23. A teaspoon of salt has a mass of approximately 8.0 g. If this quantity is added to a swimming pool that is 25 meters long, 15 meters wide, and has an average depth of 2.0 meters, how many parts per million (ppm) of salt have been added to the contents of the pool? Assume that the density of water in the pool is 1.00 g/mL.

Hint: [tex]1 \text{ m}^3 = 1000 \text{ L}[/tex]



Answer :

To determine the number of parts per million (ppm) of salt added to a swimming pool, we can follow a systematic approach step-by-step.

1. Determine the volume of the pool:
- Given:
- Length ([tex]\( L \)[/tex]) = 25 meters
- Width ([tex]\( W \)[/tex]) = 15 meters
- Average depth ([tex]\( D \)[/tex]) = 2.0 meters

The volume of the pool ([tex]\( V \)[/tex]) in cubic meters ([tex]\( \text{m}^3 \)[/tex]) is calculated by multiplying these dimensions:

[tex]\[ V = L \times W \times D \][/tex]

[tex]\[ V = 25 \, \text{m} \times 15 \, \text{m} \times 2 \, \text{m} = 750 \, \text{m}^3 \][/tex]

2. Convert the volume from cubic meters to liters:
- Recall that 1 cubic meter ([tex]\( 1 \, \text{m}^3 \)[/tex]) is equivalent to 1000 liters ([tex]\( \text{L} \)[/tex]).

[tex]\[ \text{Volume in liters} = 750 \, \text{m}^3 \times 1000 \, \text{L/m}^3 = 750,000 \, \text{L} \][/tex]

3. Calculate the total mass of water in the pool:
- Given:
- Density of water = 1.0 g/mL (which is the same as 1000 g/L since [tex]\(1 \, \text{mL} = 1 \, \text{cm}^3 \)[/tex])

The mass of water ([tex]\( \text{Mass}_{\text{water}} \)[/tex]) is calculated by multiplying the volume of water in liters by its density (g/L):

[tex]\[ \text{Mass}_{\text{water}} = 750,000 \, \text{L} \times 1000 \, \text{g/L} = 750,000,000 \, \text{g} \][/tex]

4. Determine parts per million (ppm) of salt in the water:
- Given:
- Mass of salt ([tex]\( \text{Mass}_{\text{salt}} \)[/tex]) = 8.0 g

Ppm is calculated using the formula:

[tex]\[ \text{ppm} = \left( \frac{\text{Mass}_{\text{salt}}}{\text{Mass}_{\text{solution}}} \right) \times 10^6 \][/tex]

Here, the solution is the entire mass of water in the pool:

[tex]\[ \text{ppm} = \left( \frac{8.0 \, \text{g}}{750,000,000 \, \text{g}} \right) \times 10^6 \][/tex]

Simplifying the fraction:

[tex]\[ \frac{8.0}{750,000,000} \approx 1.0666666666666666 \times 10^{-8} \][/tex]

Converting it to ppm:

[tex]\[ \text{ppm} = 1.0666666666666666 \times 10^{-8} \times 10^6 = 10.67 \][/tex]

Therefore, the number of parts per million of salt added to the swimming pool contents is approximately 10.67 ppm.