Question

$3 \cdot 92{\text{ g}}$ of ferrous ammonium sulphate (FAS) reacts completely with ${\text{50 ml}}$ $\dfrac{{\text{N}}}{{10}}{\text{ }}KMn{O_4}$ solution. The percentage purity of the sample is:
(a) ${\text{50}}$
(b) $78 \cdot 4$
(c) ${\text{28}}$
(d) ${\text{40}}$

Answer 1

Hint: Percentage purity of a substance in a sample is calculated by dividing the mass of the pure compound by the total mass of the sample given, then multiplied by ${\text{100}}$ to make it a percentage.

Complete step by step answer:
(1) Ferrous ammonium sulphate (FAS) is commonly known as Mohr’s salt. It has chemical formula ${\left( {N{H_4}} \right)_2}Fe{\left( {S{O_4}} \right)_2} \cdot 6{H_2}O$, and $KMn{O_4}$ is Potassium permanganate. It is given in the question that ferrous ammonium sulphate (FAS) is reacted with Potassium permanganate. It is a redox reaction which occurs during the titration of Mohr’s salt with $KMn{O_4}$.
(2) In the given titration, FAS acts as a reducing agent, and $KMn{O_4}$ acts as an oxidising agent. In this titration, no external indicator is used because $KMn{O_4}$ acts as a self-indicator. $KMn{O_4}$ is dark purple in colour, at the endpoint of the titration, it changes into pink if the solution is colourless initially.
(3) Since, it is a titration, and it is given in the question that FAS reacts completely with $KMn{O_4}$ solution, which means that number of gram equivalents of FAS will be equal to the number of gram equivalents of $KMn{O_4}$.
(4) Number of gram equivalents is given by the formula: $N \times V$, where ${\text{N}}$ is the normality of the solution, and ${\text{V}}$ is the volume of the solution.
Equivalent of $KMn{O_4}$ used will be = Normality of $KMn{O_4}$ solution $ \times $ Volume of the $KMn{O_4}$ solution.
Volume of $KMn{O_4}$ given is ${\text{50 ml = }}\dfrac{{50}}{{1000}}{\text{ L}}$
Normality of $KMn{O_4}$ is given $\dfrac{N}{{10}}{\text{ = }}0 \cdot 1{\text{ N}}$
So, number of gram equivalents of $KMn{O_4}$ = $\dfrac{{0 \cdot 1 \times 50}}{{1000}}$
$ \Rightarrow $ number of gram equivalents of $KMn{O_4}$ = $0 \cdot 005{\text{ gram equivalents}}$
(5) By the above explanation at point number (3), the number of gram equivalents of FAS will be equal to $0 \cdot 005{\text{ gram equivalents}}$.
So, weight of FAS required = $0 \cdot 005 \times 392$
$ \Rightarrow $ weight of FAS required = $1 \cdot 96{\text{ g}}$
(6) Now, we have to find the percentage purity of FAS using the formula:
Percentage purity = $\dfrac{{{\text{Mass of pure FAS in the sample}}}}{{{\text{Total mass of the sample}}}} \times 100$
Given mass of the sample is $3 \cdot 92{\text{ g}}$.
$ \Rightarrow $ Percentage purity = $\dfrac{{1 \cdot 96}}{{3 \cdot 92}} \times 100$
$ \Rightarrow $ Percentage purity = $50\% $

Hence, option (a) ${\text{50}}$ is the correct answer.

Note:
Purity, and yield are two different things. Yield is the amount of substance obtained from the reaction, and purity is the actual amount of substance in a sample. Percentage yield is given by:
${\text{Percentage yield = }}\dfrac{{{\text{Actual yield}}}}{{{\text{Theoretical yield}}}} \times 100\% $