Question

A 5.85% (w/w) Fe(NO3)3 (241.86 g/mol) solution has a density of 1.059 g/mL. Calculate
(a) The molar analytical concentration of Fe(NO3)3 in this solution.
(b) The molar NO3 - concentration in the solution.
(c) The mass in grams of Fe(NO3)3 contained in each liter of this solution.

Answer 1

Let the Volume of Solution be 1 L = 1000mL

 

a)

Step 1: Finding mass of Solution from Density.

Density of Solution = Mass of Solution / Volume of Solution

Mass of Solution = Density x Volume = 1.059g/mL x 1000mL = 1059 g

 

Step 2: Finding Mass of Fe(NO3)3 from Mass % and thus Moles of Fe(NO3)3

Mass % of Component = (Mass of Component / Total Mass of Solution) x 100

Mass % of Fe(NO3)3 = (Mass of Fe(NO3)3 / Total Mass of Solution) x 100

6.42 = (Mass of Fe(NO3)3 / 1059g ) x 100

Mass of Fe(NO3)3 = (6.42 x 1059g) / 100 = 68g

Moles of Fe(NO3)3 = Mass of Fe(NO3)3 / Molar Mass of Fe(NO3)3 = 68g / 241.86g/mol = 0.2811 mol

 

Thus,

 

Solution contains 0.2811 moles of Fe(NO3)3 per Liter of Solution.

 

b) Molar Concentration of NO3-

Balanced Chemical Equation,

Fe(NO3)3 -----> Fe3+ + 3NO3+

From Stoichiometry and balanced Chemical Equation,

1 mole of Fe(NO3)3 dissociates to 3 moles of NO3+

No. Of moles of NO3+ = 3 x No. of Moles of Fe(NO3)3 = 3 x 0.2811 = 0.8433 mol

Molar Concentration of NO3+ = No. of Moles of NO3+ / Volume of Solution in L

Molar Concentration of NO3+ = 0.8433 mol / 1 L = 0.8433 mol/L = 0.8433 M

 

c)

As solved in Problem a)

Mass of Fe(NO3)3 contained in each Liter of Solution = 68 g/L

a. Solution contains 0.2811 moles of Fe(NO3)3 per Liter of Solution.

b. Molar Concentration of NO3+ = 0.8433 mol / 1 L = 0.8433 mol/L = 0.8433 M

c. Mass of Fe(NO3)3 contained in each Liter of Solution = 68 g/L