Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

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

##"16 mol kg"^(-1)##

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

This means that in order to calculate , 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 ##"560-g"## sample by using the compound's **molar mass**.

In this case, acetone has a molar mass of ##"58.08 g mol"^(-1)##, which tells you that **one mole** of acetone has a mass of ##"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

##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 ##"0.620 kg"## of solvent hold a total of ##9.642## **moles** of acetone, it follows that ##"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 ##"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 .