Question

Question #48627

Answer 1

`"316 g FeCl"_3`

Explanation:

The idea here is that you need to convert the volume of the sample to mass by using the solution's known density.

Notice that the density of the solution is expressed in grams per milliliter, `"g mL"^(-1)`, but that the volume of the sample is given to you in liters.

In order to be able to use the density to find the mass of the solution, you must convert the volume of the sample to milliliters. You can convert liters to milliliters by using the conversion factor

`color(blue)(ul(color(black)("1 L" = 10^3"mL")))`

In your case, the sample will have a volume of

`2.50 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = 2.50 * 10^3"mL"`

Now, the density of the solution tells you the mass of unit of volume of solution. More specifically, a density of `"1.149 g mL"^(-1)` means that every `"1 mL"` of this solution has a mass of `"1.129 g"`.

This means that your sample will have a mass of

`2.50 * 10^3 color(red)(cancel(color(black)("mL"))) * "1.149 g"/(1color(red)(cancel(color(black)("mL")))) = "2872.5 g"`

In order to find the mass of iron(III) chloride present in this sample, you must use the fact that the solution is `11%` iron(III) chloride by mass.

This means that every `"100 g"` of solution contain `"11 g"` of iron(III) chloride. In your case, the sample will contain

`2872.5 color(red)(cancel(color(black)("g solution"))) * "11 g Fe Cl"_3/(100color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("316 g FeCl"_3)))`

I'll leave the answer rounded to three sig figs, but keep in mind that you only have two significant figures for the percent concentration of the solution.