6.42% w/w Fe(NO3)3 (241.86 g/mol) density 1.059 g/mL

6.42% w/w Fe(NO3)3 (241.86 g/mol) density 1.059 g/mL

 

 

A 6.42% (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.

 

---------------------------------------------------

 

(a) the molar analytical concentration of Fe(NO3)3 in this solution.

 

몰농도 = (wt% / 몰질량) × 밀도 × 10

( 참고 https://ywpop.tistory.com/4241 )

 

= (6.42 / 241.86) × 1.059 × 10

= 0.281 M

 

 

 

 

(b) the molar NO3^- concentration in the solution.

 

Fe(NO3)3(aq) → Fe^3+(aq) + 3NO3^-(aq)

 

Fe(NO3)3 : NO3^- = 1 : 3 계수비(= 몰수비) 이므로,

[NO3^-] = 3 × 0.281 = 0.843 M

 

 

 

 

(c) the mass in grams of Fe(NO3)3 contained in each liter of this solution.

 

(0.281 mol/L) (1 L) (241.86 g/mol)

= (0.281) (1) (241.86) = 67.96 g

= 68.0 g