QJEEAdv22Ch27a7: JEE Advanced 2022 Chemistry - Class 11 JEEAdv22Ch27a7 Solution: The treatment of an aqueous solution of of with excess KI…

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QJEEAdv22Ch27a7: JEE Advanced 2022 Chemistry - Class 11 JEEAdv22Ch27a7 Solution: The treatment of an aqueous solution of of with excess KI…

The treatment of an aqueous solution of $3.74 \mathrm{~g}$ of $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ with excess KI results in a brown solution along with the formation of a precipitate. Passing $\mathrm{H}_{2} \mathrm{~S}$ through this brown solution gives another precipitate $\mathbf{X}$ . The amount of $\mathbf{X}$ (in $g$ ) is ___________.

[Given: Atomic mass of $\mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{~K}=39, \mathrm{Cu}=63, \mathrm{I}=127$ ]

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Accepted Answer

Number of moles of $C u (N O_{3})_{2} = \frac{3.74}{187} = 0.02$

$\underset{0.02}{2 C u {(N O_{3})}_{2}} + 4 K I \to C u_{2} I_{2} ↓ + \underset{0.01}{I_{2}} + 4 K N O_{3}$

$\underset{0.01}{I_{2}} + K I \to \underset{0.01 (B r o w n s o l u t i o n)}{K I_{3}}$

$\underset{0.01}{K I_{3}} + H_{2} S \to K I + \underset{0.01 (X)}{S} ↓ + 2 H I$

Number of moles of sulphur precipitated (X) = 0.01

Mass of sulphur precipitates (X) = 0.01 $\times$ 32 = 0.32 gm

QJEEAdv22Ch27a7: JEE Advanced 2022 Chemistry - Class 11 JEEAdv22Ch27a7 Solution: The treatment of an aqueous solution of of with excess KI…

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