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

##### 1 Answer

#### 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 manymoles of soluteyou have presentthemass of solventexpressed inkilograms

Now, you can determine how many moles of acetone, **molar mass**.

In this case, acetone has a molar mass of **one mole** of acetone has a mass of

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(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 **moles** of acetone, it follows that

#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

#"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**.