Question

Question 38
Compute the number of ions present in 5.85 g of sodium chloride.

Answer 1

Molar mass of NaCl = 23 + 35.5 = 58.5 g
Each NaCI particle is equivalent to one $N{a}^{+}$ one $C{l}^{-}$
Number of ions = 2
Number of ions in 5.85 g of $NaCl=\frac{2\times 6.022\times {10}^{23}\times 5.85}{58.5}=1.2042\times {10}^{23}$ ions