Question 380d4

Jun 28, 2017

$7 \cdot {10}^{11}$

Explanation:

An interesting approach to use here would be to convert the molar mass of magnesium from grams per mole, ${\text{g mol}}^{- 1}$, to picograms per mole, ${\text{pg mol}}^{- 1}$.

Now, magnesium has a molar mass of ${\text{24.305 g mol}}^{- 1}$. As you know

$\text{1 g} = {10}^{12}$ $\text{pg}$

This means that the molar mass of magnesium is equal to

24.305 color(red)(cancel(color(black)("g"))) "mol"^(-1) * (10^12color(white)(.)"pg")/(1color(red)(cancel(color(black)("g")))) = 2.4305 * 10^(13) ${\text{pg mol}}^{- 1}$

This means that $1$ mole of magnesium has a mass of $2.4305 \cdot {10}^{12}$ $\text{pg}$. You can thus say that your sample will contain

3 color(red)(cancel(color(black)("pg"))) * "1 mole Mg"/(2.4305 * 10^12color(red)(cancel(color(black)("pg")))) = 1.234 * 10^(-12)# $\text{moles Mg}$

Finally, to find the number of atoms of magnesium present in the sample, use the fact that $1$ mole of magnesium must contain $6.022 \cdot {10}^{23}$ atoms of magnesium $\to$ this is given by Avogadro's constant.

You can thus say that your sample will contain

$1.234 \cdot {10}^{- 12} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Mg"))) * (6.022 * 10^(23)color(white)(.)"atoms Mg")/(1color(red)(cancel(color(black)("mole Mg")))) = color(darkgreen)(ul(color(black)(7 * 10^11color(white)(.)"atoms Mg}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of magnesium.