Question

14. How many atoms are present in:
(a) 0.1 mol of sulphur
(b) 18.g of water (H20)

(c) 0.44 g of carbon dioxide (CO2)​

Answer 1

Answer:

a) 8atoms

b)6.023*10^23atoms

c)0.44*6.023*10^23

Answer 2

Answer:

Explanation:

(a)0.1 mol of sulpher

= mol x No

=0.1 x 6.022x10^23

=0.6022 x 10^23

• Now we have to multiple the result 8 times .cause , sulpher is S8

0.6022 x 10^23 x 8

=4.82 x 10^23

b) 18g of water (H2O)

18 g of water = 1 mol (cause every atom weights 6.022 x 10^23)

18g of water = 1mole

1 mol = 6.022x10^23