Question

The number of gram-molecules of oxygen in1L of air which contains 21% oxygen by volume under stanndard conditions is:

Answer 1

21% oxygen means 21 ml oxygen in 100 ml of solution (air)
Therefore, volume of oxygen in 100 ml (IL) solution = (21/100) x 1000
= 210 ml

22400 ml is occupied by 1 mole of a gas at standard conditions
210 ml will be occupied by 210/22400 = 0.009 moles of oxygen gas

Therefore,
Number of gram-molecules of oxygen = 0.009 ( 1mole = 1 gm molecule)