$\mathrm{Co}{({\mathrm{NH}}_{3})}_{\mathrm{x}}{\mathrm{Cl}}_{3}$ has 0.1 molal. 100% dissociation$∆{\mathrm{T}}_{\mathrm{f}}=0.558({\mathrm{K}}_{\mathrm{f}}=1.86)$. Then formula of the compound is (JEE mains memory based shift-1 23/01/2025)

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Correct answer: c) $\left[\mathrm{Co}{({\mathrm{NH}}_{3})}_{5}\mathrm{Cl}\right]C{l}_{2}$. ΔTf = Kf ⋅ m ⋅ i and i = n ( 100% dissociation) Where: •ΔTf = Freezing point depression •Kf = Freezing point depression constant •m = Molality of the solution •i = Van’t Hoff factor

\(\mathrm{Co}{({\mathrm{NH}}_{3})}_{\mathrm{x}}{\mathrm{Cl}}_{3}\) has 0.1 molal. 100% dissociation\(∆{\mathrm{T}}_{\mat…

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