Question

A mixture of CO and CO2 is found to have a density of 1.5gL at 303Kelvin and 730torr.What is the composition of the mixture?

Answer 1

When we find a mixture of CO2 and CO then d equals 1.50 litres or gram. P equals 730/760 atm and therefore T is 303K.

730 / 760 equals 1.5/ m which when multiplied by 0.0821 × 303 results in 38.35 which is molecular weight of the mixture of CO2 and CO.

Answer 2

Answer: The composition of CO is 32.68% and $CO_{2}$ is 67.32% in the mixture.

Explanation:

Molecular mass of CO = (12.01 + 16) g/mol = 28.01 g/mol

Molecular mass of $CO_{2}$ = $(12.01 + 16 \times 2) g/mol$ = 44.01 g/mol

Let the mixture contains x% of CO and (1 - x)% of $CO_{2}$.

The average molar mass can be calulated by adding the percentages of both CO and $CO_{2}$ as follows.

Average molar mass = $(28.01 \times x\%+44.01 \times (1 -x)\%) g/mol$

                                  = (44.01 - 0.16 x) g/mol

Now, the data given is as follows.

P = 730 torr,  d = 1.5 g/L and  T = 303 K.

It is known that 760 torr = 1 atm. Therefore, converting 730 torr into atm as follows.

                   730 torr = $\frac{1}{760} \times 730$

                                 = 0.961 atm

Using ideal gas equation, the formula derived for calculation is as follows.

                 Molar mass = $\frac{dRT}{P}$

   (44.01 - 0.16 x) = $\frac{1.5 g/L \times (0.082 atmL/mol K) \times 303K}{0.961 atm}$

      (44.01 - 0.16 x) = 38.781

                 -0.16 x = 38.78 - 44.01

                  0.16 x = 5.22

                          x = 32.68 g/mol

                 (100 - x) = 100 - 32.68 = 67.32

Therefore, the composition of CO is 32.68% and $CO_{2}$ is 67.32% in the mixture.