Question:
Compare the number of ions present in $5.85 \mathrm{~g}$ of sodium chloride.
Solution:
Gram formula mass of $\mathrm{NaCl}=23+35.5=58.5 \mathrm{~g}$
$58.5 \mathrm{~g}$ of $\mathrm{NaCl}$ have ions $=2 \times \mathrm{N}_{\mathrm{A}}$
$5.85 \mathrm{~g}$ of $\mathrm{NaCl}$ have ions $=\frac{2 \times \mathrm{N}_{\mathrm{A}} \times(5.85 \mathrm{~g})}{(58.5 \mathrm{~g})}$
$=\frac{2 \times 6.022 \times 10^{23}}{10}=1.2042 \times 10^{23}$ ions.