Question

Find that mass of Na2CO3 which contains as many Oxygen atoms as present in 3.6 gm C6H12O6 (Atomic masses are :Na = 23, O = 16, H = 1, C = 12)

Answer 1

Answer:

we need to find out no . of O atoms in C6H12O6

No.of atoms=no. of molecules × atomicity

=moles of C6H12O6 × avagadro const. × atomicity ( no. of times the atom appears )

=3.6/180×(6×10^23)×6

=36/1800×(6×10^23)×6

=2/100×(6×10^23)×6

=72×10^21

now we need to find weight of Na2CO3 that contains 72×10^21 O atoms

so we use the same formula ,..let w be weight that is to be found

no.of atoms=no. of moles × avagadro const. × atomicity

72×10^21=(w/23×2+12+16×3) × 6×10^23 × 3

72×10^21/3×6×10^23=(w/46+12+48)

4/100=w/106

4×106/100=w

w=4×106/100

w=424/100

w=4.24 gm