Question

Compute the number of ions present in 5.85 g of sodium chloride

Answer 1

5.85 g of NaCl = 5.85/ 58.5= 0.1moles  

or 0.1 moles of NaCl particle 

Each NaCl particle is equivalent to one Na+ one Cl^– 

⇒ 2 ions 

Total moles of ions = 0.1 × 2 

                                0.2 moles 

No. of ions= 0.2 × 6.022 ×1023 

⇒   1.2042 X10²³ ions