Question

An impure sample of sodium oxalate $(N{a_2}{C_2}{O_4})$ weighing \[0.20g\] is dissolved in aqueous solution of \[{H_2}S{O_4}\] and solution is titrated at \[{70^\circ }C\] , requiring $45mL$ of $0.02M$ $KMn{O_4}$ solution. The end point is overrun and back titration is carried out with $10mL$ of $0.1M$ oxalic acid solution. Find the \[\% \] purity of $N{a_2}{C_2}{O_4}$ in sample.
A. \[75\]
B. $83.75$
C. $90.25$
D. None of these

Answer 1

Hint: To solve this question, first we will write the balanced reaction according to the question and then find the n-factor of both the given core compound. And then we will find the milli equivalent of $N{a_2}{C_2}{O_4}$. After that we will find their weight, and with the help of its weight, we can find the percentage purity.

Complete step by step answer:
According to the question, when the sodium oxalate is dissolved in aqueous solution of \[{H_2}S{O_4}\] , then the reaction is:
$N{a_2}{C_2}{O_4} + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}{C_2}{O_4}$
Now, the solution is titrated with $KMn{O_4}$ solution, that will give:
${H_2}{C_2}{O_4} + KMn{O_4} \to M{n^{ + 2}} + {K_2}O + 2C{O_2} + {H_2}O$
Therefore, the n-factor for $N{a_2}{C_2}{O_4}$ is 2.
And, the n-factor for $KMn{O_4}$ is 5.
Now,
$
  Meq.\,of\,N{a_2}{C_2}{O_4} = Meq.\,of\,KMn{O_4}\,reacted \\
= 45 \times 0.02 \times 5 - 10 \times 0.1 \times 2 \\
 = 2.5 \\
 $
Now,
$
Weight\,of\,N{a_2}{C_2}{O_4} = \dfrac{{Meq.\,of\,N{a_2}{C_2}{O_4}}}{{n - factor\,of\,N{a_2}{C_2}{O_4}\,}} \times 134 \\
= \dfrac{{2.5 \times {{10}^{ - 3}}}}{2} \times 134 \\
 = 0.1675gm \\
 $
After obtaining the weight of the given compound, we can find the percentage purity of that compound:
$
  \therefore \% \,purity = (\dfrac{{weight\,of\,N{a_2}{C_2}{O_4}}}{{total\,weight}} \times 100)\% \\
 = \dfrac{{0.1675}}{{0.2}} \times 100 = 83.75\% \\
 $

So, the correct answer is Option B.

Note: Percentage purity indicates the amount of pure and impure substance present in a sample. The percentage purity can be calculated by
% purity = Mass of pure substance in sampleMass of sample $ \times$100%