Calculate the number of moles present in 9 g of ethane
Calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 K and 3 atm pressure.

Explanation

Mass of ethane = 9 g
Molar mass of ethane `C_(2)H_(6)`= 30 g `mol^(-1)` .
No. of moles = `("Mass")/("Molar mass") = 9/30 =`0.3 mol.
Molar volume of oxygen = 22400 ml.
22400 ml of oxygen contains `6.023 xx 10^(23)` molecules.
`:.` 224 ml of oxygen contain `(6.023 xx10^(23))/222400 xx 224` at 1 atm.pressure
For 3 atm.pressure = `(60.23x10^(23))/100 = 6.023 xx 10^(21)` molecules `xx` 3
= `18.069 xx 10^(21)` molecules (or) `1.8069 xx 10^(22)` molecules