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Question:

A gas has a volume $255\ \mathrm{mL}$, a pressure $733\ \mathrm{mmHg}$, a temperature $548\ \mathrm{^\circ C}$, and a mass $1.22~\mathrm{g}$. What is the molar mass?

My attempt:

$$pV = nRT$$

Convert $733\ \mathrm{mmHg}$ to $\mathrm{atm} = 733~\mathrm{mmHg} \cdot 1\ \mathrm{atm}/760~\mathrm{mmHg} = 0.9645\ \mathrm{atm}$

Convert $\mathrm{^\circ C}$ to $\mathrm{K}$: $548\ \mathrm{^\circ C} + 273.15\ \mathrm{K} = 821.15~\mathrm{K}$

Convert $\mathrm{mL}$ to $\mathrm{L}$: $255~\mathrm{mL} \times \frac{1~\mathrm{L}}{1000~\mathrm{mL}} = 0.255~\mathrm{L}$

$$0.9645\ \mathrm{atm} \times 0.255~\mathrm{L}= n \times \frac{0.8206~\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}} 821.15~\mathrm{K}$$

$$\frac{0.2459}{673.84} = n \times \frac{673.84}{673.84}$$

$$0.000364\ \mathrm{mol} = n$$

$$M = \frac{m}{n} = \frac{1.22~\mathrm{g}}{0.000364~\mathrm{mol}} = 3351.648$$

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The mistake you made was the value of $R$. If pressure and volume are taken in $\mathrm{atm}$ and $\mathrm{L}$, then the value of $R$ is $0.0821$. (Not $0.82..$).

Now, $$n=\frac{0.9645\times 0.255}{0.0821\times 821.15}=0.0036$$

$$M=\frac{1.22}{0.0036}=334.412\mathrm{g}$$

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