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JEE Advanced · Chemistry · 1. Mole Concept

The treatment of an aqueous solution of 3.74 g of \(\text{Cu} \left(\text{NO} _3\right)_2\) with excess KI results in a brown solution along with the formation of a precipitate. Passing \(\text{H} _2 \text{S}\) through this brown solution gives another precipitate X . The amount of X (in g ) is _____________ [Given: Atomic mass of \(\text{H} =1, \text{N}=14, \text{O} =16, \text{S}=32, \text{K}=39,\) \(\text{Cu} =63, \text{I} =127]\)

  1. A 0.32
  2. B 0.33
  3. C 0.34
  4. D 0.35
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Answer & Solution

Correct Answer

(A) 0.32

Step-by-step Solution

Detailed explanation

Number of moles of CuNO32=3.74187=0.02 2CuNO32+4KICu2I2+I2+4KNO3
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