# An imaginary ideal gas has a density of 3 g/L at STP. What is the molar mas of this gas?

Oct 25, 2015

$\text{70 g/mol}$

#### Explanation:

You know that you're dealing with a sample of an unknown gas that has a density of $\text{3 g/L}$ at STP, or Standard Temperature and Pressure.

In order to find the gas' molar mass, you need to know two things

• the mass of this sample of gas
• the number of moles of gas present in this sample

Now, STP conditions imply a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

You will have to use the ideal as law

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

in order to try to find a relationship between the gas' density and its molar mass.

You know that molar mass is defined as mass per mole, so you can say that

${M}_{\text{m" = m/n implies n = m/M_"m}}$

Replace the number of moles in the ideal gas law equation to get

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

You also know that density is defined as mass per unit of volume

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

Notice what happens if you divide both sides of the ideal gas law equation by $V$

$\frac{P \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}} = \frac{m}{M} _ \text{m} \cdot \frac{R T}{V}$

P = underbrace(m/V)_(color(green)("density")) * (RT)/M_"m" = rho * (RT)/M_"m"

Finally, isolate the molar mass on one side of the equation to get

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

Now all you have to do is plug in your values - remember that

$R = 0.082 \left(\text{atm" * "L")/("mol" * "K}\right)$

and that the pressure must be expressed in atm and the temperature in Kelvin!

M_"m" = 3"g"/color(red)(cancel(color(black)("L"))) * (0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))

${M}_{\text{m" = "68.085 g/mol}}$

SInce you only gave one sig fig for the density of the gas, the answer will be

M_"m" = color(green)("70 g/mol")