# Given a 6.174*g mass of sodium sulfate that is dissolved in a 1*L volume, what is the concentration in terms of "parts per million" with respect to sodium ion?

Jul 31, 2017

With respect to sodium ion, we have a $\text{2000 ppm}$ concentration.

#### Explanation:

$\text{1 ppm}$ specifies $1 \cdot m g$ of solute per litre of solution......

We has $\frac{\frac{6.174 \cdot g}{142.04 \cdot g \cdot m o {l}^{-} 1}}{1 \cdot L} = 0.0435 \cdot m o l \cdot {L}^{-} 1$ with respect to sodium sulfate....

And thus $2 \times 0.04346 \cdot m o l \cdot {L}^{-} 1 = 8.693 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$, with respect to sodium ion. Why did I double the concentration of sodium sulfate?

And thus, with respect to sodium ion, we have a concentration by mass of ........

$8.693 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1 \times 22.99 \cdot g \cdot m o {l}^{-} 1 = 2.000 \cdot g \cdot {L}^{-} 1$.

But $\text{1 ppm}$ $=$ $1 \times {10}^{-} 3 \cdot g \cdot {L}^{-} 1$, i.e. $1 \cdot m g \cdot {L}^{-} 1$, and thus we have.....

$2.000 \cdot g \cdot {L}^{-} 1 \times {10}^{3} \cdot m g \cdot {g}^{-} 1 = 2000 \cdot m g \cdot {L}^{-} 1 \equiv 2000$ $\text{ppm}$.