A) \[{{K}_{4}}[Fe{{(CN)}_{6}}]\]
B) \[{{K}_{2}}[Fe{{(CN)}_{4}}]\]
C) \[{{K}_{3}}[Fe{{(CN)}_{6}}]\]
D) \[Fe{{\left( CN \right)}_{4}}\]
Correct Answer: C
Solution :
[c] \[\Delta {{T}_{f}}(normal)={{K}_{f}}m=1.86\times 0.5=0.93{}^\circ \]Assuming 100% ionization, i = n = No. of ions per molecule \[=\frac{Observed\,\Delta {{T}_{f}}}{Normal\,\Delta {{T}_{f}}}=\frac{3.72}{0.93}=4\] \[{{K}_{3}}[Fe{{(CN)}_{6}}]\xrightarrow{{}}3{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{3-}}\] No. of ions = 4You need to login to perform this action.
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