# Question 64a0e

May 26, 2016

${\text{7.3 g L}}^{- 1}$

#### Explanation:

You don't actually need to know the mass of bromine to determine the density of the gas, all you need to know are the conditions for pressure and temperature.

Your starting point here will be the ideal gas law equation, which looks like this

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

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

You can rewrite the ideal gas law equation to include the mass of the gas by using molar mass, which is defined as the mass of one mole

${M}_{M} = \frac{m}{n}$

This is equivalent to

$n = \frac{m}{M} _ M$

Plug this into the ideal gas law equation to get

$P V = \frac{m}{M} _ M \cdot R T$

As you know, density is defined as mass per unit of volume.

$\rho = \frac{m}{V}$

Rearrange the above equation to get

$P \cdot {M}_{M} = \frac{m}{V} \cdot R T$

which is of course equivalent to

$P \cdot {M}_{M} = \rho \cdot R T$

This gets you

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \rho = \frac{P}{R T} \cdot {M}_{M} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Bromine has a molar mass of

M_("M Br"_2) = "159.81 g mol"^(-1)

Convert the pressure to atm and the temperature to Kelvin by using

color(purple)(|bar(ul(color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|))) " " and " "color(purple)(|bar(ul(color(black)("1 atm " = " 101.325 kPa")color(white)(a/a)|)))

Plug in your values to find

rho = (115/101.325 color(red)(cancel(color(black)("atm"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 30)color(red)(cancel(color(black)("K")))) * 159.81"g"/color(red)(cancel(color(black)("mol")))#

$\rho = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{7.3 g L}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.