# Question 52adc

Jul 12, 2016

Here's what I got.

#### Explanation:

The number of molecules present in $\text{10 L}$ of carbon dioxide, ${\text{CO}}_{2}$, actually depends on the conditions you have for pressure and temperature.

Since you didn't specify these conditions, I will assume that you're working at STP, Standard Temperature and Pressure.

STP conditions are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. Under these conditions, one mole of any ideal gas occupies $\text{22.7 L}$ $\to$ this is known as the molar volume of a gas at STP.

So, if one mole occupies $\text{22.7 L}$, it follows that the number of moles that occupy $\text{10 L}$ is equal to

10 color(red)(cancel(color(black)("L CO"_2))) * overbrace("1 mole CO"_2/(22.7color(red)(cancel(color(black)("L CO"_2)))))^(color(darkgreen)("molar volume of a gas at STP")) = "0.4405 moles CO"_2

To convert this to number of molecules, use the fact that one mole of a covalent compound contains $6.022 \cdot {10}^{23}$ molecules of said compound.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"molecules} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

In your case, the sample of carbon dioxide will contain

0.4405 color(red)(cancel(color(black)("moles CO"_2))) * (6.022 * 10^(23)"molec. CO"_2)/(1color(red)(cancel(color(black)("mole CO"_2)))) = 2.65 * 10^(23)"molec. CO"_2

I'll leave the answer rounded to two sig figs< but keep in mind that you only have one sig figs for the volume of the gas

"no. of molecules of CO"_2 = color(green)(|bar(ul(color(white)(a/a)color(black)(2.7 * 10^(23)"molecules")color(white)(a/a)|)))#

SIDE NOTE More often than not, you'll find STP conditions defined as a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

This is the old definition of STP for which one mole of any ideal gas occupies $\text{22.4 L}$. If this is the value of the molar volume of a gas at STP given to you, simply redo the calculations using $\text{22.4 L}$ instead of $\text{22.7 L}$.