Any one ????
The freezing point of an aqueous solution of KCN is 0.1892 mol Kg-1 of solvent was found to be -0.7040C. On adding 0.095 mol of Hg(CN)2, the freezing point of the solution was found to be -0.5300C. If the complex-formation takes place according to the following equation.
Hg(CN)2+nKCN = Kn[Hg(CN)n+2] Kf=1023
What is the formula of the complex? Kf(H2O) (molal freezing const)is 1.86 Kg mol-1 K
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4 Answers
gagar.iitk
·2008-12-28 04:08:15
n=2
try to write the eq
Hg(CN)2+nCN- = [Hg(CN)- n+2 ]
assume Hg(CN)2 is limiting reagent then u can calculate it
assuming CN- as limiting reagent will lead to u absurd ans
abhijeetkumar kumar
·2008-12-28 04:26:57
hey what is this molal freezing constant or just the freezing point!
0.1892 mol Kg-1!
as its unit is i think wrong!