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The freezing point of a 0.08 molal solut...

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)`

A

0.04

B

0.02

C

0.01

D

0.2

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The correct Answer is:
A
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The freezing point of a 0.08 molal aqueous solutions of NaHSO_(4) is 0.372^(@)C . Calculate the dissociation constant for the reaction HSO_(4)^(-)n hArr H^(+)+SO_(4)^(-2) Assuming no hydrolysis ( k_(t) of water = 1.86 k/kg mole)

The freezing point depression of a 0.109M aq. Solution of formic acid is -0.21^(@)C . Calculate the equilibrium constant for the reaction, HCOOH (aq) hArr H^(+)(aq) +HCOO^(Theta)(aq) K_(f) for water = 1.86 kg mol^(-1)K

Knowledge Check

  • The freezing point of a 0.05 molal solution of a non-electrolyte in water is [K_(f)=1.86K//m]

    A
    `-1.86^(@)C`
    B
    `-0.93^(@)C`
    C
    `-0.093^(@)C`
    D
    `0.93^(@)C`

  • 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`

  • The freezing point of 0.05 m solution of glucose in water is (K1 = 1.86°C m^(-1) )

    A
    0.093°C
    B
    -1.86°C
    C
    -0.093 °C
    D
    -0.93°C

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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) ).

    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)) .

    The freezing point depression of 0.1 m NaCI solution is 0.372^(@)C . What conlusion would you draw about the state in aqueous soluion ? (K_(f) "for water"=1.86 "K kg mol"^(-1)) .

    The freezing point depression of 0.1 m NaCl solution is 0.372^(@)C . What conclusion would you draw about its molecular state ? K_(f) for water is "1.86 K kg mol"^(-1) .

    The freezing point of 0.08 molal NaHSO_(4) is -0.345^(@)C . Calculate the percentage of HSO_(4)+O ions that transfers a proton to water. Assume 100% ionization of NaHSO_(4) and K_(t) for H_(2)O = 1.86 K molality^(-1) .

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