# If 2.31 g of the vapor of a volatile liquid is able to fill a 498 ml flask at 100 degrees C and 775 mmHg, how do you calculate the molar mass of the liquid? Also, how do you calculate the density of the vapor under these conditions?

Aug 18, 2016

The molar mass of the gas is 139 g/mol, and its density is 4.64 g/L.

#### Explanation:

We can use the Ideal Gas Law to solve both of these problems:

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

Molar mass

Since $n = \text{mass"/"molar mass} = \frac{m}{M}$, we can write the Ideal Gas Law as

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

We can rearrange this to get

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

$m = \text{2.31 g}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(100 + 273.15) K" = "373.15}$
P = 775 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "1.020 atm"
$V = \text{0.498 L}$

M = ("2.31 g" × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 373.15 color(red)(cancel(color(black)("K"))))/("1.020" color(red)(cancel(color(black)("atm"))) × 0.498 color(red)(cancel(color(black)("L")))) = "139 g/mol"

∴ The molar mass is 139 g/mol.

Density

We have seen that .

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

We can rearrange this to

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

But $\text{density"= "mass"/"volume}$ or color(brown)(|bar(ul(color(white)(a/a)color(black)(ρ = m/V)color(white)(a/a)|)))" "

PM = ρRT and

color(blue)(|bar(ul(color(white)(a/a)ρ = (PM)/(RT)color(white)(a/a)|)))" "

$P = \text{1.020 atm}$
$M = \text{46.01 g/mol}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(100 + 273.15) K" = "373.15 K}$

ρ = (1.020 color(red)(cancel(color(black)("atm"))) × "139.3 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.082 06" color(red)(cancel(color(black)("atm")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 373.15 color(red)(cancel(color(black)("K")))) = "4.64 g/L"