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.
Answer (integer)
4
Solution
<p>Equivalent of Oxalic acid $=$ Equivalents of $\mathrm{NaOH}$</p>
<p>$$\begin{aligned}
& 50 \times 0.5 \times 2=25 \times \mathrm{M} \times 1 \\
& \mathrm{M}_{\mathrm{NaOH}}=2 \mathrm{M} \\
& \mathrm{W}_{\mathrm{NaOH}} \text { in } 50 \mathrm{ml}=2 \times 50 \times 40 \times 10^{-3} \mathrm{~g}=4 \mathrm{g}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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