# A gram of salt added to pure (distilled) water introduces around $2 \times 10^{22}$ ions? [closed]

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.

• Have you heard of Avogadro's number and the concept of molar mass? Apr 28 '20 at 1:04
• They rounded down slightly: exactly 1 g of NaCl would have added $\mathrm{2.06088 \times 10^{22}}$ ions.
– Ed V
Apr 28 '20 at 1:17
• Look up Avogadro's number, as @Tyberius statews, and the definition of a mole. Not the critter. Apr 28 '20 at 2:09

$$\frac{1 \ \text{g}}{58.44 \ \text{g mol^{-1}}} = 0.017 \ \text{mol}$$
$$(0.017 \ \text{mol})(6.022 \times 10^{23} \ \text{mol^{-1}}) = 1.02 \times 10^{22}$$
• 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. Apr 28 '20 at 4:26