Question

1. a sample has a nitrate concentration of 5.42 mg/l (5.42 ppm no₃⁻). calculate its molar concentration (in moles/l). (this primarily involves unit conversions, including using the molar mass of nitrate to convert mass to moles.) 3. a sample has a phosphate concentration of 0.00075 m. calculate the po₄³⁻ concentration in ppm.

Explanation:

Step1: Recall molar - mass of $PO_4^{3 - }$

The molar mass of $P$ is approximately $30.97\ g/mol$ and the molar mass of $O$ is approximately $16.00\ g/mol$. For $PO_4^{3 - }$, $M = 30.97+4\times16.00=30.97 + 64.00=94.97\ g/mol\approx95\ g/mol$.

Step2: Convert molarity to ppm

1 ppm = 1 mg/L. Given molarity $c = 0.00075\ M$. This means there are $0.00075\ mol$ of $PO_4^{3 - }$ per liter.
The mass of $PO_4^{3 - }$ in 1 L is $m=n\times M$, where $n = 0.00075\ mol$ and $M = 95\ g/mol$. So $m=0.00075\ mol\times95\ g/mol = 0.07125\ g$.
Since $1\ g=1000\ mg$, $m = 0.07125\times1000\ mg=71.25\ mg/L = 71.25\ ppm$.

Answer:

$71.25\ ppm$