Question

A special gas mixture used in a bacterial growth chamber contains 1% by weight of $C{O_2}$ and 99% of ${O_2}$. What is the partial pressure (in atm) of each gas at a total pressure of 0.977 atm?
(A) ${P_{{O_2}}} = 0.970atm,{P_{C{O_2}}} = 0.1atm$
(B) ${P_{{O_2}}} = 0.977atm,{P_{C{O_2}}} = 0.00711atm$
(C) ${P_{{O_2}}} = 0.1atm,{P_{C{O_2}}} = 0.9atm$
(D) Can’t predict

Answer 1

Hint: The formula of dalton’s partial pressure is as below.
\[P = {p_{mix}}x{\text{ }}\]
P is the partial pressure of the gas and ${p_{mix}}$ is the pressure of the mixture and x is the mole fraction of the given gas in mixture.

Complete step by step solution:
We will first find the mole fraction of both the gases. Then we will use the law of Dalton to find the partial pressure of both the gases.
- Molecular weight of $C{O_2}$ = Atomic weight of C + 2(Atomic weight of O) = 12+ 2(16) = 44$gmmo{l^{ - 1}}$
- Molecular weight of ${O_2}$ = 2(Atomic weight O) = 2(16) = 32$gmmo{l^{ - 1}}$
Now, we assume that the mixture of gases has a weight of 100g. So, the weight of ${O_2}$ will be 99g and the weight of $C{O_2}$ will be 1g.
So, ${\text{Moles = }}\dfrac{{{\text{Weight}}}}{{{\text{Molecular weight}}}}$ …….(1)
Now, for oxygen gas, equation (1) becomes
\[{\text{Moles = }}\dfrac{{99}}{{32}} = 3.0937\]
For $C{O_2}$ , we can write equation (1) as
\[{\text{Moles = }}\dfrac{1}{{44}} = 0.02272\]
Now, we can find the mole fraction of both the gases in the mixture. We know that
\[{\text{Mole fraction = }}\dfrac{{{\text{Moles of gas}}}}{{{\text{Total moles in mixture}}}}{\text{ }}....{\text{(2)}}\]
For ${O_2}$ , we can write equation (2) as
\[{\text{Mole fraction = }}\dfrac{{3.0937}}{{3.0937 + 0.02272}} = \dfrac{{3.0937}}{{3.11642}} = 0.9927\]
For $C{O_2}$ , we can write equation (2) as
\[{\text{Mole fraction = }}\dfrac{{0.02272}}{{3.0937 + 0.02272}} = 0.0073\]
Now, we know Dalton's partial pressure law. It’s formula is given below.
\[P = {p_{mix}}x{\text{ }}...{\text{(3)}}\]
 Here, P is the partial pressure of the gas and ${p_{mix}}$ is the pressure of the mixture and x is the mole fraction of the given gas in mixture.
For ${O_2}$ , we can write equation (3) as
\[{P_{{O_2}}} = 0.977 \times 0.9927 = 0.9698atm\]
For $C{O_2}$, we can write equation (3) as
\[{P_{C{O_2}}} = 0.0073 \times 0.977 = 0.0071atm\]

Thus, we can conclude that the correct answer is (B).

Note: Remember that we need to multiply the total pressure of the mixture by the mole fraction of the gas. So, do not get confused with putting the number of moles here. We can also cross-check our answer by adding the partial pressures of both the gases which should be equal to the total pressure of the mixture.