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Monochloroethylene gas is used to make polyvinylchloride (PVC). It has a density of $\pu{2.56 g L-1}$ at $\pu{22.8 ^\circ C}$ and $\pu{101 kPa}$. What is the molar mass of monochloroethylene? What is the molar volume under these conditions?

$$PM=dRT$$

$$M=\frac{\pu{2.56 g L-1}\cdot\pu{295.8 K}\cdot\pu{0.0821 atm L mol-1 K-1}}{\pu{0.9968 atm}} = \pu{62.37 g mol-1}$$

After this I can't find molar volume. Which formula should I use?

I know $PV=nRT$, but without knowing the gas mass I can't use this formula, I believe.

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  • $\begingroup$ You need to start with PV=nRT. Let n=1 and solve for V. Then using density you can calculate the molar mass. $\endgroup$ – MaxW Jan 15 '18 at 17:11
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    $\begingroup$ Yes, I tried to do your way and got it. thanks . But i realized even if i do this my way plugging in the numbers to this equation $d=\frac MV$ i get the same result $\endgroup$ – EldarRahim Jan 15 '18 at 17:16
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    $\begingroup$ You might get the same number, but take a look at the units. The units $\mathrm{g}\cdot\mathrm{L}^{-1}$ are not the units for molecular mass $\mathrm{g}\cdot\mathrm{mol}^{-1}$ $\endgroup$ – MaxW Jan 15 '18 at 17:29
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    $\begingroup$ Right totally. İt was a typo my bad $\endgroup$ – EldarRahim Jan 15 '18 at 18:42
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You don't need to know the volume to determine the molar volume $V_\mathrm{m}$ of a substance. In your notations:

$$V_\mathrm{m} = \frac{M}{d} = \frac{\pu{62.37 g mol-1}}{\pu{2.56 g L-1}} = \pu{24.36 L mol-1}$$

Alternatively, you could've just used ideal gas law:

$$V_\mathrm{m} = \frac{V}{n} = \frac{RT}{P} = \frac{\pu{0.0821 atm L mol-1 K-1}\cdot\pu{295.8 K}}{\pu{0.9968 atm}} = \pu{24.36 L mol-1}$$

P. S. You determined molecular mass correctly, but, as MaxW pointed out, the units are wrong: you probably made a typo and put $\pu{g L-1}$ instead of $\pu{g mol-1}$.

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    $\begingroup$ @MaxW No, this is incorrect. Molar volume has SI units of $\pu{m3 mol-1}$, which are also equally good converted to $\pu{L mol-1}$ (see IUPAC Green Book, p. 47). There is a good reason it's called molar volume. You must divide both part by $n$ in your formula. $\endgroup$ – andselisk Jan 15 '18 at 17:58
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    $\begingroup$ you're right. I deleted my previous comment which was confusing. $\endgroup$ – MaxW Jan 15 '18 at 18:06
  • $\begingroup$ @MaxW No prob, I also tend to overthink quite often:) $\endgroup$ – andselisk Jan 15 '18 at 18:08
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    $\begingroup$ Thank you guys so much. Even though this might be an easy question, it feels good to have everything clear $\endgroup$ – EldarRahim Jan 15 '18 at 18:48

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