Question

An octahedral complex is prepared by mixing \[CoCl{}_3\]and \[NH{}_3\] in the molar ratio \[1:4,0.1m\] solution of this complex was found to freeze at \[0.372^\circ C\]. What is the formula of the complex? Given that molal depression \[(K{}_f)\] for water \[ = 1.86^\circ C/m\]

Answer 1

Hint: freezing point depression of nonelectrolyte solution has been determined equal to molality (m) of the solute times a proportionality constant is known as molal freezing depression constant of cryoscopic constant.

Formula used: \[\vartriangle T{}_f = K{}_f \times m\]

Complete step by step answer:
The observed value is twice of the theoretical value which means that the molecules of the given complex give out 2 ions on dissociation.
\[\vartriangle T{}_f = K{}_f \times m\]
As us use \[i\]as the value of \[K{}_f\]so we get,
\[0.372 = i \times m\]
\[m = 0.1\]
\[CoCl{}_2 + 4NH{}_3 \to Co(NH{}_3){}_4Cl\]
\[i = 2\]
Therefore, when the complex is dissociated in the solution it forms two ions.
Therefore, the complex can only be such that only one chlorine atom is outside.

Additional Information:
Cryoscopic constant is known to be used in calculating molar mass. In greek cryoscopy means freezing measurement. It is a colligative property so \[\vartriangle T\] depends on the number of solute particles which is dissolved. It is related to ebullioscopy, which helps in determination of the same value from the ebullioscopic constant (of boiling point elevation). It relates molality to freezing point depression

Note: Molality is defined as number of moles of solute present in one kg of solvent. K \[Kg/mole\] is the unit of molal depression constant. This depression constant does not depend on the nature of solute of but depends on the nature of solvent. Molal elevation is related to the elevation of boiling point and also molal depression constant.