# 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?

May 14, 2018

$21 \frac{g}{m o l}$

#### Explanation:

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

$P$ = $686 m m H g$

$R$ = $\frac{62.36367 m m H g \cdot L}{m o l \cdot K}$

$T$ = $293 K$

• Plug into the equation

$\frac{n}{V}$ = (686 mmHg)/((62.36367 mmHg*L)/ (mol*K)(293K)

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

$\frac{n}{V}$ = $0.03754 \frac{m o l}{L}$

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

$\frac{0.28 g}{0.354 L}$ = $0.79096 \frac{g}{L}$

• Take the $\frac{g}{L}$ and divide by $\frac{m o l}{L}$

$\frac{0.79096 \frac{g}{\cancel{L}}}{0.03754 \frac{m o l}{\cancel{L}}}$ = $21 \frac{g}{m o l}$