Question

how many ions na+ and cl- present in 30 g

Answer 1

Answer:

1.025×NA

Mole of 30gm NaCl is=30/58.5=0.512

Total molecule in 0.512mole of NaCl is = 0.512×NA

ion of Na and Cl in one molecule of NaCl = 2

So total no. of ion = 0.512×NA×2=1.025NA