The boiling point of 0.1 Molal K4[Fe(CN)6] solution will be ( given Kb for water = O.52°C kg/mol)

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Arun · Answer

Dear student K4 [Fe (CN)6] gives 5 ions(i) ∆Tb= i kb m= 5X 0.52 X0.1=0.26, Tb=100+0.26=100.26 °C Regards Arun (askIITians forum expert)

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The boiling point of 0.1 Molal K4[Fe(CN)6] solution will be ( given Kb for water = O.52°C kg/mol)

The boiling point of 0.1 Molal K4[Fe(CN)6] solution will be ( given Kb for water = O.52°C kg/mol)

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The boiling point of 0.1 Molal K4[Fe(CN)6] solution will be ( given Kb for water = O.52°C kg/mol)

The boiling point of 0.1 Molal K4[Fe(CN)6] solution will be ( given Kb for water = O.52°C kg/mol)

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