1 liter of a mixture of carbon monoxide and carbon dioxide has a mass of 1.43. Determine the composition in volume fractions?

Given:
V mixture (CO2, CO) = 1 l
m mixture (CO2, CO) = 1.43 g

To find:
φ (CO2) = -?
φ (CO) = -?

Decision:
1) n mixture (CO2, CO) = V mixture (CO2, CO) / Vm = 1 / 22.4 = 0.045 mol;
2) Let n (CO2) = (x) mol, then n (CO) = (0.045 – x) mol;
3) m (CO2) = n (CO2) * M (CO2) = x * 44 = (44x) g;
4) m (CO) = n (CO) * M (CO) = (0.045 – x) * 28 = (1.26 – 28x) g;
5) m mixture (CO2, CO) = m (CO2) + m (CO);
1.43 = 44x + 1.26 – 28x;
x = 0.011;
n (CO2) = x = 0.011 mol;
6) V (CO2) = n (CO2) * Vm = 0.011 * 22.4 = 0.246 l;
7) φ (CO2) = V (CO2) * 100% / V mixture (CO2, CO) = 0.246 * 100% / 1 = 24.6%;
8) φ (CO) = 100% – φ (CO2) = 100% – 24.6% = 75.4%.

Answer: The volume fraction of CO2 is 24.6%; CO – 75.4%.