Question

22. If 3.01x10^20 molecules are removed from 98mg of H2SO4 then the number of moles of H2SO4 left are

Answer 1

Answer:

Explanation:In case 3.01 ,*10^20 atoms are expelled from 98 mg of H2So4 then the quantity of moles of H2SO4 left are Molecular mass of H2SO4=2+32+64=98g consequently 98 g of sulphuric corrosive will contain 6.02*10^23 particles 6.02*10^23=1mole 3.01*10^23=1/2mole 0.5 moles of H2SO4 are leff.

Sub-atomic mass H=2. S=32. O=64 Moles= mass/sub-atomic mass = 9.8/2+32+64 = 9.8/98 = 98/98*10 = 0.1 Therefore 0.1 moles are in 9.8 gm of H2SO4.

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