Question

A mixture contains 0.2 g of oxygen gas and 0.4 g of
carbon dioxide gas. The ratio of oxygen atoms present
in oxygen gas to that in carbon dioxide gas is
(1) 3 : 16 (2) 11 : 8
(3) 3 : 4 (4) 11 : 16

Answer 1

ans 4
n(O2)=0.2/32=1/160
n(CO2)=0.4/44=1/110
Now
n(O2)/n(CO2)=no.of oxygen atoms/no.of CO2 atoms
N O2/N CO2=1/160/1/110=110/160=11:16

Answer 2

answer : option (4) 11 : 16

explanation : mass of oxygen gas = 0.2g

number of mole of Oxygen gas = mass of oxygen gas/molar mass of oxygen gas

= 0.2/32 = 1/160

so, number of mole of oxygen atoms = 2 × number of mole of oxygen gas

[ as you know, in O2 molecule , there are two oxygen atoms ]

= 2 × 1/160 = 1/80

similarly, mass of carbon dioxide = 0.4

number of mole of carbon dioxide = mass of carbon dioxide/molar mass of carbon dioxide

= 0.4/44 = 1/110

so, number of mole of oxygen atoms in carbon dioxide = 2 × 1/110 = 1/55

[ as you know, there are two oxygen atoms present in one molecule of CO2 ]

now, the ratio of oxygen atoms present in oxygen gas to that in carbon dioxide gas is (1/80)/(1/55) = 55/80 = 11/16

hence, option (4) is correct