Question

A 0.001 molal solution of $\left[Pt\right(N{H}_3{)}_{4}C{l}_4]$ in water has a freezing point depression of ${0.0054}^{0}C$. If $K_f$ for water is 1.80 K.kg.mol${}^{-1}$, the correct formulation of the above molecule is:

• A. $\left[Pt\right(N{H}_3{)}_{4}C{l}_3]Cl$
• B. $\left[Pt\right(N{H}_3{)}_{4}C{l}_2]C{l}_2$
• C. $\left[Pt\right(N{H}_3{)}_{4}C{l}_2]C{l}_3$
• D. $\left[Pt\right(N{H}_3{)}_{4}C{l}_4]$

Answer 1

The correct option is B $\left[Pt\right(N{H}_3{)}_{4}C{l}_2]C{l}_2$

$\Delta T_f=i\times K_f\times m$

$⟹0.0054=i\times 1.80\times 0.001$.

$⟹i=\frac{0.0054}{1.80\times 0.001}=3$.

$⟹1+\alpha (n-1)=3$, where $\alpha =1$.

$⟹n-1=2$

$⟹n=3$

$⟹$ Number of ions in the complex should be 3.

So, option B is the correct answer.