# Question 5dfc9

Jul 29, 2017

$1.37 \times {10}^{25}$ ${\text{molecules CO}}_{2}$

#### Explanation:

We're asked to find the number of ${\text{CO}}_{2}$ molecules in $1.0$ ${\text{kg CO}}_{2}$.

To do this, we can first use the molar mass of ${\text{CO}}_{2}$ ($44.01$ $\text{g/mol}$) to find the number of moles of ${\text{CO}}_{2}$:

1.0cancel("kg CO"_2)((10^3cancel("g CO"_2))/(1cancel("kg")))((1color(white)(l)"mol CO"_2)/(44.01cancel("g CO"_2))) = 22.72 ${\text{mol CO}}_{2}$

Now, we can use Avogadro's number ($6.022 \times {10}^{23}$ ${\text{mol}}^{-} 1$) to find the number of ${\text{CO}}_{2}$ molecules:

22.72cancel("mol CO"_2)((6.022xx10^23color(white)(l)"molecules CO"_2)/(1cancel("mol CO"_2)))

= color(red)(ul(1.37xx10^25color(white)(l)"molecules CO"_2#