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A complex is prepared by mixing CoCl(3) ...

A complex is prepared by mixing `CoCl_(3)` and `NH_(3)` in the molar ratio of `1 : 4`. 0.1 M solution of this complex was found to freeze at `-0.372^(@)C`. What is the formula of the complex ? Given that molal depression constant of water `(K_(f))=1.86^(@)C//m`

Text Solution

AI Generated Solution

To determine the formula of the complex formed by mixing CoCl₃ and NH₃ in a 1:4 molar ratio, we will follow these steps: ### Step 1: Calculate the depression in freezing point (ΔTf) The depression in freezing point can be calculated using the formula: \[ \Delta T_f = K_f \times m \] where: ...
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A complex is prepared by mixing CoCl_(3) & NH_(3) 0.1 M solution of the complex was found to freeze at -0.372^(º)C . Total geometrical isomers of complex and x. (Molar depression constant of water = 1.86^(º)C//m ) Report your answer by multiplying x with 6.

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Knowledge Check

  • A complex is prepared by mixing CoCl_3 and Nh_3 , 0.1 M solution of the complex was found to freeze at -0.372^(@)C . The formula of the complex is _________. [Molar depression constant of water = 1.86^(@)C//m ]

    A
    `[Co(NH_3)_6]Cl_3`
    B
    `[Co(NH_3)_5Cl]Cl_2`
    C
    `[Co(NH_3)_4Cl_2]Cl`
    D
    `[Co(NH_3)_3Cl_3]`

  • What will be the freezing point of a 0.5 m KCl solution ? The molal freezing point constant of water is 1.86^@C m^(-1) .

    A
    `-1.86^@C`
    B
    `-0.372^@C`
    C
    `-3.2^@C`
    D
    `0^@C`