# What is the molality of a solution of 560 g of acetone, CH_3COCH_3, in 620 g of water?

Apr 3, 2016

${\text{16 mol kg}}^{- 1}$

#### Explanation:

You can calculate a solution's molality by keeping track of how many moles of solute you get in one kilogram of solvent.

This means that in order to calculate molality, you essentially need to know

• how many moles of solute you have present
• the mass of solvent expressed in kilograms

Now, you can determine how many moles of acetone, ("CH"_3)_2"CO", you get in that $\text{560-g}$ sample by using the compound's molar mass.

In this case, acetone has a molar mass of ${\text{58.08 g mol}}^{- 1}$, which tells you that one mole of acetone has a mass of $\text{58.08 g}$.

This means that your sample contains

560 color(red)(cancel(color(black)("g"))) * "1 mole acetone"/(58.08color(red)(cancel(color(black)("g")))) = "9.642 moles acetone"

Now, your goal when finding molality is to find the number of moles of solute per kilogram of solvent. Convert the mass of water, which is your solvent, from grams to kilograms by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" = 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will get

620color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.620 kg"

So, if $\text{0.620 kg}$ of solvent hold a total of $9.642$ moles of acetone, it follows that $\text{1.0 kg}$ of solvent will hold

1 color(red)(cancel(color(black)("kg solvent"))) * "9.642 moles"/(0.620color(red)(cancel(color(black)("kg solvent")))) = "15.55 moles"

This means that your solution will have a molality of $\text{16 molal}$

"molality" = b = color(green)(|bar(ul(color(white)(a/a)"16 mol kg"^(-1)color(white)(a/a)|)))

The answer is rounded to two sig figs.