# Question f9eac

Sep 8, 2016

$1.363 \cdot {10}^{24} \text{ Na"^(+)"cations}$

#### Explanation:

If by amount of sodium ions you mean how many of them you have in that sample of sodium carbonate, ${\text{Na"_2"CO}}_{3}$, then all you have to do here is use two conversion factors

• one to take you from grams to moles
• one to take you from moles to number of ions

Now, before moving forward, notice that one mole of sodium carbonate contains

• two moles of sodium cations, $2 \times {\text{Na}}^{+}$
• one mole of carbonate anions, $1 \times {\text{CO}}_{3}^{2 -}$

So, use the molar mass of sodium carbonate to calculate the number of moles present in that sample

120.0 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"CO"_3)/(106.0color(red)(cancel(color(black)("g")))) = "1.132 moles Na"_2"CO"_3

This means that your sample will contain

1.132 color(red)(cancel(color(black)("moles Na"_2"CO"_3))) * ("2 moles Na"^(+))/(1color(red)(cancel(color(black)("mole Na"_2"CO"_3))))

$= {\text{2.264 moles Na}}^{+}$

Now all you have to do is use Avogadro's number

$\textcolor{p u r p \le}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 mole" = 6.022 * 10^(23)"particles}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to calculate the number of sodium ions present in that many moles

2.264 color(red)(cancel(color(black)("moles Na"^(+)))) * (6.022 * 10^(23)" Na"^(+)"cations")/(1color(red)(cancel(color(black)("mole Na"^(+)))))#

$= \textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{1.363 \cdot {10}^{24} \text{ Na"^(+)"cations}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to four sig figs.