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For a solution of 0.89 g of mercurous c...

For a solution of 0.89 g of mercurous chloride in 50 g of `HgCl_(2)(l)` the freezing point depression id `1.24^(@)`C. `K_(f)` for `HgCl_(2)` is 34.3 . What is the state of mercurous chloride in `HgCL_(2)`? (Hg-200, Cl-35.5).

A

as `Hg_(2)Cl_(2)` molecules

B

as "HgCl molecules"

C

as `Hg^(+)` and `Cl^(-)` ions

D

as `Hg_(2)^(2+)` and ` Cl^(-)` ions

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • For a solution of 0.849g of mercurous chloride in 50g of HgCl_(2)(l) the freezing point depression is 1.24^(@)C. K_(f) for HgCl_(2) is 34.3 . What is the state of mercurous chloride in HgCl_(2) ? (Hg-200,CI-35.5)

    A
    as `Hg_(2)Cl_(2)` molecules
    B
    as `HgCl` molecules
    C
    as `Hg^(+)` and `Cl^(-)` ions
    D
    as `Hg_(2)^(2+)` and`Cl^(-)` ions

  • When 40 g of substance is dissolved in 1000 g g of water, its freezing point is depressed by 1.86^@C. K_f for water is 1.86^@C mol^-1 then find the molar mass of the solute.

    A
    4
    B
    10
    C
    40
    D
    400

  • If a solution containing 0.072 g atom of sulphur in 100 g of a solvent (k_f=7.0) gave a freezing point depression of 0.84 @_C , the molecular formula of sulphur in the solutions is :

    A
    `S_6`
    B
    `S_7`
    C
    `S_8`
    D
    `S_9`

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    45 g of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 650 g of water. The freezing point depression (in K) will be (K_(f)"for water "=1.86 "K kg mol"^(-1)) .

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    The freezing point of an aqueous solution of 0.1 m Hg_(2)Cl_(2) will be : (If Hg_(2)Cl_(2) is 80% ionised in the solution to give Hg_(2)^(2+) and Cl^(-) :

    2.56 f of suphur in 100 g of CS_(2) has depression in freezing point of 0.010^(@) K_(f)=0.1^(@) (molal)^(-1) . Hence atomicity of sulphurin the solution is

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