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0.1 M K(4)[Fe(CN)(6)] is 50% ionised in ...

0.1 M `K_(4)[Fe(CN)_(6)]` is 50% ionised in aqueous medium. What will be its van't Hoff factor ?

Text Solution

Verified by Experts

The correct Answer is:
3
`alpha=(i-1)/(n-1)`
`0.5=(i-1)/(5-1)`
`i=3`]
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Knowledge Check

  • The electrolyte solutions show abnormal colligative porperties.To account for this effect we define a quantity called the Van't Hoff factor given by i=("Actual number of particles in solution after dissociation")/("Number of formula units initially dissolved in solution") i=1 ("for non-electrolytes") igt1 ("for electrolytes, undergoing dissociation") ilt1 ("for solutes, undergoing association") Answer the following questions: 0.1 M K_(4)[Fe(CN)_(6)] is 60% ionized. What will be its Van't Hoff factor?

    A
    `1.4`
    B
    `3.4`
    C
    `2.4`
    D
    `4.4`

  • If van't Hoff factor i=1,then

    A
    it is dissociation
    B
    It is association
    C
    Both 1 and 2
    D
    Neither dissociation nor association

  • K_(2)HgI_(4) is 55% ionized in aqueous solution. The value of Van't Hoff factor is

    A
    2.1
    B
    4.3
    C
    1.9
    D
    3.7

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    If Hg_(2)Cl_(2) is 100% ionised, then find out its van't Hoff factor.

    K_(2)HgI_(4) is 50% ionised in aqueous solution. Which of the following are correct ?

    K_(2)HgI_(4) is 50% ionised in aqueous solution. Which of following are correct ?

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