Question

(16) The no. of atoms present in 4.25 g of NH3
(a) 3.011x1023 (b) 1.505 x 1023
(c) 6.022x1023 (d) 1.505 x 1019​

Answer 1

Answer:

Complete step by step answer:

Therefore, 4.25g of ammonia = 6.023×102317×4.25= 1.505×1023 molecules of ammonia. Now each molecule of ammonia has four atoms, one atom of nitrogen and three atoms of hydrogen. Hence, the total number of atoms in 1.505×1023 molecules of ammonia = 4×1.505×1023= 6.023×1023 atoms.