Question

A. Calculate the molarity of a solution containing 53.6 mg of Iron (iii) nitrate dissolved in 50.0 mL of water. Assume the volume of water equals the volume of solution.

I got 4.43 x 10^-3 M

B. Calculate the mass percent of Iron(iii) nitrate in the solution described in A. Assume the solution has a density of 1.00g/mL

I got 0.1%

C. Convert your answer in part C to parts of Iron(iii) nitrate per million parts solution,

I got 1072

Answer 1

1. Moles of Fe(NO3)3 = mass ( in grams ) / molar mass

= 53.6*10^-3 / 242

= 0.000221488 moles

Molarity = moles / volume ( in litre)

= 0.000221488 / ( 50*10^-3)

= 4.43*10^-3 M

2. Mass of water = density * volume

= 1.00*50 = 50 grams

Total mass of Solution = 50 + 0.0536 = 50.0536 grams

Mass % = mass of Fe(NO3)3 / total mass of Solution *100

= (0.0536 / 50.0536)*100

= 0.1071%

3. 1% = 10000 ppm

0.1071% = 1071 ppm

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