Home
Class 12
CHEMISTRY
If the freezing point of a 0.01 molal aq...

If the freezing point of a `0.01` molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is `-0.0558^(@)C`, the number of chloride (s) in the coordination sphere of the complex if `[K_(f)` of water `=1.86 K kg mol^(-1)]`

Text Solution

Verified by Experts

The correct Answer is:
1

`Delta T_(f) = I K_(f) m`
Given m = 0 .01 molal `Delta T_(f) = 0 - (0.0558^(@)) = 0.0558^(@)C`
`k_(f) = 1.86 K kg mol^(-1)`
`I = (Delta T_(f))/(k_(f) xx m) = (0.0558)/(1.86 xx 0.01) = 3`
Since three ions are produced by the complex , the molecular formual of the complex is `[Co(NH_(3))_(5)Cl]Cl_(2)` .
Thus, the number of `Cl^(-1)` ions in the coordination sphere is only one .
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|14 Videos
  • SOLUTIONS

    MODERN PUBLICATION|Exercise COMPETITION FILE (MARTIX MATCH TYPE QUESTIONS)|2 Videos
  • SOLID STATE

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|13 Videos
  • SURFACE CHEMISTRY

    MODERN PUBLICATION|Exercise COMPETITION FILE (OBJECTIVE TYPE QUESTIONS) (MATCHING LIST TYPE QUESTIONS)|2 Videos

Similar Questions

Explore conceptually related problems

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

The freezing point of a 0.05 molal solution of a non-electrolyte in water is: ( K_(f) = 1.86 "molality"^(-1) )

What will be the freezing point of 0.2 molal aqueous solution of MgBr_(2) ? If salt dissociates 40% in solution and K_(f) for water is 1.86 KKg mol^(-1)

The lowering in freezing point of 0.75 molal aqueous solution NaCI (80% dissociated) is [Given, K_f , for water 1.86 K kg mol^(-1)]