if the components of air are N2-78% ;O2-21%; Ar-0.9% and Co2-0.1% by volume (or mole) what would be the molecular weight of air?

Good and a little bit difficult

Plug in the mol weights of the compounds in 9.78×14(N2)+0.21×16(O2)+0.009×40(Ar) +0.001×(12+16 ×2)(co2)

To calculate the molecular weight of air, you need to consider the molecular weights of each component gas and their respective percentages.

Let's start by considering the mole percentages given in the question:
- N2: 78% by volume (or mole)
- O2: 21% by volume (or mole)
- Ar: 0.9% by volume (or mole)
- CO2: 0.1% by volume (or mole)

Now, we can calculate the molecular weights of each gas component:
- N2 (nitrogen gas): molecular weight = 28 g/mol
- O2 (oxygen gas): molecular weight = 32 g/mol
- Ar (argon gas): molecular weight = 40 g/mol
- CO2 (carbon dioxide gas): molecular weight = 44 g/mol

Next, we multiply the mole percentages by the respective molecular weights of each component:
Molecular weight of N2 = (78/100) * 28 = 21.84 g/mol
Molecular weight of O2 = (21/100) * 32 = 6.72 g/mol
Molecular weight of Ar = (0.9/100) * 40 = 0.36 g/mol
Molecular weight of CO2 = (0.1/100) * 44 = 0.044 g/mol

Finally, we add up the molecular weights of all the components to find the molecular weight of air:
Molecular weight of air = Molecular weight of N2 + Molecular weight of O2 + Molecular weight of Ar + Molecular weight of CO2
= 21.84 g/mol + 6.72 g/mol + 0.36 g/mol + 0.044 g/mol
= 29.964 g/mol

Therefore, the molecular weight of air is approximately 29.964 g/mol.

just plug in the mol wts of the compounds in

0.78*N2 + 0.21*O2 + 0.009*Ar * 0.001*CO2