# What is the molar mass of a gas if a 40.0-gram sample of the gas occupies 11.2 liters of space at STP?

##### 1 Answer
Mar 11, 2016

$M = 80.0 \frac{g}{m o l}$

#### Explanation:

For this solution one would use the variation of the Ideal Gas Law
$P M = \mathrm{dR} T$
$P = P r e s s u r e$ in $a t m$
$M = M o l a r M a s s$ in $\frac{g}{m o l}$
$d = D e n s i t y$ in $\frac{g}{L}$
$R = 0.0821 \frac{a t m L}{m o l K}$ the gas law constant
$T = T e m p e r a t u r e$ $K$

The density can be found by dividing the mass by the volume.

$d = \frac{m a s s}{v o l u m e} = \frac{g}{L}$

$m a s s = 40.0 g$
$v o l u m e = 11.2 L$

$d = \frac{40.0 g}{11.2 L}$
$d = 3.57 \frac{g}{L}$

The STP values (standard temperature and pressure) are 1 atm and 273 K.

$P = 1 a t m$
M = ?
$d = 3.57 \frac{g}{L}$
$R = 0.0821 \frac{a t m L}{m o l K}$
$T = 273 K$

$P M = \mathrm{dR} T$
$\left(1 a t m\right) \left(M\right) = \left(3.57 \frac{g}{L}\right) \left(0.0821 \frac{a t m L}{m o l K}\right) \left(273 K\right)$
$\left(M\right) = \frac{\left(3.57 \frac{g}{\cancel{L}}\right) \left(0.0821 \frac{\cancel{a t m} \cancel{L}}{m o l \cancel{K}} \left(273 \cancel{K}\right)\right)}{1 \cancel{a t m}}$
$M = 80.0 \frac{g}{m o l}$