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The freezing point of an aqueous solution of KCN containing 0.189 mol/kg was -0.704 C .On adding 0.095 mol of Hg(CN)2, the freezing point of the solution bacame -0.53 C .What will be new 'i' factor of the resulting ? [assuming that the Hg(CN)2 and KCN produced the complex].

Asked by Balbir 30th July 2019, 4:06 PM
Answered by Expert
Answer:
Given:
 
Molality  of KCN solution = 0.1892 mol/kg
 
 
 
We know,
 
ΔTf = Kf × m × i
 
straight i equals space fraction numerator increment straight T subscript straight f over denominator straight K subscript straight f cross times straight m end fraction

space equals fraction numerator increment straight T subscript straight f over denominator straight K subscript straight f cross times straight m end fraction

equals fraction numerator 0.704 over denominator 1.86 cross times 0.1892 end fraction

space equals space 2.0
Answered by Expert 31st July 2019, 3:06 PM
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