A gram of salt added to pure (distilled) water introduces around $2 \times 10^{22}$ ions? I am currently studying Practical Electronics For Inventors, Fourth Edition, by Scherz and Monk. Chapter 2.5.2 Resistivity and Conductivity, claims the following:


  Adding an ionic compound in the form of common salt ($\ce{NaCl}$) to water increases the ion concentration within solution -- $\ce{NaCl}$ ionizes into $\ce{Na^+}$ and $\ce{Cl^-}$. A gram of salt added introduces around $2 \times 10^{22}$ ions. These ions act as charge carriers, which in turn effectively lowers the solution's resistance to below an ohm per meter. If we use the solution as a conductor between a battery and a lamp -- via electrodes placed in solution -- there is ample current to light the lamp.


How did the authors calculate that a gram of salt added introduces around $2 \times 10^{22}$ ions?

I would greatly appreciate it if someone would please take the time to clarify this.
Using dimensional analysis, we get

$$\frac{1 \ \text{g}}{58.44 \ \text{g mol$^{-1}$}} = 0.017 \ \text{mol}$$

$$(0.017 \ \text{mol})(6.022 \times 10^{23} \ \text{mol$^{-1}$}) = 1.02 \times 10^{22}$$