Verified Solution • 2025–26 Session moderate

QJEEM24Ch37f0: JEE Main 2024 Chemistry - Class 11 JEEM24Ch37f0 Solution: If of oxalic acid is required to neutralise of solution, the…

If 50 mL50 \mathrm{~mL} of 0.5M0.5 \mathrm{M} oxalic acid is required to neutralise 25 mL25 \mathrm{~mL} of NaOH\mathrm{NaOH} solution, the amount of NaOH\mathrm{NaOH} in 50 mL50 \mathrm{~mL} of given NaOH\mathrm{NaOH} solution is ______ g.

Step-by-Step Verified Solution

Equivalent of Oxalic acid == Equivalents of NaOH\mathrm{NaOH}

amp;50×0.5×2=25×M×1amp;MNaOH=2Mamp;WNaOH in 50ml=2×50×40×10−3 g=4g\begin{aligned}& 50 \times 0.5 \times 2=25 \times \mathrm{M} \times 1 \\& \mathrm{M}_{\mathrm{NaOH}}=2 \mathrm{M} \\& \mathrm{W}_{\mathrm{NaOH}} \text { in } 50 \mathrm{ml}=2 \times 50 \times 40 \times 10^{-3} \mathrm{~g}=4 \mathrm{g}\end{aligned}

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