JEE Main · 2024 · Shift-IeasySOL-080

We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1 M, 0.01 M and 0.001 M,…

Solutions · Class 12 · JEE Main Previous Year Question

Question

We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1 M, 0.01 M and 0.001 M, respectively. The value of van't Hoff factor for these solutions will be in the order.

Options
  1. a

    iA<iB<iCi_A < i_B < i_C

  2. b

    iA<iC<iBi_A < i_C < i_B

  3. c

    iA=iB=iCi_A = i_B = i_C

  4. d

    iA>iB>iCi_A > i_B > i_C

Correct Answera

iA<iB<iCi_A < i_B < i_C

Detailed Solution

Strategy:\n> The van't Hoff factor (ii) for a strong electrolyte like \ceNaCl\ce{NaCl} is theoretically constant (i=2i=2). However, as concentration increases, interionic attractions and ion-pair formation increase, which effectively reduces the number of free particles and thus decreases the actual measured ii.\n\nStep 1: Compare concentrations\n- Solution A: 0.1 M (Highest concentration)\n- Solution B: 0.01 M\n- Solution C: 0.001 M (Most dilute)\n\nStep 2: Relate concentration to dissociation/ion-pairing\n- In Solution C (very dilute), \ceNaCl\ce{NaCl} is most "freely" dissociated. ii is closest to 2.\n- In Solution A (more concentrated), ion interactions are strongest. ii is lowest.\n\nConclusion:\nThe van't Hoff factor value follows the order: iA<iB<iCi_A < i_B < i_C.\n\ntextAnswer:(1)\boxed{\\text{Answer: (1)}}

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We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1 M,… (JEE Main 2024) | Canvas Classes