Question

One molal aqueous solution of $PdC{l}_4.6H_{2}O$ has a 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${}^{-1}$]. The salt is a hydrated complex.

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

Answer 1

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

The depression in the freezing point $\Delta T_f=273.15-269.28=3.87$ K.

$\Delta T_f=i{K}_{f}m$

$3.87=i\times 1.86\times 1$

The vant hoff factor $i=2$

Thus, 1 molecule of $PdC{l}_4.6H_{2}O$ dissociates to give two ions.

Hence, the molecular formula of hydrated salt is $\left[Pd\right(H_{2}O{)}_{3}C{l}_3]Cl.3H_{2}O$

Hence, the correct option is $\text{B}$