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 :

• A. 40 g
• B. 20 g
• C. 80 g
• D. 4 g

Answer 1

Answer: (D)

4 g

Answer 2

${H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O }$
${ m_{eq} \; of \; H_2 C_2O_4 = m_{eq} \; NaOH}$
$50 \times 0.5 \times 2 = 25 \times M_{NaOH} \times 1$
${ \therefore \; M_{NaOH} = 2 M }$
Now $1000 \,ml$ solution = $2$ $\times$ $40$ gram $NaOH$
$\therefore$ $50$ ml solution = $4$ gram $NaOH$