Concentration 15.8 mg/L
Final Volume 100.0 mL
What is the concentration of phosphate in the original lake water sample if the unknown lake water sample in the option has been diluted by taking 25.0 mL of lake water and diluting it to the indicated final volume?
I don't know what "in the option" means. Is that 15.8 mg/L the concn in the 100 mL? If so then
15.8 x 100/25 = ? mg/L in the lake.
Don't you have times 100/25 by 1000mL/1L to get it to mg/L?
To find the concentration of phosphate in the original lake water sample, you need to use the information provided about the diluted sample:
Concentration of the diluted sample: 15.8 mg/L
Final volume of the diluted sample: 100.0 mL
The dilution factor is calculated by dividing the final volume by the initial volume. In this case, the initial volume is 25.0 mL, and the final volume is 100.0 mL.
Dilution factor = final volume / initial volume
Dilution factor = 100.0 mL / 25.0 mL
Dilution factor = 4
Now, to find the concentration of the original lake water sample, you can use the dilution factor and the concentration of the diluted sample.
Original concentration = diluted concentration * dilution factor
Original concentration = 15.8 mg/L * 4
Original concentration = 63.2 mg/L
Therefore, the concentration of phosphate in the original lake water sample is 63.2 mg/L.