# What is the molality, molarity, and mole fraction of FeCl_3 in a 22.2 mass % aqueous solution?

## $d = 1.280 g$$/ m L$.

Aug 13, 2017

$\text{molality} = 1.76$ $m$

$\text{molarity} = 1.75$ $M$

$\text{mole fraction} = 0.0307$

#### Explanation:

• the molality

• the molarity

• the mole fraction (of ${\text{FeCl}}_{3}$)

of a 22.2% by mass ${\text{FeCl}}_{3}$ solution.

$\text{ }$

MOLALITY:

The equation for molality is

$\text{molality" = "mol solute"/"kg solvent}$

If we assume a $100$-$\text{g}$ sample of solution, we would have

• $22.2$ ${\text{g FeCl}}_{3}$

• $77.8$ $\text{g H"_2"O}$

The number of moles of ${\text{FeCl}}_{3}$ is found using the molar mass:

22.2cancel("g FeCl"_3)((1color(white)(l)"mol FeCl"_3)/(162.2cancel("g FeCl"_3))) = color(red)(ul(0.137color(white)(l)"mol FeCl"_3

The number of kilograms of the solvent (water) is

77.8cancel("g H"_2"O")((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = color(green)(ul(0.0778color(white)(l)"kg H"_2"O"

And so the molality is

"molality" = (color(red)(0.137color(white)(l)"mol FeCl"_3))/(color(green)(0.778color(white)(l)"kg H"_2"O")) = color(blue)(ulbar(|stackrel(" ")(" "1.76color(white)(l)m" ")|)

$\text{ }$

MOLARITY:

The equation for molarity is

$\text{molarity" = "mol solute"/"L solution}$

To find the molarity of a solution with its molality known, we need to know the density of the solution (which is given to us as $1.280 \textcolor{w h i t e}{l} \text{g/mL}$).

Since we assumed a $100$-$\text{g}$ sample of solution, we can find the volume of the solution:

100cancel("g soln")((1cancel("mL soln"))/(1.280cancel("g soln")))((1color(white)(l)"L soln")/(10^3cancel("mL soln"))) = color(red)(ul(0.0781color(white)(l)"L solution"

We found the moles of solute earlier, so the molarity is thus

"molarity" = (0.137color(white)(l)"mol FeCl"_3)/(color(red)(0.0781color(white)(l)"L solution")) = color(blue)(ulbar(|stackrel(" ")(" "1.75color(white)(l)M" ")|)

$\text{ }$

MOLE FRACTION:

The equation for the mole fraction of ${\text{FeCl}}_{3}$ is given by

${\chi}_{\text{FeCl"_3) = ("mol FeCl"_3)/("mol FeCl"_3 + "mol H"_2"O}}$

Again, we already know the number of moles of iron(III) chloride, so we just need to find the moles of water, using its molar mass and gram-mass found earlier:

77.8cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = color(red)(ul(4.32color(white)(l)"mol H"_2"O"

The mole fraction is thus

chi_("FeCl"_3) = (0.137color(white)(l)"mol FeCl"_3)/(0.0137color(white)(l)"mol FeCl"_3 + color(red)(4.32color(white)(l)"mol H"_2"O")) = color(blue)(ulbar(|stackrel(" ")(" "0.0307" ")|)