# Question 358b9

Jun 9, 2017

$2 \cdot {10}^{23}$

#### Explanation:

You don;t really need to know the conditions under which the gas is being kept to answer this question, so you can ignore the part about it being kept under STP conditions.

The first thing to do here is to convert the mass of oxygen gas to moles. To do that, use its molar mass

8 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.25 moles O"_2#

Now, you know that $1$ mole of oxygen gas must contain $6.022 \cdot {10}^{23}$ molecules of oxygen gas, as given by Avogadro's constant, which essentially acts as the definition of a mole.

This means that your sample will contain

$0.25 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles O"_2))) * (6.022 * 10^(23)color(white)(.)"molecules O"_2)/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)(2 * 10^(23)color(white)(.)"molecules O}}_{2}}}}$

The answer is rounded to one significant figure, the number of sig figs you have for the mass of oxygen gas.