# Question 1a4dd

May 6, 2016

${\text{44 g mol}}^{- 1}$

#### Explanation:

The idea here is that you can use the ideal gas law equation to find a relationship between the density of the gas and its molar mass.

So, the ideal gas law equation 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

As you know, the number of moles present in a sample of gas, $n$, can be expressed using the mass of that sample, $m$, and the molar mass of the gas, ${M}_{M}$

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{n = \frac{m}{M} _ M} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug this into the ideal gas law equation to get

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

Since you're interested in find the molar mass of the gas, rearrange to isolate ${M}_{M}$ on one side of the equation

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

Notice that this equation contains a term that expresses the mass of the sample per unit of volume, i.e. the density, $\rho$, of the gas.

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\rho = \frac{m}{V}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that you have

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

Before plugging in your values, make sure that you convert the units for pressure and temperature to the units used in the expression of the ideal gas constant.

You will thus need to convert the pressure from mmHg to atm and the temperature from degrees Celsius to Kelvin by using

color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 760 mmHg")color(white)(a/a)|)))" " and " " color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))

Also, notice that the density of the gas is expressed in grams per cubic decimeter, ${\text{g dm}}^{- 3}$. As you know, this is equivalent to grams per liter, ${\text{g L}}^{- 1}$, since

color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = "1 dm"^3)color(white)(a/a)|)))" "

The molar mass of the gas will be equal to

M_M = 1.43 "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 + 27)color(red)(cancel(color(black)("K"))))/(608/760color(red)(cancel(color(black)("atm"))))#

${M}_{M} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{44 g mol}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$