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In which case, van’t Hoff factor i remain unchanged ?
  • a)
    PtCl4 reacts with aq. KCl
  • b)
    aq. ZnCl2 reacts with aq. NH3
  • c)
    aq. FeCl3 reacts with aq. K4[Fe(CN)]6
  • d)
    KMnO4 reduced t o MnO2 in alkaline medium
Correct answer is option 'B'. Can you explain this answer?
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In which case, van’t Hoff factor i remain unchanged ?a)PtCl4 reacts w...
I Remains unchanged when number of ions before and after complex ion remains constant.
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In which case, van’t Hoff factor i remain unchanged ?a)PtCl4 reacts w...
Explanation:
The van't Hoff factor, denoted by "i", is a measure of the extent to which a solute dissociates in a solution. It is defined as the ratio of the moles of particles formed after dissociation to the moles of solute dissolved.

In the given options, we need to identify the case in which the van't Hoff factor remains unchanged. Let's analyze each option:

a) PtCl4 reacts with aq. KCl:
- In this case, PtCl4 reacts with KCl to form a complex ion, [PtCl6]2-, and KCl remains undissociated.
- The van't Hoff factor for PtCl4 is 1, as it does not dissociate, and the van't Hoff factor for KCl is 2, as it dissociates into K+ and Cl- ions.
- Therefore, the overall van't Hoff factor for this reaction is 1 + 2 = 3, which is not unchanged.

b) aq. ZnCl2 reacts with aq. NH3:
- In this case, ZnCl2 reacts with NH3 to form a complex ion, [Zn(NH3)4]2+, and NH3 remains undissociated.
- The van't Hoff factor for ZnCl2 is 3, as it dissociates into Zn2+ and 2 Cl- ions, and the van't Hoff factor for NH3 is 1, as it does not dissociate.
- Therefore, the overall van't Hoff factor for this reaction is 3 + 1 = 4, which is not unchanged.

c) aq. FeCl3 reacts with aq. K4[Fe(CN)]6:
- In this case, FeCl3 reacts with K4[Fe(CN)]6 to form a precipitate of Fe4[Fe(CN)6]3 and KCl remains undissociated.
- The van't Hoff factor for FeCl3 is 4, as it dissociates into Fe3+ and 3 Cl- ions, and the van't Hoff factor for K4[Fe(CN)]6 is 1, as it does not dissociate.
- Therefore, the overall van't Hoff factor for this reaction is 4 + 1 = 5, which is not unchanged.

d) KMnO4 reduced to MnO2 in alkaline medium:
- In this case, KMnO4 is reduced to MnO2 in an alkaline medium.
- Both KMnO4 and MnO2 do not dissociate, so their van't Hoff factors are both 1.
- Therefore, the overall van't Hoff factor for this reaction is 1 + 1 = 2, which remains unchanged.

Conclusion:
Among the given options, the case in which the van't Hoff factor remains unchanged is when KMnO4 is reduced to MnO2 in an alkaline medium (option d).
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In which case, van’t Hoff factor i remain unchanged ?a)PtCl4 reacts with aq. KClb)aq. ZnCl2 reacts with aq. NH3c)aq. FeCl3 reacts with aq. K4[Fe(CN)]6 d)KMnO4 reduced t o MnO2 in alkaline mediumCorrect answer is option 'B'. Can you explain this answer?
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In which case, van’t Hoff factor i remain unchanged ?a)PtCl4 reacts with aq. KClb)aq. ZnCl2 reacts with aq. NH3c)aq. FeCl3 reacts with aq. K4[Fe(CN)]6 d)KMnO4 reduced t o MnO2 in alkaline mediumCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In which case, van’t Hoff factor i remain unchanged ?a)PtCl4 reacts with aq. KClb)aq. ZnCl2 reacts with aq. NH3c)aq. FeCl3 reacts with aq. K4[Fe(CN)]6 d)KMnO4 reduced t o MnO2 in alkaline mediumCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In which case, van’t Hoff factor i remain unchanged ?a)PtCl4 reacts with aq. KClb)aq. ZnCl2 reacts with aq. NH3c)aq. FeCl3 reacts with aq. K4[Fe(CN)]6 d)KMnO4 reduced t o MnO2 in alkaline mediumCorrect answer is option 'B'. Can you explain this answer?.
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