Question

9. The number of molecules in 4.25 g of ammonia is
approximately​

Answer 1

Answer:

1.505 × 10^23

Explanation:

No.of molecules = Given Wt. / Gram Molecular Weight × Avogadro Number

= 4.25 / 14 + 3 × 6.023 × 10^23

= 4.25/17 × 6.023 × 10^23

= 0.25 × 6.023 × 10^23

= 1.505 × 10^23

Answer 2

Answer:

0.25 mole

Explanation:

molecular weight of Ammonia = 17 g

17 g contais 1 mole of Ammonia

so 4.25 g contais (4.25÷17)= 0.25 mole