Question

Given,25.0 g of $FeS{{O}_{4}}.7{{H}_{2}}O$ was dissolved in water containing dilute ${{H}_{2}}S{{O}_{4}}$ and the volume was made up to 1.0 L. 25 mL of this solution required 20 mL of a $\dfrac{N}{10}\text{ KMn}{{\text{O}}_{4}}$ solution for complete oxidation. The percentage of $FeS{{O}_{4}}.7{{H}_{2}}O$ in the acid solution is:
(A) 78%
(B) 98%
(C) 89%
(D) 79%

Answer 1

To solve this question we first need to find out the molar mass of iron (II) sulfate heptahydrate ($FeS{{O}_{4}}.7{{H}_{2}}O$). The weight of one mole of a sample is known as its molar mass. The weight of one mole of an atom is known as its atomic mass. Its SI unit is g/mol.
The atomic mass of Fe = 55.845, the atomic mass of S = 32.065, the atomic mass of O = 15.9994, the atomic mass of H = 1.00794.
So, the molar mass of $FeS{{O}_{4}}.7{{H}_{2}}O$ will be

& {{M}_{FeS{{O}_{4}}.7{{H}_{2}}O}}={{M}_{Fe}}+{{M}_{S}}+(4\times {{M}_{O}})+(7\times [(2\times {{M}_{H}})+{{M}_{O}}]) \\\ & {{M}_{FeS{{O}_{4}}.7{{H}_{2}}O}}=55.845+32.065+(4\times 15.9994)+(7\times [(2\times 1.00794)+15.9994]) \\\ & {{M}_{FeS{{O}_{4}}.7{{H}_{2}}O}}=278.01456\text{ g/mol} \\\ \end{aligned}$$ **Complete answer:** We can see that 20 mL of $\dfrac{N}{10}\text{ KMn}{{\text{O}}_{4}}$ solution is required for complete oxidation of 25 mL solution of 1 L solution of 25.0 g of $FeS{{O}_{4}}.7{{H}_{2}}O$ dissolved in water containing dilute ${{H}_{2}}S{{O}_{4}}$. So, the milliequivalents of $FeS{{O}_{4}}.7{{H}_{2}}O$ in 25 mL = the milliequivalents of $\dfrac{N}{10}\text{ KMn}{{\text{O}}_{4}}$ used. The milliequivalents of $\dfrac{N}{10}\text{ KMn}{{\text{O}}_{4}}$ used = $20\text{ }mL\times \dfrac{N}{10}=2\text{ }milliequivalents$. Hence the milliequivalents of $FeS{{O}_{4}}.7{{H}_{2}}O$ in 25 mL = 2 milliequivalents. Now, the milliequivalents of $FeS{{O}_{4}}.7{{H}_{2}}O$ in 1000 mL = $\dfrac{1000}{25}\times 2=80\text{ }milliequivalents$. Now, we know that 1000 mL of the solution contains 80 milliequivalents of $FeS{{O}_{4}}.7{{H}_{2}}O$ i.e., $$\dfrac{{{m}_{FeS{{O}_{4}}.7{{H}_{2}}O}}}{\dfrac{{{M}_{FeS{{O}_{4}}.7{{H}_{2}}O}}}{n}}\times 1000=80$$ Where n = n-factor of $FeS{{O}_{4}}.7{{H}_{2}}O$ = 1. So the mass of $FeS{{O}_{4}}.7{{H}_{2}}O$ in 1000 mL solution will be $$\begin{aligned} & {{m}_{FeS{{O}_{4}}.7{{H}_{2}}O}}=\dfrac{80\times 278.01456}{1000} \\\ & {{m}_{FeS{{O}_{4}}.7{{H}_{2}}O}}=22.24\text{ g} \\\ \end{aligned}$$ Now, the percentage of $FeS{{O}_{4}}.7{{H}_{2}}O$ in the 25 mL solution will be $$\begin{aligned} & \Rightarrow \dfrac{{{m}_{FeS{{O}_{4}}.7{{H}_{2}}O}}}{{{V}_{sol}}}\times 100 \\\ & \Rightarrow \dfrac{22.24}{25}\times 100\cong 88.96 \\\ & \Rightarrow \cong 89 \\\ \end{aligned}$$ **So, the percentage of $FeS{{O}_{4}}.7{{H}_{2}}O$ in the acid solution is approximately option (C) 89%.** **Note:** It should be noted that we have used milliequivalents instead of milligrams to solve this question. We have done so because the acid solution of $FeS{{O}_{4}}.7{{H}_{2}}O$ in dilute ${{H}_{2}}S{{O}_{4}}$ of water is an electrolytic solution in which the molecules undergo dissociation and ions undergo oxidation and reduction.