JEE Main 2024 — 29 January, Shift 2 — Question 79

Oxalic acid: $\mathrm{H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O}$

Moles of oxalic acid used $\mathrm{= 0.050 \times 0.5 = 0.025\ mol}$

Equivalents of oxalic acid $\mathrm{= 2 \times 0.025 = 0.050\ mol}$

Hence, moles of $\mathrm{NaOH}$ in $25\ \mathrm{mL}$ $\mathrm{= 0.050\ mol}$

Molarity of $\mathrm{NaOH}$ $\mathrm{= \dfrac{0.050}{0.025} = 2.0\ M}$

Moles of $\mathrm{NaOH}$ in $50\ \mathrm{mL}$ $\mathrm{= 0.050 \times 2.0 = 0.10\ mol}$

Mass of $\mathrm{NaOH}$ $\mathrm{= 0.10 \times 40 = 4.0\ g}$