Question

Compound $PdCl_{4}.6H_{2}O$ is a hydrated complex; 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex

( $K_{f}$ for water = 1.86 K kg $mol^{- 1}$ )

• A. $\left[ Pd \left( H _{2} O \right)_{6}\right] Cl _{4}$
• B. $\left[ Pd \left( H _{2} O \right)_{4} Cl _{2}\right] Cl _{2} \cdot 2 H _{2} O$
• C. $\left[ Pd \left( H _{2} O \right)_{3} Cl _{3}\right] Cl .3 H _{2} O$
• D. $\left[ Pd \left( H _{2} O \right)_{2} Cl _{4}\right] .4 H _{2} O$

Answer 1

Answer: (C)

$\Delta T=i\times K_{f}\times m$

$\left(\right.273-269.28\left.\right)=i\times 1.86\times 1$

$3.72=i\times 1.86$

$i=2$

$\alpha =\frac{i - 1}{n - 1}$

$1=\frac{2 - 1}{n - 1}$

or $n=2$

Thus, the complex should give two ions in the solution, i.e., the complex will be $\left[ Pd \left( H _{2} O \right)_{3} Cl _{3}\right] Cl .3 H _{2} O$.