How many $N a^{+}$ ions are present in 50 mL of a 0.5 M solution of $N a C l$ ?

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Solution

Number of moles of $N a C l = \frac{M V}{1000}$
$= \frac{0.5 \times 50}{1000} = 0.025$
$N a C l \to N a^{+} + C l^{-}$
Number of moles of $N a^{+}$ = Number of moles of NaCl
$= 0.025$
Number of ions of $N a^{+} = 0.025 \times 6.023 \times 10^{23}$
$= 1.505 \times 10^{22}$

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