Question

A mixture of CO and $C{{O}_{2}}$ is found to have a density of 1.5 g /L at ${{30}^{\circ }}C$ and 730 torr. What is the composition of the mixture?

Answer 1

We can solve this question by using the ideal gas equation formula, which is the relation between the pressure, volume, and temperature of the gas, and the formula is PV = RT. Since the volume of the mixture is not given, use the relation between density, molecular mass, and volume. By calculating the molecular mass we can find the composition of carbon monoxide and carbon dioxide.

Complete answer:
We are given a mixture of gas in which two gases are there, i.e., carbon monoxide (CO) and carbon dioxide ($C{{O}_{2}}$), and the temperature of the mixture is ${{30}^{\circ }}C$at pressure 730 torr.
We can solve this question by using the ideal gas equation formula, which is the relation between the pressure, volume, and temperature of the gas, and the formula is:
PV = RT
R is gas constant.
Since the volume of the mixture is not given, use the relation between density, molecular mass, and volume. The volume can be written as:
$V=\dfrac{M}{d}$
The ideal gas formula will be:
$P\dfrac{M}{d}=RT$
We can find the molecular mass of the mixture by putting all the values:
$M=\dfrac{dRT}{P}$
$M=\dfrac{1.5\text{ x 0}\text{.0821 x 303 }}{\dfrac{730}{760}}$
$M=38.85$
Now let us assume that x moles of CO are present and (1 – x) moles of $C{{O}_{2}}$ is present. We know the molecular mass of carbon monoxide is 28 and the molecular mass of carbon dioxide is 44. We can write:
$x\text{ x 28 + (1 - }x)\text{ x 44 = 38}\text{.85}$
$x=0.3218$
Hence, percentage of carbon monoxide present is = 0.3218 x 100 = 32.18%
Percentage of carbon dioxide present is = 1 - 0.3218 = 0.6782 x 100 = 67.82%

Note:
While solving this question, we have to take all the quantities in the standard, i.e., the temperature should be converted into K by adding the given temperature with 273, and the pressure should be divided by 760 torrs.