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Depression in freezing point of 0.1 mola...

Depression in freezing point of 0.1 molal solution of HF is `-0.201^(@)C`. Calculate percentage degree of dissociation of HF. `(K_(f)=1.86 K kg mol^(-1))`.

Text Solution

Verified by Experts

We know
`DeltaT=ixxK_(f)xxm`
`0.201=ixx1.86xx0.1`
`i=1.0806`
The degree of dissociation HF may be calculated as
`alpha=(i-1)/(n-1)=(1.0806-1)/(2-1)=0.0806`
Percentage dissociation `= alphaxx100=0.0806xx100`
`=8.06`
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Depression in freezing of 0.10 molal solution of HF is -0.201^(@)C. Calculate the percentage degree of dissociation of HF. ( K_(f)="1.86 K kg mol"^(-1) ).

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Knowledge Check

  • The freezing point of 0.2 molal K_(2)SO_(4) is -1.1^(@)C . Calculate van't Hoff factor and percentage degree of dissociation of K_(2)SO_(4).K_(f) for water is 1.86^(@)

    A
    `97.5`
    B
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  • In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

    A
    274.674 K
    B
    271.60 K
    C
    273 K
    D
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  • The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

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    `2.7%`
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    In a 0.5 molal solution of KCl, KCl is 50% dissociated. The freezing point of solution will be (K_f = 1.86 "k kg mol"^(-1)) :

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