If 50 mL of 0.5 M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is ______ g.

Equivalent of Oxalic acid = Equivalents of NaOH 50 x 0.5 x 2=25 x M x 1 \ M_ NaOH=2 M \ W_ NaOH in 50 ml=2 x 50 x 40 x 10^-3 g=4 g \

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If 50 mL of 05 M oxalic acid is required to neutralise ... | null (Chemistry) | DivineJEE

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Chemistry

Exam

Chemistry

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

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Correct Answer:

Explanation

Equivalent of Oxalic acid

=

Equivalents of

\mathrm{NaOH}

\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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