Experts pls.explain.following quest Q-Calculate the weight of carbon monoxide having same number of oxygen atoms as are present in 88 g of carbon dioxide.
Dear student,
Molar mass of CO2 is 44 g/mol
Therefore the number of moles of CO2 present in 88g = mass/molar mass of CO2
= 88 / 44 = 2 moles
As there are 6.023 X 1023 molecules present in one mole of a compound, therefore 2 moles of CO2 will contain 2 X 6.023 X 1023 molecules = 12.046 X 1023 molecules of CO2.
Now 1 molecule of CO2 contains 2 atoms of oxygen. Therefore 12.046 X 1023 molecules of CO2 will contain 2 X 12.046 X 1023 atoms of oxygen. Thus
12.046 X 1023 molecules of CO2 = 2.4092 X 1024 atoms of oxygen
So 2.4092 X 1024 atoms of oxygen are present in 88g of carbon dioxide
Now the molar mass of CO = 28 g/mole
There are 6.023 X 1023 molecules of CO present in 28 g of CO. As one molecule of CO contains 1 atom of oxygen, therefore there are 6.023 X 1023 atoms of oxygen present in 6.023 X 1023 molecules of CO.
So
6.023 X 1023 atoms of Oxygen → 28 g of Oxygen
1 atom of oxygen → 28 / [ 6.023 X 1023] g
therefore 2.4092 X 1024 atoms of oxygen → 28 X [ 2.4092 X 1024] / [ 6.023 X 1023] g
= 112 g of CO
Thus 112 g of CO will have the same number of oxygen atoms as present in 88g of CO2
Regards