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Given that 0.28 g of dry gas occupies a volume of 354 mL at a temperature of 20°C and a pressure of 686 mmHg, how do you calculate the molecular weight of the gas?

1 Answer

May 14, 2018

$21 \frac{g}{m o l}$ $21 g/(mol)$

Explanation:

We start with the Ideal Gas Law equation $P V$ $PV$ = $n R T$ $nRT$
Rearrange the equation to $\frac{n}{V}$ $n/V$ = $\frac{P}{R T}$ $P/(RT)$
Convert the $20^{\circ} C$ $20^@C$ to Kelvin $(K)$ $(K)$ by adding $273$ $273$

$P$ $P$ = $686 m m H g$ $686 mmHg$

$R$ $R$ = $\frac{62.36367 m m H g \cdot L}{m o l \cdot K}$ $(62.36367 mmHg*L)/ (mol*K)$

$T$ $T$ = $293 K$ $293K$

Plug into the equation

$\frac{n}{V}$ $n/V$ = $\frac{686 m m H g}{\frac{62.36367 m m H g \cdot L}{m o l \cdot K} (293 K)}$ $(686 mmHg)/((62.36367 mmHg*L)/ (mol*K)(293K)$

Multiply $62.36367$ $62.36367$ by $293$ $293$ then divide by $686$ $686$

$\frac{n}{V}$ $n/V$ = $0.03754 \frac{m o l}{L}$ $0.03754 (mol)/L$

Convert $354 m L$ $354mL$ to $L$ $L$ by dividing $354 m L$ $354mL$ by $1000$ $1000$
Now calculate the density $d$ $d$ = $\frac{M}{V}$ $M/V$

$\frac{0.28 g}{0.354 L}$ $(0.28 g) / (0.354 L)$ = $0.79096 \frac{g}{L}$ $0.79096 g/L$

Take the $\frac{g}{L}$ $g/L$ and divide by $\frac{m o l}{L}$ $(mol)/L$

$(0.79096 g/cancelL) / (0.03754 (mol)/cancelL)$ = $21 g/(mol)$

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