Question

A 10.69% (weight/weight) Fe(NO_​3​)_​3 (241.86 g/mol) solution has a density of 1.059g solution/mL solution. Calculate:

a.) The molar concentration of ​Fe(NO_​3​)_​3 in this solution.

b.) The mass in grams of Fe(NO_​3​)_​3 contained in one liter of this solution.

c.) The molar concentration of NO_​3​^​- in the solution.

Answer 1

a)
consider 1000 mL of solution
Volume = 1000 mL * 1L
density = 1.059 g/mL
mass of solution = density * volume
= 1.059*1000
= 1059 g

mass of Fe(NO3)3 = 10.69% of 1059
=10.69*1059/100
=113.21 g

molar mass = 241.86 g/mol
number of moles = mass/molar mass
= 113.21/241.86
=0.4681 mol
volume = 1 L
Molar concentration = number of moles/volume
= 0.4681 M

b)
already calculated above, mass = 113.21 g

c)
1 molecule of Fe(NO3)3 has 3 molecule of NO3-
[NO3-] = 3*[Fe(NO3)3]
= 3*0.4681 M
=1.404 M