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.
Answers
Answer:
here's your answer
Explanation:
From the given,
Molar mass of Fe(NO_{3})_{3}Fe(NO
3
)
3
= 241.86 g/mol
Density of Fe(NO_{3})_{3}Fe(NO
3
)
3
= 1.059 g/mol
w/w of Fe(NO_{3})_{3}Fe(NO
3
)
3
= 6.42%
Let's calculate the molarity of Iron(III) nitrate.
\bold{Molarity\,of\,Fe(NO_{3})_{3}}MolarityofFe(NO
3
)
3
:
Density = \frac{Mass}{Volume}=\frac{1.059}{10^{-3}L }Density=
Volume
Mass
=
10
−3
L
1.059
w/w=\frac{Given}{weight}=\frac{6.42}{100}w/w=
weight
Given
=
100
6.42
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}MolarityofFe(NO
3
)
3
=Density×w/w×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}=0.281MMolarityofFe(NO
3
)
3
=
10
3
1.059
×
100g
6.42
×
241.86g/mol
1molFe(NO
3
)
3
=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}}MolarconcentrationofNO
3
−
=Numberofnitrateions×molarityofFe(NO
3
)
3
= 3\times (0.281)M=0.843M=3×(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}=68gMassofFe(NO
3
)
3
=
1L
0.281mol
×
Mol
241.86
=68g
Therefore, mass in grams of Iron(III)nitrate contains 68 g of salt in each litre of solution.