Question

Compound $PdC{l}_4\cdot 6H_{2}O$ is a hydrate complex; $1$ molal aqueous solution of it has freezing point of $269.28K$. Assming $100\%$ ionization of complex, calculate the molecular formula of the complex: $(K_f$ for water $=1.86K.kg/mol$ )

• A. $\left[Pd{\left(H_{2}O\right)}_{3}C{l}_3\right]Cl\cdot 3H_{2}O$
• B. $\left[Pd{\left(H_{2}O\right)}_6\right]C{l}_4$
• C. $[Pd\left(H_2{O}_{4}C{l}_3\right]C{l}_2\cdot 2H_{2}O$.
• D. $\left[Pd{\left(H_{2}O\right)}_{2}C{l}_4\right]\cdot 4H_{2}O$

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 $\left[Pt{\left(H_{2}O\right)}_{3}C{l}_3\right]Cl\cdot 3H_{2}O$.