Question

A special gas mixture used in bacterial growth
chamber contains 1% by weight of Co, and 99%
of O2. What is the partial pressure (in atm) of each
gas at a total pressure of 0.977 atm?
(1) Po, = 0.970 atm; Pco, = 0.1 atm
(2) Po, = 0.970 atm; Pco, = 0.00711 atm
(3) Po, = 0.1atm; Pco, = 0.9 atm
(4) Can't predict
​

Answer 1

Answer : The correct option is, (2)

Explanation :

As per question,

Mass of $CO_2$ = 1 g

Mass of $O_2$ = 99 g

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $CO_2$ and $O_2$.

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}=\frac{1g}{44g/mole}=0.023mole$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{99g}{32g/mole}=3.094mole$

Now we have to calculate the mole fraction of $CO_2$ and $O_2$.

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }O_2}=\frac{0.023}{0.023+3.094}=0.00738$

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }CO_2+\text{Moles of }O_2}=\frac{3.094}{0.023+3.094}=0.993$

Now we have to partial pressure of $CO_2$ and $O_2$.

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of gas

$p_T$ = total pressure of gas

$X$ = mole fraction of gas

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.00738\times 0.977atm=0.00711atm$

and,

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.993\times 0.977atm=0.970atm$

Therefore, the partial pressure of $CO_2$ and $O_2$ are, 0.00711 atm and 0.970 atm respectively.