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

Hint: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.