Question

Compound $PdC{l}_4.6H_{2}O$ is a hydrated complex,1 molal aqueous solution of it has freezing point 269.28K. Assuming 100% ionization of complex, calculate the molecular formula of the complex.

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

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

Answer 1

The correct option is C $\left[Pd\right(H_{2}O{)}_{3}C{l}_3]Cl.3H_{2}O$

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

$\Delta T_f=273-269\cdot 28$

$\Delta T_f=3\cdot 72$

$3\cdot 72=1\cdot 86\times 1\times i$

$i=2$

$i=1+\alpha (n-1)=2$

$n-1=1$

$n=2$ (no. of ions)

In option [C], number of ions is two, hence, $\left[Pd\right(H_{2}O\left)C{l}_3\right]Cl.3H_{2}O$ is right complex.

Option C is correct.