A 0.001 molal solution of [Pt(NH3)4cl4] in water had a freezing point depressi

Discussion Forum : Solutions

Question -

A 0.001 molal solution of $[P t (N H_{3})_{4} c l_{4}]$ in water had a freezing point depression of ${0.0054}^{\circ}$ . If $K_{f}$ for water is 1.80, the correct formulation for the above molecule is

Options:

A .

[P t (N H_{3})_{4} C l_{3}] C l

B .

[P t (N H_{3})_{4}] C l_{4}

C .

[P t (N H_{3})_{4} C l_{2}] C l_{3}

D .

[P t (N H_{3})_{4} C l_{2}] C l_{2}

Answer: Option D
:
D

$△ T = i K_{f} m$
$\Rightarrow i = \frac{△ T_{f}}{K_{f} m} = \frac{0.0054}{1.80 \times 0.001} = 3$
Thus complex produces three ions per molecule. hence possible correct formula is
$[P t (N H_{3})_{4} C l_{2}] C l_{2}$

Was this answer helpful ?

Next Question

Discussion Forum : Solutions

Submit Your Solution hear: