Question

49.50 grams of Fe(NO3)3 are dissolved in enough water to prepare 550.0 ml of solution

calculate the concentration of iron(III) nitrate in the solution in molarity.

how many milliliters of this solution would be required to supply 0.084 moles of nitrate ions

Answer 1

Molarity(M) = w/mwt*(1000/v in mL)

Mwt = molarmass of Fe(NO3)3 = 241.86 g/mol

w = mass of Fe(NO3)3 required = 49.5 g

v = volume of solution = 550 mL

   Molarity = (49.5/241.86)*(1000/550)

Molarity of Fe(NO3)3(M) = 0.372 M

1 mol Fe(NO3)3 = 3 mol NO3^-

Molarity of NO3^- = 0.372*3 = 1.116 M

volume of solution required = n/M = 0.084/1.116

                            = 0.0753 L

                            = 75.3 ml