# Question 0730c

Jul 28, 2017

Molarity: $2 M$ (given)

Molality: $1.85 m$

Mole Fraction of Acetic Acid: $0.0323$

#### Explanation:

We're asked to find the molarity and mole fraction of acetic acid in a solution.

Well, the molarity is given as $2 M$...so I will also find the molality of the solution.

We know the density of the $\text{CH"_3"COOH}$ solution is $1.2$ $\text{g/mL}$, which is the same as color(green)(1200 color(green)("g/L".

Let's assume we have $1$ $\text{L}$ of solution..

Then, there are $2$ $\text{mol CH"_3"COOH}$ in the solution, because the given molarity value says there are two moles of solute per liter of solution.

Using the molar mass of acetic acid, we can find the number of grams of acetic acid present (which we'll use to find the quantity of water (solvent) present):

2cancel("mol CH"_3"COOH")((60.052color(white)(l)"g CH"_3"COOH")/(1cancel("mol CH"_3"COOH"))) = color(red)(120.104 color(red)("g CH"_3"COOH"

Now, using the given density of the solution, we can find the number of grams of solution; we can then subtract the grams of solute from that to find the mass of water:

Since we assumed $1$ $\text{L soln}$, according to the density value there are color(green)(1200 color(green)("g soln", so the mass of water present is

color(green)(1200color(white)(l)"g soln") - color(red)(120.104color(white)(l)"g CH"_3"COOH") = color(purple)(1080 color(purple)("g H"_2"O"

or

color(purple)(1.080color(white)(l)"kg H"_2"O"

Thus, the molality of the solution is

"molality" = "mol solute"/"kg solvent" = (2color(white)(l)"mol CH"_3"COOH")/(color(purple)(1.080color(white)(l)"kg H"_2"O")) = ul(1.85m

Lastly, let's find the mole fraction of acetic acid of the solution.

What we can do is calculate the number of moles of water present using its molar mass ($18.015$ $\text{g/mol}$) and mass present (color(purple)(1080 color(purple)("g"):

color(purple)(1080)cancel(color(purple)("g H"_2"O"))((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = color(orange)(59.9 color(orange)("mol H"_2"O"

The mole fraction of acetic acid is thus

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

= (2color(white)(l)"mol CH"_3"COOH")/(2color(white)(l)"mol CH"_3"COOH" + color(orange)(59.9color(white)(l)"mol H"_2"O")) = color(blue)(0.0323#