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) weighing 0.20g is dissolved in aqueous solution of H2SO4 and solution is titrated at 70∘C , requiring 45mL of 0.02M KMnO4 solution. The end point is overrun and back titration is carried out with 10mL of 0.1M oxalic acid solution. Find the % purity of Na2C2O4 in sample.
A. 75
B. 83.75
C. 90.25
D. None of these
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 Na2C2O4. 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 , then the reaction is:
Na2C2O4+H2SO4→Na2SO4+H2C2O4
Now, the solution is titrated with KMnO4 solution, that will give:
H2C2O4+KMnO4→Mn+2+K2O+2CO2+H2O
Therefore, the n-factor for Na2C2O4 is 2.
And, the n-factor for KMnO4 is 5.
Now,
Meq.ofNa2C2O4=Meq.ofKMnO4reacted =45×0.02×5−10×0.1×2 =2.5
Now,
WeightofNa2C2O4=n−factorofNa2C2O4Meq.ofNa2C2O4×134 =22.5×10−3×134 =0.1675gm
After obtaining the weight of the given compound, we can find the percentage purity of that compound:
∴%purity=(totalweightweightofNa2C2O4×100)% =0.20.1675×100=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 ×100%