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.
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?
2 answers
Don't you have times 100/25 by 1000mL/1L to get it to mg/L?
0 answers