JEE Advanced 2021 Paper 1 Chemistry: Paragraph 3

The amount of iodate is $n_{\mathrm{IO_3^-}}=\dfrac{0.1783}{214.0}=8.332\times10^{-4}\ \mathrm{mol}$. Each iodate gives three $\mathrm{I_2}$, so $n_{\mathrm{I_2}}=2.4995\times10^{-3}\ \mathrm{mol}$. Each $\mathrm{I_2}$ reacts with two thiosulfate ions, so $n_{\mathrm{S_2O_3^{2-}}}=4.999\times10^{-3}\ \mathrm{mol}$; effectively one iodate corresponds to six thiosulfate ions, an equivalence of $6\ \mathrm{mol}$ per mole of iodate. The molarity is $\dfrac{4.999\times10^{-3}}{0.02500}=0.1999\approx0.20\ \mathrm{mol\,L^{-1}}$. The common error is using only the 1:3 ratio between iodate and iodine and forgetting the 1:2 stoichiometry of the thiosulfate titration, which would give $0.10\ \mathrm{M}$ (or an equally wrong $0.60\ \mathrm{M}$ if the factor of six is applied twice). Always trace the stoichiometry through the intermediate iodine to avoid this.

In the second titration, two $\mathrm{Cu^{2+}}$ ions liberate one $\mathrm{I_2}$, and one $\mathrm{I_2}$ consumes two $\mathrm{S_2O_3^{2-}}$, so effectively one $\mathrm{Cu^{2+}}$ corresponds to one thiosulfate ion. The amount of thiosulfate used is $0.20\times25.60\times10^{-3}=5.12\times10^{-3}\ \mathrm{mol}$, so $n_{\mathrm{Cu}}=5.12\times10^{-3}\ \mathrm{mol}$. The mass of copper is $5.12\times10^{-3}\times63.5=0.3251\ \mathrm{g}$. The mass percentage in the $0.5000\ \mathrm{g}$ sample is $\dfrac{0.3251}{0.5000}\times100=65.02\%$. The remaining $34.98\%$ is zinc, which is typical of a common brass. The data are consistent because the zinc, being oxidised only by nitric acid and not interfering with iodide after neutralisation, does not affect the titre. Dividing the thiosulfate amount by two (treating the ratio as $\mathrm{Cu^{2+}}:\mathrm{S_2O_3^{2-}}=1:2$) is the usual mistake and would give only $32.5\%$.