PPM Calculations

The average mug of tea holds 300 mL of water. If there are 4.2 grams of sugar in a teaspoon, what is the concentration of the sugar in parts per million?

Answer explanation

4.2 g x 1000000 / 300 g

= 14000 ppm

Contaminated water was found with 22 ppm of mercury within it. What would the mass of mercury in 100 mL of the water be?

Answer explanation

Rearrange equation to give:

ppm x mass of solution / 1,000,000 = mass of solute

22ppm x 100g / 1000000 = 2.2 x 10^-3, or 0.0022g

Sulfur dioxide is a pollutant gas which contributes towards acid rain. A 6000 dm³ sample found a volume of 3.7 dm³ of sulfur dioxide in the air. Calculate its concentration in ppm. Give your answer to 3 significant figures.

Answer explanation

3.7/6000 x 1000000

= 616.6666666....7

= 617 (3 s.f.)

Tags

NGSS.HS-PS1-2

NGSS.HS-PS1-5

A scientist collecting water samples found 1.024 grams of impurities in 10 litres of water. Calculate the impurity concentration in ppm.

Answer explanation

1.024/10,000 g x 1000000

= 102.4 ppm

Tags

NGSS.MS-PS1-2

Calculate the total volume of the sample if the concentration of CO₂ was 116 ppm, which occupied a volume of 2.4 dm³.

Give your answer to 4 significant figures.

Answer explanation

Rearrange equation first by multiplying both sides by mass of solution, then by dividing both sides by ppm to give:

mass of solution = mass of solute/ppm x 1000000

In this case we can swap volume for mass.

2.4 dm³ x 1000000 / 116 ppm = 20689.655

= 20690 dm³ (4 s.f.)

Calculate the mass of magnesium hydroxide in a 250 mL water solution where the Mg(OH)₂ concentration is 300 ppm.

Answer explanation

Rearrange equation by multiplying both sides by the mass of solution, and then by dividing both by 1000000.

300 x 250 / 1000000 = 0.075 g

Tags

NGSS.HS-PS1-5

NGSS.HS-PS1-7

Calculate the atmospheric concentration of NO₂ if 0.85 dm³ of it is found within a 120 dm³ sample of air.

Give your answer as an integer.

Answer explanation

ppm = 0.85 / 120 x 1000000

ppm = 7083.33333....

ppm = 7083 (to nearest integer)

The concentration of calcium ions in a water supply determines whether the water is hard (Ca²⁺ concentration > 120 ppm) or soft (Ca²⁺ concentration < 60 ppm). Concentrations between these values are moderate.

If 0.05 g of calcium ions were found in a 750 mL sample, what would the concentration be?

Answer explanation

concentration = 0.05 / 750 x 1000000

concentration = 66.6ppm

Tags

NGSS.HS-ESS2-5

NGSS.HS-PS1-5

The concentration of calcium ions in a water supply determines whether the water is hard (Ca²⁺ concentration > 120 ppm) or soft (Ca²⁺ concentration < 60 ppm). Concentrations between these values are moderate.

If 0.02 g of calcium ions were found in a 500 mL sample, what would the concentration be?

Answer explanation

concentration = 0.02 / 500 x 1000000

concentration = 40 ppm

Tags

NGSS.HS-ESS2-5

The concentration of calcium ions in a water supply determines whether the water is hard (Ca²⁺ concentration > 120 ppm) or soft (Ca²⁺ concentration < 60 ppm). Concentrations between these values are moderate.

If 0.004 g of calcium ions were found in a 10 mL sample, what would the concentration be?

Answer explanation

concentration = 0.004 / 10 x 1000000

concentration = 400 ppm

Tags

NGSS.HS-ESS2-5

NGSS.HS-PS1-5