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JEE Main 202125 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Evening Shift
Subject
Chemistry
Chapter
Redox Reactions

JEE Main 2021 · 25 February · Evening Shift · Chemistry · Redox Reactions

Consider titration of NaOH solution versus 1 . 25M oxalic acid...

Consider titration of NaOHNaOH solution versus 1.25M1 . 25M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5mL4 . 5mL
(ii) 4.5mL4 . 5mL
(iii) 4.4mL4 . 4mL
(iv) 4.4mL4 . 4mL
(v) 4.4mL4 . 4mL
If the volume of oxalic acid taken was 10.0mL10 . 0mL then the molarity of the NaOHNaOH solution is ___MM. (Rounded-off to the nearest integer)

Correct answer

0

Step-by-step solution

VNaOH=4.4ml{V}_{NaOH} = 4 . 4ml

eq. of NaOH=NaOH = eq. of H2C2O4{H}_{2}{C}_{2}{O}_{4}

or, M×4.4×1=1.25×10×2M \times 4 . 4 \times 1 = 1 . 25 \times 10 \times 2

or, M=5.68MM = 5 . 68M

∴∴ Nearest integer answer is 66.

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