Question

Compound $PdC{l}_4.6H_{2}O$ 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 $mo{l}^{-1}$ )

• A. $\left[Pd{\left(H_{2}O\right)}_6\right]C{l}_4$
• B. $\left[Pd{\left(H_{2}O\right)}_{4}C{l}_2\right]C{l}_2\cdot 2H_{2}O$
• C. $\left[Pd{\left(H_{2}O\right)}_{3}C{l}_3\right]Cl.3H_{2}O$
• D. $\left[Pd{\left(H_{2}O\right)}_{2}C{l}_4\right].4H_{2}O$

Answer 1

Answer: (C)

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

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

$3.72=i\times 1.86$

$i=2$

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

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

or $n=2$

Thus, the complex should give two ions in the solution, i.e., the complex will be $\left[Pd{\left(H_{2}O\right)}_{3}C{l}_3\right]Cl.3H_{2}O$.