# What is the molar mass of a gas that has a density of 1.02 g/L at 0.990 atm pressure and 37°C?

May 30, 2016

${\text{26.3 g mol}}^{- 1}$

#### Explanation:

Your starting point here will be the ideal gas law equation

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

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas, $\rho$, under those conditions for pressure and temperature, and its molar mass, ${M}_{M}$.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass $m$ of this gas, you can express its molar mass as the ratio between $m$ and $n$, the number of moles it contains

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{M}_{M} = \frac{m}{n}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass $m$ of this gas, you can express its density as the ratio between $m$ and the volume it occupies

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\rho = \frac{m}{V}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

Plug equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ into the ideal gas law equation to get

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

Rearrange to get

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

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

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

Finally, use equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to write

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

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

M_M = 1.02 "g"/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 + 37)color(red)(cancel(color(black)("K"))))/(0.990color(red)(cancel(color(black)("atm"))))

${M}_{M} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{26.3 g mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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