Question

A 6.42%(w/w) Fe(NO3)3 (241.86 g/mol)solution has a density of 1.059g/mol .caculate.
The molar analytical concentration of Fe(NO3)3in this solution
The molar equilibrium concentration of NO3- in this solution
The mass in grams of Fe(NO3)3 contained in each litre of this solution

Answer 1

Molar equilibrium concentration of nitrate ions is 0.843 M.

Mass in grams of Iron(III)nitrate contains 68 g of salt in each liter of solution.

Step by step explanation:

From the given,

Molar mass of $Fe(NO_{3})_{3}$ = 241.86 g/mol

Density of $Fe(NO_{3})_{3}$ = 1.059 g/mol

w/w of $Fe(NO_{3})_{3}$ = 6.42%

Let's calculate the molarity of Iron(III) nitrate.

$\bold{Molarity\,of\,Fe(NO_{3})_{3}}$:

$Density = \frac{Mass}{Volume}=\frac{1.059}{10^{-3}L }$

$w/w=\frac{Given}{weight}=\frac{6.42}{100}$

one mole of Iron(III) nitrate contains 241.86 g of substance.

$\bold{Molarity of Fe(NO_{3})_{3}=Density\times w/w \times moles}$

$Molarity of Fe(NO_{3})_{3}=\frac{1.059}{10^{3}}\times \frac{6.42}{100g}\times \frac{1 mol\,Fe(NO_{3})_{3}}{241.86g/mol}=0.281M$

In Iron(III) nitrate have three nitrate ions.

$\bold{Molar\,concentration\,ofNO_{3}^{-}=Number\,of\,nitrate\,ions \times molarity\,of\,Fe(NO_{3})_{3}}$$= 3\times (0.281)M=0.843M$

Therefore, molar equilibrium concentration of nitrate ions is 0.843 M

One liter of Iron(III)nitrate contains 0.281 M of salt.

$Mass\,of\,Fe(NO_{3})_{3}=\frac{0.281\,mol}{1L}\times \frac{241.86}{Mol}=68g$

Therefore, mass in grams of Iron(III)nitrate contains 68 g of salt in each litre of solution.

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Answer 2

Answer:

Explanation:

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.