Redox Reactions Question 7
Question 7 - 2024 (29 Jan Shift 2)
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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Answer (4)
Solution
Equivalent of Oxalic acid $=$ Equivalents of $\mathrm{NaOH}$
$50 \times 0.5 \times 2=25 \times \mathrm{M} \times 1$
$\mathrm{M}_{\mathrm{NaOH}}=2 \mathrm{M}$
$\mathrm{W}_{\mathrm{NaOH}}$ in $50 \mathrm{ml}=2 \times 50 \times 40 \times 10^{-3} \mathrm{~g}=4 \mathrm{~g}$
