mole fraction of an individual gas component in an ideal gas mixture can be expressed in terms of the component's partial pressure or moles of the component.
xi=pip=nin{x_i} = \dfrac{{{p_i}}}{p} = \dfrac{{{n_i}}}{n}