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JEE MainChemistryToughEquivalent Concept

A 0.2 g sample containing copper (II) was analyzed
eudiometrically, where copper (II)is reduced tocopper (I) by iodide
ions. 2Cu2++4I−⟶2CuI+I22{Cu}^{2 +} + 4I^{-} \longrightarrow 2CuI + I_{2}
If 20 mL of 0.1 M Na2S2O3 solution is required for titration of the
liberated iodine, then the percentage of copper in the sample will be

Show answer and solution
2 moles of Cu2+ = 1 mole of I2
= 2 moles of hypo.
so moles of hypo used = 20 × 10-3 × 0.1 = 2 m moles = moles of copper
hence
%of copper=2×10−3×63.50.2×10%=63.5%\% of\ copper = \frac{2 \times 10^{- 3} \times 63.5}{0.2} \times 10\% = 63.5\%

Source: Grass-G / CH-16 Chemistry Paper 25 Nov. Fisrt.docx

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