Question

Question #36d85

Answer 1

`"0.14 g"`

Explanation:

The idea here is that a solution's parts per million concentration, ppm, tells you the number of grams of solute present in exactly

`10^6 = 1,000,000`

parts of solution. Since the mass of the solvent is much, much bigger than the mass of the solute, which I'll assume is the chromium(III) cation, `"Cr"^(3+)`, you can assume that the mass of the solution is equal to the mass of the solvent, i.e. of water.

You know that `"1 L"` of water has a mass of `"2.5 kg"`, so you can say that your solution will contain

`2.5 color(red)(cancel(color(black)("L"))) * (1.00color(red)(cancel(color(black)("kg"))))/(1color(red)(cancel(color(black)("L")))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 2.5 * 10^3` `"g"`

of water, the solvent. This means that you assume that the mass of the solution is equal to `2.5 * 10^3` `"g"`.

Now, your solution is `"57 ppm"` chromium(III) cations, which means that you get `"57 g"` of chromium(III) cations for every `10^6` `"g"` of solution.

You can thus say that your solution will contain

`2.5 * 10^3 color(red)(cancel(color(black)("g solution"))) * overbrace("57 g Cr"^(3+)/(10^6color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 57 ppm Cr"^(3+))) = color(darkgreen)(ul(color(black)("0.14 g Cr"^(3+))))`

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution and for its ppm concentration.