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50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount (in g) of NaOH in 50 mL of the given sodium hydroxide solution is :

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\begin{aligned} &\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{H}_{2} \mathrm{O}\\ &\mathrm{m}_{\mathrm{eq}} \text { of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}=\mathrm{m}_{\mathrm{eq}} \mathrm{NaOH}\\ &50 \times 0.5 \times 2=25 \times M_{N a O H} \times 1\\ &\therefore \mathrm{M}_{\mathrm{NaOH}}=2 \mathrm{M}\\ &\text { Now } 1000 \mathrm{ml} \text { solution }=2 \times 40 \ \mathrm{gram} \ \mathrm{NaOH}\\ &\therefore 50 \mathrm{ml} \text { solution }=4 \ \mathrm{gram} \ \mathrm{NaOH} \end{aligned}

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