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.
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.