Question

A mixture of carbon monoxide and carbon dioxide is found to have a density of $1.7{\text{ }}g/l$ at S.T.P. The mole fraction of carbon monoxide is:
A. $0.37$
B. $0.40$
C. $0.30$
D. $0.50$

Answer 1

In this type of question first we calculate the Molar mass of the gas mixture in terms of density by using the formulae - $\left[ {M = \dfrac{{\rho RT}}{P}} \right]$
where, M= Molar mass of mixture,
$\rho$ = density of gas mixture
$R$ = gas constant
$T$ = temperature
$P$ = Pressure
Then we assume the mole fraction of $CO = \left( x \right)$ to determine its the Mole fraction by average molar mass of a mixture of gases

Complete answer:
To find the molar mass of a mixture of gases, you need to take into account the molar mass of each gas in the mixture, as well as their relative proportion. So the average molar mass of a mixture of gases is equal to the sum of the mole fractions of each gas, multiplied by their respective molar masses:
$M = {\text{ }}\sum N{\text{ }} \times {\text{ Mi}}$
where, $N$= mole fraction and $Mi$= Molar mass of gas.
Here given,
Density of a mixture of gas = $1.7{\text{ }}g/l$
At S.T.P
Temperature = $273K$
Pressure = $1{\text{ }}atm$
R= $0.082{\text{ }}L/atm/mole/K$
Now first we calculate the Molar mass of the gas mixture $\left( {CO{\text{ }} + {\text{ }}C{O_2}} \right)$ in terms of density,
The formula used for molar mass of a mixture of gas in term of density is,
$\left[ {M = \dfrac{{\rho RT}}{P}} \right]$ -------- formulae (1)
where M= Molar mass of mixture,
Now put all the given values in above formula, we get
$\Rightarrow M = \dfrac{{(1.7{\text{ }}g/l){\text{ }} \times {\text{ (}}0.082{\text{ }}L{\text{ }}atm/mole/K){\text{ }} \times (273K)}}{{1atm\;\;}}{\text{ }}$
$\Rightarrow M =$ $38.102{\text{ }}g/mole$
Now let us assume the mole fraction of$CO = x$
So the mole fraction of $C{O_2}\; = {\text{ }}\left( {1 - x} \right).$
By using the formula of Average molar mass of a mixture is to determine the value of $x$
$\Rightarrow M = {\text{ }}\sum N{\text{ }} \times {\rm{ M_i}}$ ........... (2)
Given, Molar mass of $CO{\text{ }} = {\text{ (12 + 16) = }}28{\text{ }}g/mole$
Molar mass of $C{O_2}\; = (12 + 2 \times 16) = 44{\text{ }}g/mole$
Now put all the given values in above formula (2), we get
$M = {\text{ }}\sum N{\text{ }} \times {\rm{ M_i}}$
$\Rightarrow 38.102 = 28x{\text{ }} + {\text{ }}44{\text{ }}\left( {1 - x{\text{ }}} \right)$
By rearranging the terms, we get the value of ($x$)
$\Rightarrow x = 0.368$
Hence, the mole fraction of CO = $x = 0.368$ $\approx 0.37$

**So the option (A) is correct.

Note:**
Molar mass $\left( M \right)$ is equal to the mass of one mole of a particular element or compound; as such, molar masses are expressed in units of grams per mole $\left( {g{\text{ }}mo{l^{-1}}} \right)$ and are often referred to as molecular weights.