Question

caluclate the no of atoms of carbon, oxygen , hydrogen in 0.05 mols of glugose

Answer 1

Answer:

C=0,025

O=0,025

Explanation:

glucose  C6H12O6 so

C=0,025

O=0,025

Answer 2

Answer:

Molecular formula =C 6 H 12 O 6

Molar mass =6×12+12×1+16×6=180gmol

−1

Moles in 18g of sample=

180

18

=0.1mole

So,moles of carbon present in sample=0.1×6=0.6

Moles of hydrogen=0.1×12=1.2

Moles of oxygen=6×0.1=0.6

No. of atoms present in carbon=0.6×6.022×10

23

=3.6×10

23

No. of atoms present in hydrogen=1.2×6.022×10

23

=7.2×10

23

No. of atoms present in oxygen=0.6×6.022×10

23

=3.6×10

23