Question

A sample from a local stream is found to have 3.0 ppm nitrates (in the form of sodium nitrate). How many grams of sodium nitrate would there be in 2.5 L of the stream (assuming a density of 1.0 g/mL)?

Answer 1

Answer:

0.01028  grams of sodium nitrate would there be in 2.5 L of the stream.

Explanation:

The ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

The ppm concentration of nitrates = 3.0 ppm

Mass of nitrates = x

Mass of steam= m

Volume of steam = V = 2.5 L = 2500 ml ( 1 L = 1000 mL)

Density of steam = d = 1.0 g/mL

$M=d\times V=1.0 g/mL\times 2500 mL = 2500 g$

Putting values in above equation, we get:

$3.0=\frac{x}{2500 g}\times 10^6$

$x=0.0075 g$

Mass of nitrate = 0.0075 g

Moles of nitrate = $\frac{0.0075 g}{62 g/mol}=0.0001210 mol$

1 mole of nitrate ion is present in 1 mole of sodium nitrate.

Then 0.0001210 moles of nitrate will be present in :

$\frac{1}{1}\times 0.0001210 mol=0.0001210 mol$ of sodium nitrate;

Mass of 0.0001210 moles of sodium nitrate :

0.0001210 mol × 85 g/mol = 0.01028 g

0.01028  grams of sodium nitrate would there be in 2.5 L of the stream.