# How many moles are present in 7.25xx10^25 "carbon dioxide molecules"?

Feb 2, 2017

Well one mole of stuff, any stuff, including carbon dioxide, specifies $6.022 \times {10}^{23}$ individual items of that stuff.

#### Explanation:

And thus we work out the quotient:

$\left(7.2 \times {10}^{25} \cdot \text{carbon dioxide molecules")/(6.022xx10^23*"carbon dioxide molecules} \cdot m o {l}^{-} 1\right)$

$\cong 120 \cdot m o l$ $\text{carbon dioxide}$.

This is dimensionally consistent, because we get an answer with units $\frac{1}{m o {l}^{-} 1} = \frac{1}{\frac{1}{\text{mol}}} = m o l$ as required.

Do you agree?

What is the mass of this quantity of carbon dioxide?