How many atoms of oxygen and hydrogen are present in 0.15 mole of water ?

Explanation

`underset(("1 mol"))(H_(2)O)underset(("2 gram atoms"))(-=2H)+underset(("1 gram atom"))(O)`
(i) No. of hydrogen atoms
1 mole of water `(H_(2)O)` has hydrogen (H) = 2 gram atoms
0.15 mole of water `(H_(2)O)` has hydrogen `(H)=((0.15"mol"))/(("1 mol"))xx("2 gram atoms")=0.30` gram atom
No. of hydrogen atoms present `= 0.30 xx 6.022 xx 10^(23) = 1.81 xx 10^(23)` atoms
(ii) No. of oxygen atoms
1 mole of water `(H_(2)O)` has oxygen (O) = 1 gram atom
0.15 mole of water `(H_(2)O)` has oxygen `(O) = (("0.15 mol"))/(("1 mol"))xx("1 gram atom")=0.15` gram atom
No. of oxygen atoms present `= 0.15 xx6.022xx10^(23)=9.03xx10^(22)` atoms.