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.

  • 2
    $\begingroup$ Have you heard of Avogadro's number and the concept of molar mass? $\endgroup$ – Tyberius Apr 28 at 1:04
  • $\begingroup$ They rounded down slightly: exactly 1 g of NaCl would have added $\mathrm{2.06088 \times 10^{22}}$ ions. $\endgroup$ – Ed V Apr 28 at 1:17
  • $\begingroup$ Look up Avogadro's number, as @Tyberius statews, and the definition of a mole. Not the critter. $\endgroup$ – DrMoishe Pippik Apr 28 at 2:09

Using dimensional analysis, we get

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

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

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  • 5
    $\begingroup$ You basically have it. The last little detail is that, when dissolved, NaCl breaks down into 2 ions, $\ce{Na+}$ and $\ce{Cl-}$. Also, in your first equation, the 1 is 1 gram. $\endgroup$ – Tyberius Apr 28 at 4:26

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