# Question d7ebc

Jul 11, 2017

$0.25$ ${\text{mol O}}_{2}$

$1.5 \times {10}^{23}$ ${\text{molecules O}}_{2}$

#### Explanation:

NOTE: I'll have to make the assumption that the conditions are *standard temperature and pressure*.

If this is true then

One mole of an (ideal) gas at standard temperature and pressure conditions occupies a volume of $22.41$ $\text{L}$.

We can use this to convert from liters of ${\text{O}}_{2}$ to moles (treating oxygen as an ideal gas). We need to convert from milliliters to liters:

5600cancel("mL")((1color(white)(l)"L")/(10^3cancel("mL"))) = color(red)(5.6 color(red)("L"

Using the above mole-liter conversion:

5.6cancel("L")((1color(white)(l)"mol O"_2)/(22.41cancel("L O"_2))) = color(blue)(0.25 color(blue)("mol O"_2

Now, we can use Avogadro's number ($6.022 \times {10}^{23}$) to convert from moles to individual units (molecules) of oxygen gas:

color(blue)(0.25)cancel(color(blue)("mol O"_2))((6.022xx10^23color(white)(l)"molecules O"_2)/(1cancel("mol O"_2)))

= color(green)(1.5xx10^23 color(green)("molecules O"_2#