A 0.001 molal solution of Pt ( NH 3 ) 4 Cl 4 in water had a freezing point depression of 0.005 4 ∘ C . If K f for water is 1.80 , the correct formula for the above compound assuming its complete dissociation is

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A 0.001 molal solution of $Pt (NH_{3})_{4} Cl_{4}$ in water had a freezing point depression of $0.005 4^{\circ} C$ . If $K_{f}$ for water is 1.80 , the correct formula for the above compound assuming its complete dissociation is

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Answer:

Solution:

$Δ T_{f} = i \times K_{f} \times m$

$0.0054 = i \times 1.80 \times 0.001$

$i = 3$

$i = 1 + (n - 1) α$

$n = 3$

Therefore, the correct formula of the compound is $[Pt (N H_{3})_{4} C l_{2}] C l_{2}$ .

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Colligative Properties

⏱ 2mℹ️ Source: QB

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