Question

Question #856a9

Answer 1

`b = "0.05 mol kg"^(-1)`

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 mL"` of solution.

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

So, if `"1 mL"` of solution has a mass of `"1.055 g"`, it follows that the density of the solution is `"1.055 g mL"^(-1)`.

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 `"3.25 g"` of iron(II) chloride, `"FeCl"_2`. Since the mass of the solution is equal to the mass of the solvent, which in your case is water, plus the mas of the solute, you can say that the initial sample also 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 `"2.25-g"` sample

`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 `"10-mL"` sample.

Since you get `0.02564` moles of iron(II) chloride in the `"500-mL"` sample, you can say that the `"10-mL"` sample will contain

`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))`