Question

if 3.01 into 10 raise to power 20 molecules are removed from 98 milligram of H2 S o4 when the number of moles in H2 S o4 left are​

Answer 1

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.