Question

is 3.01 into 10 raise to power 20 molecules are removed from 98 mg of H2 S o4 then the number of moles of H2 S o4 left

Answer 1

Answer:

Molecular mass of H2SO4=2+32+64=98gm

98 gm of sulphuric acid will contain 6.02*10^23 molecule.

1 gm = 1000 mg  

So 1 mg = 1/1000 gm

So 98 mg of sulphuric acid will contain 6.02*10^20 molecule

​Total molecule present initially equal to 6.02*10^20 molecule  

3.01*10^20 molecule removed so molecule remained = Total molecule - molecule removed

= 6.02*10^20 -3.01*10^20

= 3.01*10^20 molecule will remain  

6.02*10^23 molecule = 1 mole  

1 molecule = 1/(6.02*10^23) mole  

3.01*10^20 molecule = (3.01*10^20)/ (6.02*10^23)

= 1/2000 mole = .0005 mole  

0.0005 moles of H2SO4 are left

Answer : 0.0005 mole or 5*10^(-4) mole