What volume is occupied by 4.02 * 10^22 molecules of helium gas at STP?

Feb 11, 2016

The molar volume of an ideal gas at STP is $22.4$ ${\mathrm{dm}}^{3}$, or $22.4$ $L$.

Explanation:

In one mole of gas there are Avogadro's number of particles. There are $6.022$ $\times$ ${10}^{23}$ ${\text{particles mol}}^{-} 1$.

So how many moles are there in $4.02 \times {10}^{22}$ helium particles?

It is simply the quotient: $\left(4.02 \times {10}^{22} {\text{ helium particles")/(6.022xx10^23" helium particles mol}}^{-} 1\right)$

$=$ ?? mol

The volume of course is the prior quotient mulitplied by the molar volume of an ideal gas:

$\left(4.02 \times {10}^{22} {\text{ helium particles")/(6.022xx10^23" helium particles mol}}^{-} 1\right)$ $\times$ $22.4 \cdot L \cdot m o {l}^{-} 1$

$=$ ??L