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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

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The correct Answer is:
`8.06 %`
Observed `DeltaT_(f)=0.201^(@)C`. Calculated `DeltaT_(f)-K_(f)xxm=1.86xx0.1=0.186^(@)C" "therefore" "i=(0.201)/(0.186)=1.0806`
`{:(HF,hArr,H^(+),+,F^(-),),("1 mol",,"0",,"0",),(1-alpha,,alpha,,alpha",", "Total "=1+alpha.." Hence, i"=1+alpha" or "alpha=i-1=0.0806=8.06%):}`
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Depression in freezing point of 1.10-molal solution of HF is 0.201^(@)C . Calculate percentage degree of dissoviation of HF ( K_(f) =1.856 K kg mol^(-1) ).

0.15 molal solution of NaCI has freezing point -0.52 ^(@)C Calculate van't Hoff factor . (K_(f) = 1.86 K kg mol^(-1) )

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
    `90.75`
    C
    `105.5`
    D
    `85.75`

  • 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
    none of these

  • 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:-

    A
    `2.7%`
    B
    `30%`
    C
    `10%`
    D
    `5.2%`

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    0.01 molal aqueous solution of K_(3)[Fe(CN)_(6)] freezes at -0.062^(@)C . Calculate percentage dissociation (k_(f)=1.86)

    The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C . Calculate the dissociation constant for the reaction. K_(f) for water = 1.86 K m^(-1)

    Calculate the molality of NaCl solution whose elevation in boiling point is equal to the depression in freezing point of 0.25 m sodium carbonate solution in water assuming complete dissociation of salts. ( k_(f) = 1.86 K m^(-1) , k_(b) = 0.52 K m^(-1) )

    The freezing point of 0.08 molal aq NaHSO_(4) solution is -0.372 . Calculate alpha (degree of dessociation) of HSO_(4)^(-) given alpha (degree of dissociation) of NaHSO_(4) is 100% & K_(f) for H_(2)O=1.86 (K)/(m) . Give answer after multiplying by 10.

    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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