[tomyoung]

Hi,

First i need say that i love the subject but sometimes it's to hard and i would be appreciated if someone can help me with below.
I'm having some homework to do and i'm really confused about how i should solve it.


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.

[Borek]

0.015 mol/L means 0.015 moles in 1 liter of solution, doesn't it? What is a mass of 0.015 moles of copper sulfate?

[tomyoung]

=0.015 *  187,55 ; where 187,55 it's mollar mass of copper nitrate.

Sorry but i dont know why you asking for copper sulfate.
It's any chance that you will be explain me both, i know it's pretty easy when you will get understand couple of them.

THank you.
Tom.

[Borek]

=0.015 *  187,55 ; where 187,55 it's mollar mass of copper nitrate.

OK. And what mass of copper is present in the same 1L of solution?

Sorry but i dont know why you asking for copper sulfate.

Sorry, my mistake. I thought about nitrate.

[tomyoung]

75 gm "I know that 75 ppm Cu is equal to 75 mg of CU per litre of solution".

it's right?

[tomyoung]

Ok, i it wasn't so hard i think and at the end using C1V1=C2V2 we did get the

953ppm x V1 = 75ppm x 0.005L

What about the 2nd one?

the ppm calculated from graph will equal = 0.728 and next i was thinking about using this same C1V1=C2V2 to get something like that

400*10(-6)*V1=25*10(-3)*0.728ppm = 45.5