Question #2b5ce

Jan 1, 2018

$6.0 \cdot {10}^{23}$

Explanation:

The thing to remember about an element's molar mass, which you'll find listed in the Periodic Table, is that it tells you the mass of $6.022 \cdot {10}^{23}$ atoms of that element.

This is the case because the molar mass of an element represents the mass of exactly $1$ mole of that element. And since Avogadro's constant tells you that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 mole" = 6.022 * 10^(23)color(white)(.)"particles}}}}$

you can say that the molar mass of an element tells you the mass of $6.022 \cdot {10}^{23}$ atoms of that element.

Now, the Periodic Table tells you that carbon--diamond is simply an allotrope of carbon, i.e. a form that carbon can take--has a molar mass of ${\text{12.011 g mol}}^{- 1}$.

This means that $1$ mole of carbon has a mass of $\text{12.011 g}$. In other words, $6.022 \cdot {10}^{23}$ atoms of carbon, the equivalent of $1$ mole of carbon, have a mass of $\text{12.011 g}$.

This means that $\text{12 g}$ of carbon will contain

$12 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * (6.022 * 10^(23)color(white)(.)"atoms of C")/(12.011color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(6.0 * 10^(23)color(white)(.)"atoms of C}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of carbon.