# On analysing a sample of water, a chemist finds that 100cm^3 of it contains 0.007g of calcium carbonate. What is the concentration of calcium carbonate in ppm?

Jan 11, 2017

$\text{1 ppm"-=} 1 \cdot m g \cdot {L}^{-} 1$

#### Explanation:

And thus $\text{concentration}$ $=$ $\text{Mass of solute"/"Volume}$.

$= \frac{7 \cdot m g}{0.100 \cdot L} = 70 \cdot m g \cdot {L}^{-} 1$

And thus $0.070 \cdot g \cdot {L}^{-} 1 \equiv 70 \cdot m g \cdot {L}^{-} 1 \equiv \text{70 ppm}$.

At these concentrations, we don't really have to worry too much about the density of the solution. $\text{70 ppm}$ denotes a concentration of $70 \cdot m g \cdot {L}^{-} 1$ with respect to $\text{calcium carbonate}$. Of course this will be a GREATER concentration than the concentration with respect to the metal ion. Why should this be so?

I think you can see why a concentration of $1 \cdot m g \cdot {L}^{-} 1$ is referred to as a $\text{part per million}$. A litre of water has a mass of $1000 \cdot g$, and each of these $\text{grams}$ is composed of a $1000 \cdot m g$ (i.e. $1 \cdot m g = {10}^{-} 3 g$. $1000 \times 1000 \equiv \text{million}$, hence $1 \cdot m g$ $\equiv$ $\text{1 part per million}$.