# What is the concentration in "ppm" when a 0.6068 mass of solute is dissolved in a 100*mL of water?

Jul 24, 2017

We gots approx. $\text{6000 ppm}$

#### Explanation:

At these concentrations, the densities of the solution versus that of the pure solvent are tolerably close, so I am not going to consider the density.

We gots a concentration of $\frac{0.6068 \cdot g}{0.100 \cdot L} = 6.068 \cdot g \cdot {L}^{-} 1$

$= 6.068 \cdot g \cdot {10}^{3} \cdot m g \cdot {g}^{-} 1 \cdot {L}^{-} 1$

$= 6068 \cdot m g \cdot {L}^{-} 1$

$= 6068 \cdot \text{ppm}$, because $\text{1 ppm} \equiv 1 \cdot m g \cdot {L}^{-} 1$.