A 0.001 molal solution of a complex represented as $[P t (N H_{3})_{4} C l_{4}]$ in water had freezing point depression of ${0.0053}^{o} C$ . Given, $K_{f}$ for $H_{2} O = 1.86 K m o l a l i t y^{- 1}$ . Assuming 100% ionisation of the complex, write the ionisation nature and formula of complex.

[P t (N H_{3})_{4} C l_{2}] C l_{2} \to [P t (N H_{3})_{4} C l_{2}]^{2 +} 2 C l^{-}

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[P t (N H_{3})_{4} C l_{2}] C l_{2} \to [P t (N H_{3})_{4} C l_{2}]^{+} C l^{-}

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[P t (N H_{3})_{4} C l_{2}] C l_{2} \to 2 [P t (N H_{3})_{4} C l_{2}]^{+} C l^{-}

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[P t (N H_{3})_{4} C l_{2}] C l_{2} \to [P t (N H_{3})_{4} C l_{2}]^{3 +} 3 C l^{-}

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Solution

The correct option is D $[P t (N H_{3})_{4} C l_{2}] C l_{2} \to [P t (N H_{3})_{4} C l_{2}]^{2 +} 2 C l^{-}$
Depression in boiling point is given by

$△ T_{f} = i \times$ molality $\times K_{f}$

${.0053}^{o} C = i \times .001 \times 1.86$

$i = 3$

$∴$ The formula of compound is given by $[P t (N H_{3})_{4} C l_{2}] C l_{2}$

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