JEE MAIN - Chemistry (2024 - 29th January Evening Shift - No. 23)
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
4
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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