# Question #64a0e

##### 1 Answer

#### Answer:

#### 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

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where

*universal gas constant*, usually given as

**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 = m/n#

This is equivalent to

#n = m/M_M#

Plug this into the ideal gas law equation to get

#PV = m/M_M * RT#

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

#rho = m/V#

Rearrange the above equation to get

#P * M_M = m/V * RT#

which is of course equivalent to

#P * M_M = rho * RT#

This gets you

#color(blue)(|bar(ul(color(white)(a/a)rho = P/(RT) * M_Mcolor(white)(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 = color(green)(|bar(ul(color(white)(a/a)"7.3 g L"^(-1)color(white)(a/a)|)))#

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