Question:
$50\, mL$ of $0.5\, M$ oxalic acid is needed to neutralize $25\, mL$ of sodium hydroxide solution. The amount of $NaOH$ in $50\, mL$ of the given sodium hydroxide solution is :
$50\, mL$ of $0.5\, M$ oxalic acid is needed to neutralize $25\, mL$ of sodium hydroxide solution. The amount of $NaOH$ in $50\, mL$ of the given sodium hydroxide solution is :
Updated On: Apr 3, 2026
- 40 g
- 20 g
- 80 g
- 4 g
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The Correct Option is D
Solution and Explanation
${H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O } $
${ m_{eq} \; of \; H_2 C_2O_4 = m_{eq} \; NaOH}$
$50 \times 0.5 \times 2 = 25 \times M_{NaOH} \times 1$
${ \therefore \; M_{NaOH} = 2 M } $
Now $1000 \,ml$ solution = $2$ $\times$ $40$ gram $NaOH$
$\therefore$ $50$ ml solution = $4$ gram $NaOH$
${ m_{eq} \; of \; H_2 C_2O_4 = m_{eq} \; NaOH}$
$50 \times 0.5 \times 2 = 25 \times M_{NaOH} \times 1$
${ \therefore \; M_{NaOH} = 2 M } $
Now $1000 \,ml$ solution = $2$ $\times$ $40$ gram $NaOH$
$\therefore$ $50$ ml solution = $4$ gram $NaOH$
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There are two ways of classifying the matter:
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