Question

How many atoms of each element are present in 4.9 gram of sulphuric acid??

Answer 1

No.of atom =given mass (4.9)/molar mass (HNo{subscript}3)×No(avagadro no.)
=4.9/1+14+16×3×6.022×10^23
=4.9/48+15×avagadro no.
=4.9/63×6.022×10^3