Question

32g of a sample of $FeS{O_4} \cdot 7{H_2}O$ were dissolved in dilute sulphuric acid and water its volume was made up to 1litre. $25mL$ of this solution required $20mL$ of $0.02M$ $KMn{O_4}$ solution for complete oxidation. Calculate the mass% of $FeS{O_4} \cdot 7{H_2}O$ in the sample.
A.$34.75$
B.$69.5$
C.$89.5$
D.None of these

Answer 1

We have to calculate the molarity of $FeS{O_4} \cdot 7{H_2}O$ first from the molarity of potassium permanganate and volume of $FeS{O_4} \cdot 7{H_2}O$. From the molarity, we have to calculate the moles of the solution using the volume of $FeS{O_4} \cdot 7{H_2}O$. From the moles of $FeS{O_4} \cdot 7{H_2}O$, we have to calculate the mass of $FeS{O_4} \cdot 7{H_2}O$ and later calculate the mass percentage of $FeS{O_4} \cdot 7{H_2}O$ in the sample.

Complete step by step solution:
Given data contains,
Mass of $FeS{O_4} \cdot 7{H_2}O$ in the sample is $32g$.
Volume of the solution is one litre.
Required volume of $FeS{O_4} \cdot 7{H_2}O$ is $25mL$.
Volume of $KMn{O_4}$ is $20mL$.
Molarity of $KMn{O_4}$ is $0.02M$.
The oxidation reaction of iron is written as,
$F{e^{2 + }} \to F{e^{3 + }}$
We can see that one electron is given up so the n factor for the oxidation reaction of iron is one.
Let us now determine the n factor for $KMn{O_4}$.
We know that the oxidation state of $Mn$in $KMn{O_4}$ is $+ 7.$
$Mn{O_4}^ - + 5{e^ - } + 8{H^ + } \to M{n^{2 + }} + 4{H_2}O$
We can see that five electrons are given, so the n factor of $KMn{O_4}$ is five.
Let us consider the molarity of $FeS{O_4} \cdot 7{H_2}O$ as M.
$25 \times M \times 1 = 20 \times 0.02 \times 5$
$M = \dfrac{{20 \times 0.02 \times 5}}{{25 \times 1}}$
On simplifying we get,
$\Rightarrow M = 0.08M$
So, the molarity of $FeS{O_4} \cdot 7{H_2}O$ is $0.08M$.
From the molarity and volume of $FeS{O_4} \cdot 7{H_2}O$, we can calculate the moles of $FeS{O_4} \cdot 7{H_2}O$.
We know that the formula to calculate the moles of $FeS{O_4} \cdot 7{H_2}O$ can be written as,
$Moles = Molarity \times Volume$
Let us now substitute the values of molarity and volume to get the moles of $FeS{O_4} \cdot 7{H_2}O$.
$Moles = Molarity \times Volume$
$Moles = 0.08\dfrac{{mol}}{{\not{L}}} \times 1\not{L}$
$Moles = 0.08mol$
Moles of $FeS{O_4} \cdot 7{H_2}O$ is $0.08mol$.
We can now calculate the mass of $FeS{O_4} \cdot 7{H_2}O$ in grams using the molar mass of moles of $FeS{O_4} \cdot 7{H_2}O$ and moles of moles of $FeS{O_4} \cdot 7{H_2}O$.
We know that $278g/mol$is molar mass of $FeS{O_4} \cdot 7{H_2}O$.
$g = 0.08\not{{mol}} \times \dfrac{{278g}}{{\not{{mol}}}}$
On simplifying we get,
$\Rightarrow g = 22.24g$
The mass of $FeS{O_4} \cdot 7{H_2}O$ is $22.24g$.
Let us now calculate the mass percentage.
We can calculate the mass percentage using the grams of $FeS{O_4} \cdot 7{H_2}O$ and total mass of the sample.
${\text{Mass percentage}} = \dfrac{{{\text{Grams of solute}}}}{{{\text{Grams of solution}}}} \times 100\%$
Let us now substitute the grams of the solute and grams of the solution.
${\text{Mass percentage}} = \dfrac{{{\text{Grams of solute}}}}{{{\text{Grams of solution}}}} \times 100\%$
Mass percentage$= \dfrac{{22.24g}}{{32g}} \times 100\%$
Mass percentage$= 69.5\%$
The mass percentage of $FeS{O_4} \cdot 7{H_2}O$ in the sample is $69.5\%$.
Therefore,option (B) is correct.

Note:
From molarity, we can also calculate the mass percentage, molality. An illustration of calculation of mass percentage from molarity is shown below.
Example:
We can calculate the concentration of the phosphoric acid expressed in mass percentage as below,
Given,
Molarity of the solution is $0.631M$
Density of the solution is $1.031g/ml$
The grams of the phosphoric acid are calculated using the molar mass.
Grams of phosphoric acid=$0.631mol \times \dfrac{{97.994g}}{{1mol}} = 61.8342g$
The mass of the solution is calculated from the density of solution
Mass of the solution=$1L \times \dfrac{{1.031g}}{{1ml}} \times \dfrac{{1000ml}}{{1L}} = 1031g$
The concentration of the solution is,
${\text{Mass percentage}} = \dfrac{{{\text{Grams of solute}}}}{{{\text{Grams of solution}}}} \times 100\%$
Concentration of the solution=$\dfrac{{61.8342g}}{{1031g}} \times 100\%$
Concentration of the solution =$5.9974\%$
The concentration of the solution expressed in mass percentage is $5.9974\%$.