How many moles of $N a^{+}$ ions are present in $275.0 m L$ of $0.35 M$ $N a_{3} P O_{4}$ solution ?

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Solution

The definition of concentration is
$C = \frac{n}{V}$
where concentration is is moles per litre (M), n is the number of moles (mol) and V is the volume in litres (L). Make sure to use these units to get the right answer i.e.,
$C = 0.35 M$
$L = 275.0 \cdot 10^{- 3} L$
Now rearrange the concentration formula to solve for the number of moles of $N a_{3} P O_{4}$ . I will round my answers to two significant figures , as that is the least amount given in the question.
$n (N a_{3} P O_{4}) = C \cdot V = 0.35 \cdot 275.0 \cdot 10^{- 3} = 9.6 \cdot 10^{- 2} m o l$
The number of moles is a measure of how many particles there are. The chemical formula shows that for every $N a_{3} P O_{4}$ salt particle, there are three $N a^{+}$ ions, so we need to multiply the answer by three, giving
$n (N a^{+}) = 3 \cdot n (N a_{3} P O_{4}) = 3 \cdot 9.6 \cdot 10^{- 2} = 0.29 m o l$

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