PrepRung

JEE Advanced 2023 Paper 1 · Chemistry · Qualitative and Practical Chemistry · medium

A pale green crystalline salt X dissolves in water to give a pale green solution. The solution gives a deep blue precipitate with K3[Fe(CN)6] and a…

Single correct option

A pale green crystalline salt X\mathbf{X} dissolves in water to give a pale green solution. The solution gives a deep blue precipitate with K3[Fe(CN)6]\mathrm{K_3[Fe(CN)_6]} and a white precipitate with BaCl2\mathrm{BaCl_2} that is insoluble in dilute HCl\mathrm{HCl}. X\mathbf{X} contains seven molecules of water of crystallisation per formula unit. A sample of 2.78 g2.78\ \mathrm{g} of X\mathbf{X} is strongly heated; the solid residue is a red-brown oxide and the evolved gases, in which SO2\mathrm{SO_2} and SO3\mathrm{SO_3} are formed in equal amounts, are passed into an excess of acidified aqueous BaCl2\mathrm{BaCl_2}. Use molar masses (in g mol−1\mathrm{g\,mol^{-1}}) Fe=55.85\mathrm{Fe}=55.85, S=32.06\mathrm{S}=32.06, O=16.0\mathrm{O}=16.0, H=1.008\mathrm{H}=1.008, Ba=137.33\mathrm{Ba}=137.33. The masses of the solid residue and of the white precipitate formed in BaCl2\mathrm{BaCl_2} solution are respectively
  1. A. 0.80 g0.80\ \mathrm{g} and 2.33 g2.33\ \mathrm{g}
  2. B. 0.72 g0.72\ \mathrm{g} and 1.17 g1.17\ \mathrm{g}
  3. C. 1.60 g1.60\ \mathrm{g} and 2.33 g2.33\ \mathrm{g}
  4. D. 0.80 g0.80\ \mathrm{g} and 1.17 g1.17\ \mathrm{g}
Show answer and solution

Answer: (D) 0.80 g0.80\ \mathrm{g} and 1.17 g1.17\ \mathrm{g}

The blue precipitate with K3[Fe(CN)6]\mathrm{K_3[Fe(CN)_6]} shows Fe2+\mathrm{Fe^{2+}} (Turnbull's blue); the BaCl2\mathrm{BaCl_2}-insoluble white precipitate shows sulphate. So X=FeSO4⋅7H2O\mathbf{X}=\mathrm{FeSO_4\cdot7H_2O} with M=278.0 g mol−1M=278.0\ \mathrm{g\,mol^{-1}}, i.e. 2.78278.0=0.0100 mol\dfrac{2.78}{278.0}=0.0100\ \mathrm{mol}.
Heating: 2FeSO4→ΔFe2O3+SO2+SO3\mathrm{2FeSO_4\xrightarrow{\Delta}Fe_2O_3+SO_2+SO_3}. Moles of Fe2O3=0.0050\mathrm{Fe_2O_3}=0.0050, mass =0.0050×159.7=0.80 g=0.0050\times159.7=0.80\ \mathrm{g}. Moles of SO3=0.0050\mathrm{SO_3}=0.0050 and of SO2=0.0050\mathrm{SO_2}=0.0050.
In acidified BaCl2\mathrm{BaCl_2} only SO3\mathrm{SO_3} (giving SO42−\mathrm{SO_4^{2-}} in water) precipitates BaSO4\mathrm{BaSO_4} (M=233.4M=233.4): 0.0050×233.4=1.17 g0.0050\times233.4=1.17\ \mathrm{g}; SO2\mathrm{SO_2} gives only H2SO3\mathrm{H_2SO_3}, whose BaSO3\mathrm{BaSO_3} dissolves in acid.
The 2.33 g2.33\ \mathrm{g} values count both gases as BaSO4\mathrm{BaSO_4} (wrong); 0.72 g0.72\ \mathrm{g} corresponds to FeO\mathrm{FeO} (wrong residue); 1.60 g1.60\ \mathrm{g} forgets the factor of 2 in the balanced equation. Answer: (D).
Report a mistake

Original question written in the JEE Advanced 2023 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake