Question

Question #7e852

Answer 1

Here's what I got.

Explanation:

I'm not sure what your solute is here, but I would guess that it's not hydrogen oxide, which is one of the names used for water, `"H"_2"O"`.

Maybe your solute is hydrogen peroxide, `"H"_2"O"_2`. Anyway, I'll just assume that you're dealing with a `"60-ppm"` solution of solute `"X"`.

Now, we use parts per million, or ppm, to express the concentrations of solutions that contain very, very small amounts, sometimes called trace amounts, of solute.

More specifically, we use parts per million to represent the number of grams of solute present for every

`10^6 = 1,000,000`

grams of solution. More often than not, the amount of solute is so small that you can approximate the mass of the solution with the mass of the solvent.

Assuming that water is the solvent and using `"1 g mL"^(-1)` as the density of water, you can say that a `"60-ppm"` solution of `"X"` will contain `"60 g"` of `"X"` for every `10^6` `"mL"` of solution.

By comparison, a solution's mass by volume percent concentration represents the number of grams of solute present for every `10^2` `"mL"` of solution.

This means that in order to find this solution's mass by volume percent concentration, you must determine the number of grams of `"X"` present in `10^2` `"mL"` of solution.

To do that, use the ppm concentration as a conversion factor

`10^2 color(red)(cancel(color(black)("mL solution"))) * overbrace("60 g X"/(10^6color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 60 ppm X")) = "0.006 g X"`

This means that the solution's mass by volume percent concentration is

`color(darkgreen)(ul(color(black)("% m/v = 0.006% X")))`

SIDE NOTE I would be remiss if I didn't use the fact that you mentioned hydrogen oxide as an excuse to share the famous dihydrogen monoxide hoax.

https://en.wikipedia.org/wiki/Dihydrogen_monoxide_hoax