Question

Please help!!!
3.60 mL of "Unknown Solution" were pipetted into a 100 mL volumetric flask along with the molybdate and nitric acid solutions. This was diluted to volume, mixed and measured in the spectrometer.

The absorbance of this solution was 0.285
The concentration of this solution (from standard plot) was 0.0110 mM

Calculate:

e) Concentration of the unknown solution _______0.3056________ mM
f) Concentration of the unknown solution ___9.46____________ ppm P
g) Grams of "H3PO4" in 248 mL of unknown _______________ g
I just need help wiht the last part!

Answer 1

How do you do E & F? I'm stuck on those. From there, maybe I can help with letter G

Answer 2

for e 0.0110 times 100ml divided by 3.60 ml of unknown
for F you just take the concentration of the unknown solution and multiply by the 30.97 which is the MW of Phosphate

Answer 3

Thanks so much. I figured out G. First, you take your answer to letter E (0.3056 mM) and divide it by 1000 to get moles. Then take the volume of the unknown (248 mL) and divide it by 1000 to get liters. Now multiply those two numbers (moles times liters): the product gives you the moles of H3PO4. Take the moles of H3PO4 and multiply it by the 98 g/mol (MW of H3PO4). This gives you the grams of H3PO4 in the unknown

Answer 4

Thank you so much!!!!!!!!