Question

Compound $PdCl _{4} \cdot 6 H _{2} O$ is a hydrate complex; $1$ molal aqueous solution of it has freezing point of $269.28\, K$. Assming $100 \%$ ionization of complex, calculate the molecular formula of the complex: $\left(K_{f}\right.$ for water $=1.86\, K . kg / mol$ )

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

Answer 1

Answer: (A)

$\Delta T=i \times K_{f} \times m$
$(273-269.28)=i \times 1.86 \times 1$
$3.72=i \times 1.86$
$i=2$
$\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, that is,
the complex will be $\left[ Pt \left( H _{2} O \right)_{3} Cl _{3}\right] Cl \cdot 3 H _{2} O$.