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JEE Advanced 2020Paper 1
Exam
JEE Advanced
Year
2020
Paper
Paper 1
Subject
Chemistry
Chapter
Redox Reactions

JEE Advanced 2020 · Paper 1 · Chemistry · Redox Reactions

5 . 00mL of 0 . 10M oxalic acid solution taken in a conical flask...

5.00mLof0.10M5 . 00mL of 0 . 10M oxalic acid solution taken in a conical flask is titrated against NaOHNaOH from a burette using phenolphthalein indicator. The volume of NaOHNaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOHNaOH solution?

Exp. No. Vol. of NaOH(mL)</td></tr><tr><tdstyle="text−align:center;">NaOH \left(mL\right)</td> </tr> <tr> <td style="text-align: center;">1</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">12 . 5</td></tr><tr><tdstyle="text−align:center;"></td> </tr> <tr> <td style="text-align: center;">2</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">10 . 5</td></tr><tr><tdstyle="text−align:center;"></td> </tr> <tr> <td style="text-align: center;">3</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">9 . 0</td></tr><tr><tdstyle="text−align:center;"></td> </tr> <tr> <td style="text-align: center;">4</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">9 . 0</td></tr><tr><tdstyle="text−align:center;"></td> </tr> <tr> <td style="text-align: center;">5</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">9 . 0

Correct answer

0.11

Step-by-step solution

True data for NaOHNaOH used volume =9ml= 9ml
Milli equivalent of H2C2O4={H}_{2}{C}_{2}{O}_{4} =Milli equivalent of NaOHNaOH

5×0.1×2=9××M×15 \times 0 . 1 \times 2 = 9 \times \times M \times 1

x=19=0.11Mx = \dfrac{1}{9} = 0 . 11M

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