# Question 16856

Apr 14, 2017

3.61

#### Explanation:

its quite simple if you go by a simple method of finding density at STP
we got
"Density"=("Mass")/("Volume")
we know that mas of $H B r$ is 81 units
and one mole of $H B r$ weighs 81 $g m$
and we also know that the volume of one mole of any gas is 22.4 $L$hence , we have got our data .
upon substitution with the formal equation of density we get .
$d = \frac{81}{22.4} = 3.61 g {m}^{- 3}$

Apr 14, 2017

The density of $\text{HBr}$ at STP is 3.563 g/L.

#### Explanation:

Yes, there is a gas law equation for density.

It is a variation of the Ideal Gas Law.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since

$n = \text{mass"/"molar mass} = \frac{m}{M}$,

we can write

$p V = \frac{m}{M} R T$

or

$p = \left(\frac{m}{V}\right) \frac{R T}{M}$

And density ρ = m/V,

so

p = ρ(RT)/M

This gives

ρ = (pM)/(RT)

STP is defined as 1 bar and 0 °C.

In this problem,

$p \textcolor{w h i t e}{l l} = \text{1 bar}$
$M = \text{80.91 g·mol"^"-1}$
$R \textcolor{w h i t e}{l} = \text{0.083 14 bar·L·K"^"-1""mol"^"-1}$
$T \textcolor{w h i t e}{l} = \text{273.15 K}$

ρ = (1 color(red)(cancel(color(black)("bar"))) × 80.91 "g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K")))) = "3.563 g/L"#