# Question #74d82

Jan 15, 2017

You divide molar mass by molar volume.

#### Explanation:

Molar mass for $H C l$ is (rounded) $1.0 + 35.5 = 36.5 g / m o l$
(you can get more exact figures from any source).

Molar volume at $273 K \mathmr{and} 1 a t m = 22.4 {\mathrm{dm}}^{3} / m o l$

So $H C l$ has a density of:

$\rho = \frac{36.5 g / \cancel{m o l}}{22.4 {\mathrm{dm}}^{3} / \cancel{m o l}} = 1.63 g / {\mathrm{dm}}^{3} = 1.63 k g / {m}^{3}$

Note : There are many STP conditions defined. I used the one that is most common in my country.