Question

A complex is represented as $CoC{l}_3.xN{H}_3$ its 0.1 molal solution in aq. solution shows $\Delta T_f={0.558}^{0}C.K_f$ for $H_{2}O$ is $1.86$ $Kmo{l}^{-1}kg$. Assuming $100\%$ ionisation of complex and coordination no. of $Co$ is six, calculate formula of complex.

Answer 1

Given,

Molality (m) of complex $=0.1$ molal

$△T_f={0.558}^{o}C$

$K_f=1.86Kmo{l}^{-1}kg$

$\alpha =100$%

We know, for dissociation or association

$△T_f=i\times K_f\times m$ , where $i=$ van't Hoff factor

$\Rightarrow i=\frac{△T_f}{K_f\times m}$

$\Rightarrow i=\frac{0.558}{1.86\times 0.1}=3$

$\therefore i=3$

$\therefore$ Here dissociation occurs as $i>1$

Also, no. of moles of the complex after dissociation $=3$

$\therefore 3$ moles ions are liberated after the reaction.

$\Rightarrow CoC{l}_3.xN{H}_3⟶[CoCl.xN{H}_3{]}^{2+}+2C{l}^{-}$

As coordination number of $Co$ is $6$

$\Rightarrow x=6-1$ $[\because 1$ $Cl$ is present$]$

$\Rightarrow x=5$

$\therefore$ Complex is $\left[CoCl\right(N{H}_3{)}_5]C{l}_2.$