# Question b51d7

Feb 20, 2016

${\text{Br}}_{2}$

#### Explanation:

In order to be able to determine the identity of the gas, you must figure out its molar mass.

Since you know the mass of the sample, you can use the ideal gas law equation to find the number of moles present in the sample. Dividing these two values will give you the molar mass of the gas.

So, the ideal gas law equation looks like this

$\textcolor{b l u e}{P V = n R T} \text{ }$, 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

Now, you need to pay attention to the units used in the expression of the ideal gas constant and make sure that your units match those units.

You need to have the pressure expressed in atm, the volume in liters, and the temperature in Kelvin, so use the conversion factors

$\text{1 L" = 10^3"mL}$

$\text{1 atm " = "760 torr}$

Rearrange the ideal gas law equation to solve for $n$

$P V = n R T \implies n = \frac{P V}{R T}$

$n = \left(\frac{943}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 275 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25.5)color(red)(cancel(color(black)("K}}}}\right)$

$n = \text{0.01392 moles}$

The molar mass of the gas will tell you the mass of one mole of the gas. Since you know that $0.01392$ moles have a mass of $\text{2.22 g}$, you can say that one mole will have a mass of

1color(red)(cancel(color(black)("mole"))) * "2.22 g"/(0.01392color(red)(cancel(color(black)("moles")))) = "159.5 g"#

This means that the molar mass of the gas is equal to ${\text{159.5 g mol}}^{- 1}$.

The only gas on your list that comes close to having this molar mass is bromine, ${\text{Br}}_{2}$, which has a molar mass of ${\text{159.81 g mol}}^{- 1}$.

All the other options have significantly smaller molar masses, so bromine will be your answer.