posted by Anon on .
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?
•Chemistry - DrBob222, Wednesday, December 5, 2012 at 9:45pm
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.
•Chemistry - Anon, Thursday, December 6, 2012 at 12:38am
Don't you have times 100/25 by 1000mL/1L to get it to mg/L?
I still don't know what "in the option" means; however, if the concn is 15.8 mg/L in the measured sample, then multiplying by any dilution factor leaves the concn in mg/L and you do not need to multiply by anything.
But when you times (100mL/25mL) you get 4mL. Then you times 4mL times 15.8 mg/L and it doesn't cancel out the mL.
What in the world are you talking about?
15.8 mg/L x (100 mL/25 mL) = 15.8 mg/L x 4 = 63.2 mg/L. (100 mL/25 mL) is 100 mL DIVIDED by 25 mL and that is 4. The mL of 100 and the mL of 25 cancel to leave a pure number x 15.8 mg/L to leave units of mg/L).
The units are mg/L x (mL/mL) = mg/L.
Ok. I thought 100mL/25mL=4mL. Thanks
I finally caught on to what you're doing. Your error is you think (100 ml/4 ml) = 4 mL. It isn't. It is 4. The mL in the top cancels with the mL in the bottom to leave a unitless number of 4.00