Hi,
I'm having some problems with home work and i stuck.
1. How you would prepare 5 mL of a 75 ppm Cu solution from a 0.015 mol/L solution of copper (II) nitrate. Show full details of your working
I know that 75 ppm Cu is equal to 75 mg of CU per litre of solution, so in 5ml we have: 0.075 g L(-1) x 5 x 10(-3) = 0.375 x 10(-3)g
mollar mass of Cu(NO3)2 = 187,55 g/mol,
it's really confused me the information of 0.015 mol/L solution.
2. A 400 µL sample of a water sample from a polluted river was added to a 25 mL volumetric flask and
tested for phosphate using a colourimetric method by adding the required reagents and making to the
mark with water. The absorbance of the resulting solution at 880 nm was determined as 0.462.
Using the calibration plot provided below and showing full details of your working, calculate the
phosphate concentration in the original river water sample in ppm and molar units.
I would be happy if someone could explain that so simple that next time i can do everything myself.
Thank you very much for your help
Tom.
