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JEE Main 202127 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Evening Shift
Subject
Chemistry
Chapter
Redox Reactions

JEE Main 2021 · 27 July · Evening Shift · Chemistry · Redox Reactions

10 . 0mL of 0 . 05M KMnO _ 4 solution was consumed in a titration...

10.0mL10 . 0mL of 0.05MKMnO40 . 05M{KMnO}_{4} solution was consumed in a titration with 10.0mL10 . 0mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is ........ ×10−2g/L\times {10}^{- 2}g / L. (Round off to the nearest integer)

Correct answer

0

Step-by-step solution

neqKMnO4=neqH2C2O4⋅2H2O{n}_{eq}{KMnO}_{4} = {n}_{eq}{H}_{2}{C}_{2}{O}_{4} \cdot 2{H}_{2}O

or, 10×0.051000×5=10×M1000×2\dfrac{10 \times 0 . 05}{1000} \times 5 = \dfrac{10 \times M}{1000} \times 2

∴∴ Conc. of oxalic acid solution =0.125M= 0 . 125M

=0.125×126g/L=15.75g/L= 0 . 125 \times 126g / L = 15 . 75g / L

=1575×10−2g/L= 1575 \times {10}^{- 2}g / L

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