What volume is occupied by $4.02 \cdot 10^{22}$ $4.02 * 10^22$ molecules of helium gas at STP?

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What volume is occupied by $4.02 \cdot 10^{22}$ $4.02 * 10^22$ molecules of helium gas at STP?

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Answer 1 · 2016-02-11T09:25:14

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

Explanation:

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

So how many moles are there in $4.02 \times 10^{22}$ $4.02xx10^22$ helium particles?

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

$=$ $=$ $? ? m o l$ $?? mol$

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

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

$=$ $=$ $? ? L$ $??L$

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What volume is occupied by $4.02 \cdot 10^{22}$ $4.02 * 10^22$ molecules of helium gas at STP?

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