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A $2.1\ \mathrm g$ sample of a liquid vaporizes and exerts $120\ \mathrm{mmHg}$ pressure at $1.5\ \mathrm L$ volume and $80\ \mathrm{^\circ C}$ temperature. Find the molar mass of the gas.

I first get the equation $PV=nRT$ and replace $n$ with $m/M$ since $M=m/n$ Then the equation becomes this $$PV=\frac{mRT}{M} \Rightarrow M=\frac{mRT}{PV}$$

$$\begin{align} M&=\frac{mRT}{PV}\\[6pt] \Rightarrow M&=\frac{(2.1\ \mathrm g)\times(62.364\ \mathrm{L\ mmHg/(mol\ K}))\times(80+273.13)\ \mathrm K}{(120\ \mathrm{mmHg})(1.5\ \mathrm L)}\\[6pt] \Rightarrow M&=257\ \mathrm{g/mol} \end{align}$$

But the answer key says $150\ \mathrm{g/mol}$. What am I doing wrong?

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  • $\begingroup$ Your method seems okay. Just check calculations. If it is correct, then your answer key is wrong . $\endgroup$ – user14857 Oct 30 '16 at 21:09
  • $\begingroup$ The calculation is correct, too. $\endgroup$ – Jan Oct 30 '16 at 21:27
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Sometimes (too often) answer keys are wrong. In this case we've now triple checked the calculation, and you, Jan (as implied in his comment) and myself have all come up with a molar mass of $\pu{257 g/mol}$. At this point, we have to say the answer key is wrong and you are correct.

Your answer is $\pu{1.71}$ times that given by the key, and neither that nor any other comparison appears to correlate to any obvious error on their part. It simply appears to be a misplaced or incorrectly entered answer in the key, end of story and move on. And good job getting the correct answer ;)

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