Question

A 4.55L sample of water contains .115g of sodium ions. Determine the concentration of sodium ions in ppm if the density of the solutions is 1.00g/mL.

Ok so ppm=mass of solute/mass of solution *10^-6

so we have the mass of the solute .115g
and for the solution we use 4.55 L and the given density and figure out grams. which I got 4550g

.115g/4550+.115*10^-6=2.53ppm
but the answer is supposed to be 25.3, I'm not sure what I'm doing wrong

Answer 1

Since the density is 1.00 g/mL, that means we can use 1.0 mL = 1.0 g, I remember that 1 mg/L = 1 ppm
So 0.115 g/4.55 L = 0.02527 g/L or that x 1000 = 25.27 mg/L or 25.3 ppm.

Your way:
0.115 g/4550 g + (.115 x 10^-6) =
0.115 g/4550 g + (negligible) =.
2.527 x 10^-5 g/1.0 g.
Then from g/g to g/10^6 g
2.527 x 10^-5 x 10^6/10^6 =
25.27 g/10^6 g = 25.27 ppm = 25.3 ppm
<b>I'm not sure what the 0.115 x 10^-6 does. If you are doing grams, then
0.115/4550 g + 0.115 g = 0.115g/4550 and you can choose to add the 0.115 or not. I think what you want is
0.115 x 10^6/(4550+0.115) = 25.3 ppm which does all of what I did above in one step.</b>