# Question #856a9

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you need to use the volume and the mass of the **sample** to determine the **density** of the solution. This will then allow you to determine how much *solvent* you have in the original sample.

So, the density of a solution tells you the mass of **one unit of volume** of that solution. Since you're dealing with *milliliters*, you can say that the density of the sample, which is **the same as** the density of the **original solution**, will tell you the mass of

#1 color(red)(cancel(color(black)("mL"))) * "10.55 g"/(10color(red)(cancel(color(black)("mL")))) = "1.055 g"#

So, if

Use the density of the solution to figure out the mass of the **original sample**

#500color(red)(cancel(color(black)("mL solution"))) * overbrace("1.055 g"/(1color(red)(cancel(color(black)("mL solution")))))^(color(purple)("density")) = "527.5 g"#

You know that the original sample contains

#m_"solution" = m_"solvent" + m_"solute"#

#m_"solvent" = "527.5 g" - "3.25 g" = "524.25 g water"#

Next, use iron(II) chloride's **molar mass** to determine how many moles you'd get in that

#3.25 color(red)(cancel(color(black)("g"))) * "1 mole FeCl"_2/(126.75color(red)(cancel(color(black)("g")))) = "0.02564 moles FeCl"_2#

As you know, **molality** tells you how many *moles of solute* you get **per kilogram of solvent**.

#color(blue)(b = n_"solute"/m_"solvent")#

Use the fact that

#"1 kg" = 10^3"g"#

to calculate the molality of your solution

#b = "0.02564 moles"/(524.25 * 10^(-3)"kg") = color(green)("0.05 mol kg"^(-1))#

The answer can only have one **sig fig**, since that's how many sig fig you have for the volumes of the original solution and of the sample.

**ALTERNATIVE METHOD**

You can also find the molality of the solution by calculating how many moles of solute you get in that

Since you get

#10color(red)(cancel(color(black)("mL"))) * "0.02564 moles Fe Cl"_2/(500color(red)(cancel(color(black)("mL")))) = "0.0005128 moles FeCl"_2#

The mass of iron(II) chloride in this sample will be

#10color(red)(cancel(color(black)("mL solution"))) * "3.25 g Fe Cl"_2/(500color(red)(cancel(color(black)("mL solution")))) = "0.0650 g FeCl"_2#

The mass of water in this sample will be

#m_"water" = "10.55 g" - "0.0650 g" = "10.485 g"#

The molality will once again be

#b = "0.0005128 moles"/(14.485 * 10^(-3)"kg") = color(green)("0.05 mol kg"^(-1))#