# How to calculate the concentration of the diluted solution given a source concentration in ppm?

A student adds $37.4~\mathrm{ml}$ of a $50.0~\mathrm{ppm}$ solution to a $100~\mathrm{ml}$ volumetric flask and dilutes to the mark. What is the concentration of the diluted solution?

1. $13.7~\mathrm{ppm}$
2. $7.5~\mathrm{ppm}$
3. $37.4~\mathrm{ppm}$
4. $23.6~\mathrm{ppm}$

I used $$M_1V_1 = M_2V_2$$ which worked for all the others and I assumed would work for this considering I had all of the variables except for the 2nd concentration, but those calculations gave me $18.7~\mathrm{ppm}$, which is not an answer choice.

Let us assume for a moment, that the parts-per-notation is actually unambiguous. We then know that we can convert the concentration given in ppm $c^\dagger$ into the concentration $c$ in SI units $\mathrm{mol\,L^{-1}}$, or $c^\dagger\propto c$. Then we know that the concentration is defined as the amount of substance per volume of solution. $$c = \frac n V$$
When diluting a solution the amount of substance stays the same, hence we have two equations, that we can combine. \begin{align} c_0 &= \frac{n}{V_0} &&\text{source solution}\tag1\\ c_\mathrm{dil} &=\frac{n}{V_\mathrm{dil}} &&\text{diluted solution}\tag2\\ c_\mathrm{dil} &=\frac{c_0V_0}{V_\mathrm{dil}} &&\therefore\tag{1=2} \end{align}
Which is exactly the formula you used (when I assume that $M$ stands for molarity).
Ergo your calculation result, as orthocresol already pointed out, of $c^\dagger_\mathrm{dil} = 18.7~\mathrm{ppm}$ is correct.