Question

A 5.09% (w/w) solution of Fe(NO3)3 (241.86 g/mol) has a density of 1.035 gmL. Calculate the molar analytical concentration of NO3 in the solution.

Answer 1

Let the total volume of the solution is 1000 mL.

Given that the density of the solution is 1.035 g/mL, total weight of the solution is

Weight = Volume × density = 1000 mL × 1.035 g/mL = 1035 g wez

Now, the weight percent (w/w) of the solution is 5.09%.

Weight percent is calculated as

weight of Fe (NO) weight of solution (w/w)% 3 x 100

Hence, total weight of in 1035 g of the solution can be calculated as follows

Fe NO3)3

(w/w)%-weight of Fe(NO3)3 5,09 weight of Fe(NO3)3- weight of solution 100 weight of Fe(NO3)3 1035 g x 100 5.09 x 1035 952.6815 9 100

Molar mass of = 241.86 g/mol

Fe NO3)3

Now, number of moles of present in the solution is

Fe NO3)3

52.68159 241.86 g/mol ะ 0.218 mol mass molar mass

Since, we assumed that the total solution volume is 1000 mL. Now we have 0.218 mol of
Fe NO3)3
in 1 L of solution. Thus the analytical concentration of
Fe NO3)3
is 0.218 mol/L.

Now, each mole of
Fe NO3)3
can dissociate into 3 moles of $NO_3^-$ according to following equation

Fe(NO3)3 F3NO

Hence, the analytical concentration of $NO_3^-$ must be thrice that of .

Fe NO3)3

Hence, the concentration of $NO_3^-$ in the solution is

3 × 0.218 mol/L = 0.654 mol/L