Question

What volume of 0.4M \[FeC{l_3} \cdot 6{H_2}O\] will contain 600mg of \[F{e^{3 + }}\]?
(A) 49.85mL
(B) 26.785mL
(C) 147.55mL
(D) 87.65mL

Answer 1

Hint: Formula to find molarity is \[M = \dfrac{{{\text{Weight of solute(gm)}} \times {\text{1000}}}}{{{\text{Molecular weight of solute}} \times {\text{Volume of solution(mL)}}}}\]
There is one atom of Fe present in one molecule of \[FeC{l_3} \cdot 6{H_2}O\]. Atomic weight of Fe is 55.85\[gmmo{l^{ - 1}}\]

Complete answer:
From the molecular formula of salt \[FeC{l_3} \cdot 6{H_2}O\], we can see that there is one atom of \[F{e^{3 + }}\] is present in one molecule.
Now, Atomic weight of Fe atom is 55.85\[gmmo{l^{ - 1}}\]
Molecular weight of \[FeC{l_3} \cdot 6{H_2}O\] will be = Atomic weight of Fe + 3(Atomic weight of Cl) + 6(Molecular weight of water)
= 55.85 + 106.5 + 108
So, Molecular weight of \[FeC{l_3} \cdot 6{H_2}O\] = 270.35\[gmmo{l^{ - 1}}\]
Now, we can write that
If there is 55.85gm of \[F{e^{3 + }}\] is present in 270.35gm of \[FeC{l_3} \cdot 6{H_2}O\] salt,
then 0.6gm(600mg) of \[F{e^{3 + }}\] will be present in \[\dfrac{{0.6 \times 270.35}}{{55.85}}\]=2.9043gm of \[FeC{l_3} \cdot 6{H_2}O\] salt.
So, we can say that 600mg of \[F{e^{3 + }}\] ions will be present in 2.9043gm of \[FeC{l_3} \cdot 6{H_2}O\] salt.
Let’s find the volume of the solution we will require of 0.4M \[FeC{l_3} \cdot 6{H_2}O\] that has 2.9043gm of dissolved salt.
We know that Molarity of the solution, \[M = \dfrac{{{\text{Weight of solute(gm)}} \times {\text{1000}}}}{{{\text{Molecular weight of solute}} \times {\text{Volume of solution(mL)}}}}\]
We are given that,
\[0.4 = \dfrac{{2.9043 \times {\text{1000}}}}{{270.35 \times {\text{Volume of solution(mL)}}}}\]
\[{\text{Volume of solution(mL)}} = \dfrac{{2904.3}}{{0.4 \times 270.35}}\]
Volume of the solution = 26.8568mL
So, we can say that 600mg of \[F{e^{3 + }}\] will be present in 26.8568mL of 0.4M \[FeC{l_3} \cdot 6{H_2}O\] solution.
So, the correct answer is (B) 26.785mL.

Additional Information:
Sometimes solving these types of problems, when molecular weight or atomic weights of some species are not given in the question, the value taken by us and supposed by them may vary slightly. That may result in slight change in the final answer as well. But we can easily identify this small change and find the right answer easily.

Note: Do not forget to include six water molecules into the calculation of the molecular weight of the salt. Do not put 600mg as it is in the formula of molarity. We are required to put the weight of solute in grams.