Question

A cylinder was filled with a gaseous mixture containing $CO$ and ${N_2}$(equal masses). The ratio of their partial pressure in the cylinder are:
A. 1:1
B. 1:2
C. 2:1
D. 1:3

Answer 1

In this question we will calculate the molar masses of the $CO$ and ${N_2}$. After that we will calculate the mole fraction of each gas in the cylinder. Now after this we will calculate the partial pressure of both gases. Partial pressure is the individual pressure of each gas and equal to mole fraction multiplied by total pressure.

Complete answer:
Molar mass of the C atom is 12 g.
Molar mass of an oxygen atom is 16 g.
So molar mass of CO will be 28 g
Now the molar mass of nitrogen atom is 14 g.
So molar mass of ${N_2}$ molecule is 28 g
So we have concluded that both gases have the same masses in the cylinder.
Mole fraction of CO will be$\dfrac{1}{2}$(because both gases have same mass
Similarly mole fraction of ${N_2}$ molecule will be $\dfrac{1}{2}$
$\therefore$ We all know that $Partial \,pressure = Mole\, fraction \times Total \,Pressure$
So now we will calculate the ratio of partial pressure of both gas
$\dfrac{{{P_{CO}}}}{{{P_{{N_2}}}}} = \dfrac{{{X_{CO}} \times P}}{{{X_{{N_2}}} \times P}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2}}} = 1:1$
Now we have seen that if the cylinder contains the same mass of CO and ${N_2}$ gas then the ratio of partial pressure of both gases will be 1:1.

**Therefore, the answer will be option number A.

Note:**
Mole fraction: mole fraction is defined as the ratio of number of moles of one component with the total number of moles in the constituent mixture.
Mathematically,
$Mole \,fraction=\dfrac{{mole{\text{ of one component}}}}{{{\text{number of the moles in the mixture}}}}$
As we have seen in the question both gases have the same amount so each gas will constitute 0.5 and 0.5 mole fraction.