# Question d5b78

Jun 7, 2017

$1.12 \times {10}^{25}$

#### Explanation:

At room temperature (${20}^{\circ} \text{C}$) and atmospheric pressure (101.3 kPa), 1 mol of gas will occupy ${\text{24.06 dm}}^{3}$, so ${\text{224 dm}}^{3}$ will contain

\frac{"224 dm"^3}{"24.06 dm"^3"/mol"} = "9.31 mol"# of gas.

1 mol of anything has $6.022 \times {10}^{23}$ molecules in it, so $\text{9.31 mol}$ contains

$\text{9.31 mol" xx 6.022 xx 10^23 " molecules/mol" = 5.607 xx 10^24 " molecules}$

But, each molecule of oxygen gas contains 2 oxygen atoms, as the formula for oxygen gas is ${\text{O}}_{2}$.

Therefore, there will be twice that many atoms.

So, the number of atoms present

$= 5.607 \times {10}^{24} \text{ molecules" xx 2 " atoms/molecule" = 1.12 xx 10^25 " atoms}$

If the gas is not at atmospheric pressure and room temperature, then use the ideal gas equation, $p V = n R T$ to work out the no. of moles of oxygen gas, then follow the same method.