A 0.001 molal solution of { Pt ( NH 3 ) 4 Cl 4 } in water had a freezing point depression of 0.0054°C. K f for water is 1.80 , the correct formulation for the above molecule is :

Question

A 0.001 molal solution of { Pt ( NH 3 ​ ) 4 ​ Cl 4 ​ } in water had a freezing point depression of 0.0054°C. K f ​ for water is 1.80 , the correct formulation for the above molecule is :

TestHub · Accepted Answer

Using Δ T f ​ = i K f ​ m : i = 1.80 × 0.001 0.0054 ​ = 3 Thus, the compound produces 3 ions in solution. [Pt(NH₃)₄Cl₂]Cl₂ dissociates as [ Pt ( N H 3 ​ ) 4 ​ C l 2 ​ ] 2 + + 2 C l − , giving 3 ions. ∴ Correct formulation: [Pt(NH₃)₄Cl₂]Cl₂ (Option B)

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