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Which of the following statements is true?

$11.2~\mathrm{dm^{3}}$ of nitrogen at STP

A: has a mass of $14\ \mathrm{g}$.

B: consists of $0.5\ \mathrm{mol}$ atoms.

My instinct was that the answer is B because any gas has a volume of $22.4$ liters regardless of what the gas is. If $0.5~\mathrm{mol}$ of nitrogen has a volume of $11.2~\mathrm{dm^{3}}$ at STP, then $1~\mathrm{mol}$ of nitrogen has a volume of $22.4$ liters at STP (if I am correct). It turned out the answer was A. Why is B not the correct answer?

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    $\begingroup$ B looks like a trick answer. It would be 0.5 moles of N2 molecules but not 0.5 moles of nitrogen atoms. $\endgroup$ – Jan Bos Jun 13 '15 at 17:17
  • $\begingroup$ So you thought B was correct, and consequently A was not correct, but did you stop to independently check whether A was correct or not? $\endgroup$ – Nicolau Saker Neto Jun 13 '15 at 18:20
  • $\begingroup$ Checkout WolframAlpha $\endgroup$ – Sparkler Jun 15 '15 at 14:57
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    $\begingroup$ Subtle point. In the 1980s the standard pressure part of STP was changed from 1 atmosphere = 101325 Pa to 1 bar = 100000 Pa. With this reduced standard pressure the corresponding molar volume has expanded to 22.71 L. But it seems to have been overlooked, even 30+ years on, by most texts. $\endgroup$ – Oscar Lanzi Feb 10 '18 at 14:24
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You are right that 0.5 mol of $\ce{N2}$ is 11.2 L, but read the question carefully. Option B says that 11.2 L of $\ce{N2}$ contains 0.5 mol of atoms, which is wrong, it contains 11.2 L of $\ce{N2}$ contains 0.5 mol of molecules. It seems to be a trick answer. The first option is correct as 0.5 mol of $\ce{N2}$ has a mass of 14 grams.

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The correct answer for this question is A.

Prior to start explaining the answer for your question, I would like to show you a mistake you have done. You have said

"any gas has a volume of 22.4 litres regardless of what the gas is."

This true only if the gas exist in standard temperature and pressure or (STP). And note that per one mole of any gas at STP condition occupy a volume of 22.4 litres or $22.4\ \mathrm{dm^3}$. Simply we call this as molar volume. Here is how you can derive it.

Subject $V$ in the formula $pV=nRT$, and then you can substitute values. Where $T=273.15 \ \mathrm{K}$ (value of standard temperature) and $p=101325 \ \mathrm{Pa}$ (value of standard pressure) and $R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}$

$$V_\mathrm{m}=\frac{V}{n}=\frac{RT}{p}$$

You have asked why not the answer B is correct? It is not correct. Here is why: In one mole of any compound you have $6.022\times10^{23}$ atoms

$1\ \mathrm{mol} \Rightarrow 6.022 \times 10^{23}$ atoms.

$6.022 \times 10^{23}$ is the Avogadro constant. (If you have no idea about it check this.) So here in this question you have found number of moles of nitrogen molecule, it means $\ce{N2(g)}$ as $0.5\ \mathrm{mol}$. Which means you have $0.5\times2\ \mathrm{mol}$ of atoms. So as I said before if you have one mole mean you already get $6.022 \times 10^{23}$ atoms of $\ce{N}$ atom. So you have $6.022 \times 10^{23}$ atoms or $6.022 \times 10^{23}$ nitrogen mol atoms. So now you can clearly see why answer B is incorrect. I hope my answer is helpful enough to solve your doubt.

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    $\begingroup$ The answer is not correct, in 0.5 mol of N2 you would have 6.022e23 nitrogen atoms and 1 mole of N atoms (if that is even a thing). $\endgroup$ – orthocresol Jun 14 '15 at 18:23
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    $\begingroup$ @orthocresol: Hey, as far as I know, in one mole of any compound you have 6.022e23, so how can you say " in 0.5 mol of N2 you would have 6.022e23 nitrogen atoms and 1 mole of N atoms " $\endgroup$ – On the way to success Jun 15 '15 at 0:23
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    $\begingroup$ @Onthewaytosuccess one molecule of diatomic nitrogen has 2 nitrogen atoms $\endgroup$ – orthocresol Jun 15 '15 at 0:26
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    $\begingroup$ @orthocresol: It means you are saying for 1 mol of nitrogen atom we have 6.022e23 atoms? $\endgroup$ – On the way to success Jun 15 '15 at 0:29
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    $\begingroup$ @Onthewaytosuccess yes, that is right. That is assuming that you could even have one mole of nitrogen atoms (which would, as far as I know, immediately combine to form diatomic nitrogen). $\endgroup$ – orthocresol Jun 15 '15 at 0:35

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