PrepRung
Sign inSign up free

JEE Advanced 2024 Paper 2 · Chemistry · Redox Reactions and Titrations · medium

JEE Advanced 2024 Paper 2 Chemistry: Paragraph 2

A 1.250 g1.250\ \mathrm{g} sample containing impure CuSO4⋅5H2O\mathrm{CuSO_4\cdot5H_2O} (the impurities are inert and non-reacting) is dissolved in water and an excess of KI\mathrm{KI} is added. The reaction 2Cu2++4I−→2CuI(s)+I2\mathrm{2Cu^{2+}+4I^-\to2CuI(s)+I_2} occurs quantitatively, and the liberated iodine requires 40.0 mL40.0\ \mathrm{mL} of 0.100 M0.100\ \mathrm{M} Na2S2O3\mathrm{Na_2S_2O_3} for the starch end point (I2+2S2O32−→2I−+S4O62−\mathrm{I_2+2S_2O_3^{2-}\to2I^-+S_4O_6^{2-}}). Atomic masses: Cu =63.5=63.5, I =127=127, S =32=32, O =16=16, H =1=1 (so the molar mass of CuSO4⋅5H2O\mathrm{CuSO_4\cdot5H_2O} is 249.5 g mol−1249.5\ \mathrm{g\,mol^{-1}}).

Question 1

Numerical answer

The percentage purity (w/w) of CuSO4⋅5H2O\mathrm{CuSO_4\cdot5H_2O} in the sample, correct to two decimal places, is
Show answer and solution

Answer: 79.84

Moles of thiosulfate =0.0400 L×0.100 M=4.00×10−3=0.0400\ \mathrm{L}\times0.100\ \mathrm{M}=4.00\times10^{-3} mol. Each mole of I2\mathrm{I_2} reacts with 2 mol thiosulfate, and 2 mol Cu2+\mathrm{Cu^{2+}} liberate 1 mol I2\mathrm{I_2}, so 2 mol Cu2+\mathrm{Cu^{2+}} correspond to 2 mol thiosulfate, i.e. n(Cu2+)=n(S2O32−)=4.00×10−3n(\mathrm{Cu^{2+}})=n(\mathrm{S_2O_3^{2-}})=4.00\times10^{-3} mol. Mass of CuSO4⋅5H2O=4.00×10−3×249.5=0.998\mathrm{CuSO_4\cdot5H_2O}=4.00\times10^{-3}\times249.5=0.998 g. Purity =0.9981.250×100=79.84%=\dfrac{0.998}{1.250}\times100=79.84\%.

Question 2

Numerical answer

The mass of CuI\mathrm{CuI} precipitated (in g), correct to two decimal places, is
Show answer and solution

Answer: 0.76

From the stoichiometry 2Cu2++4I−→2CuI+I2\mathrm{2Cu^{2+}+4I^-\to2CuI+I_2}, each mole of Cu2+\mathrm{Cu^{2+}} gives one mole of CuI, so n(CuI)=4.00×10−3n(\mathrm{CuI})=4.00\times10^{-3} mol. Molar mass of CuI =63.5+127=190.5 g mol−1=63.5+127=190.5\ \mathrm{g\,mol^{-1}}. Mass =4.00×10−3×190.5=0.762=4.00\times10^{-3}\times190.5=0.762 g, i.e. 0.760.76 g to two decimal places. (Iodide serves both as precipitant of Cu(I) and as reductant, so only half of the iodide ends up as I2\mathrm{I_2}.)
Report a mistake

Original question written in the JEE Advanced 2024 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake