Question

a sample of pure air contains 80% N2 10% O2 co2 5% and 5% argon by moles the average molecular weight of sample is​

Answer 1

This % means 80 g H2 and 20 g O2 in 100 g Sol.

80% H2 means moles of H2= 80÷2=40 moles

20% O2 means moles of O2=20÷32=0.625 moles

Total moles in mix.= 40.625 moles

40.625 moles of have mass =100 g

so 1 mole has mass= molecular mass= 100÷40.625=2.46 g

Answer 2

Answer:

29.8 g/mol

Explanation:

As the percentage composition is given, we will simply multiply them with the molecular weight of respective gases and add them together. And then we will divide the result with hundred as we are taking percentages.

Step1: [80*(MW of N2) + 10*(MW of O2) + 5*(MW of CO2) + 5*(MW of Ar)]/100

Step2: [80*28 + 10*32 + 5*44 + 5*40]/100

Step3: [2240+ 320 + 220 + 200]/100

Step4: [2980]/100

So, average molecular weight of this sample of air will we 29.8 g/mol

Hope it helps!!!