Question

A mixture of CO and CO2 is found to have density of 1.50 g L^-1 at 20^0C & 740 mm pressure. Calculate the composition of the mixture.

Answer 1

Answer : The composition of CO and $CO_2$ is 43.38 and 56.62 respectively.

Solution : Given,

Density = 1.50 g/L

Temperature = $20^oC=273+20=293K$         $(1^oC=273K)$

Pressure = 740 mmHg = $\frac{740}{760}=0.9737atm$        (1 atm = 760 mmHg)

Molar mass of CO = 28 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the average molecular mass of the given mixture.

$M=\frac{\rho RT}{P}$

where,

M = average molecular mass

$\rho$ = density

R = Gas constant = 0.0821 Latm/mole/K

T = temperature

P = pressure

Now put all the given values in above formula, we get the average molecular mass of mixture.

$M=\frac{(1.50g/L)\times (0.0821Latm/mole/K)\times (293K)}{0.9737atm}=37.06g/mole$

Now we have to calculate the composition of given mixture.

Let the mole % of CO in mixture = x

and the mole %  of $CO_2$ = (100 - x)

$\text{ Average molar mass}=\frac{[(x)\times 28]+[(100-x)\times 44]}{100}=37.06$

By rearranging the term, we get the value of 'x'

x = 43.38

Therefore, the composition of CO = x = 43.38

and the composition of $CO_2$ = (100 - x) = (100 - 43.38) = 56.62