Question

The average molar mass of a mixture of CO (g) and CO2 (g) is 37. The mole percentage of CO (g) in the mixture is approximately?

Answer 1

Answer:  The mole percentage of CO (g) in the mixture is approximately = 43.75 %

Explanation:

• Formula of average molecular mass = $\frac{(n1 * Molar mass 1) + (n2 * Molar mass 2)}{n1 +n2}$

          where, n1 = number of moles of specie 1

                      n2 = number of moles of specie 2

• Suppose n1 & n2 be the number of moles of CO & CO2 respectively.
• Molar mass of CO = 28 g/mol
• Molar mass of CO2 = 44 g/mol
• Applying formula:-

           37 =  [(n1  x  28 g/mol) + (n2  x  44 g/mol)]/(n1 +n2)

           On solving:-

                n1 = 7n2/9

• Now mole % = [n1/(n1+ n2)]  x 100

                              = [ (7n2/9)/(7n2/9  +  n2)]   x   100

                              = 43.75%