Compare the number of ions present in

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.

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