JEE Main · 2019 · Shift-IIeasySOL-110

K2HgI4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:

Solutions · Class 12 · JEE Main Previous Year Question

Question

K2HgI4\mathrm{K_2HgI_4} is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:

Options
  1. a

    1.6

  2. b

    1.8

  3. c

    20

  4. d

    22

Correct Answerb

1.8

Detailed Solution

Strategy:\n> Identify the number of ions produced by the dissociation of the complex and then apply the van't Hoff formula (i=1+(n1)αi = 1 + (n-1)\alpha) using the given degree of ionisation.\n\nStep 1: Identify ions (nn)\n\ceK2HgI42K++[HgI4]2\ce{K2HgI4 \rightleftharpoons 2K+ + [HgI4]^2-}\nThe complex ion behaves as a single unit. Total ions = 2+1=32 + 1 = 3.\n\nStep 2: Calculate } i \text{\nα=40%=0.40\alpha = 40\% = 0.40.\ni=1+(31)0.40=1+2(0.40)=1.8i = 1 + (3 - 1)0.40 = 1 + 2(0.40) = 1.8\n\ntextAnswer:(2)\boxed{\\text{Answer: (2)}}

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K2HgI4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is: (JEE Main 2019) | Canvas Classes