Solveeit Logo

Question

Question: An impure sample of sodium oxalate \((N{a_2}{C_2}{O_4})\) weighing \[0.20g\] is dissolved in aqueous...

An impure sample of sodium oxalate (Na2C2O4)(N{a_2}{C_2}{O_4}) weighing 0.20g0.20g is dissolved in aqueous solution of H2SO4{H_2}S{O_4} and solution is titrated at 70C{70^\circ }C , requiring 45mL45mL of 0.02M0.02M KMnO4KMn{O_4} solution. The end point is overrun and back titration is carried out with 10mL10mL of 0.1M0.1M oxalic acid solution. Find the %\% purity of Na2C2O4N{a_2}{C_2}{O_4} in sample.
A. 7575
B. 83.7583.75
C. 90.2590.25
D. None of these

Explanation

Solution

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 Na2C2O4N{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 H2SO4{H_2}S{O_4} , then the reaction is:
Na2C2O4+H2SO4Na2SO4+H2C2O4N{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 KMnO4KMn{O_4} solution, that will give:
H2C2O4+KMnO4Mn+2+K2O+2CO2+H2O{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 Na2C2O4N{a_2}{C_2}{O_4} is 2.
And, the n-factor for KMnO4KMn{O_4} is 5.
Now,
Meq.ofNa2C2O4=Meq.ofKMnO4reacted =45×0.02×510×0.1×2 =2.5  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,
WeightofNa2C2O4=Meq.ofNa2C2O4nfactorofNa2C2O4×134 =2.5×1032×134 =0.1675gm  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:
%purity=(weightofNa2C2O4totalweight×100)% =0.16750.2×100=83.75%  \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 × \times100%