for 100 % dissociation of k4(fe(CN)6,van't Hoff factors i=5
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Solution
Dear student, For the 100 % dissociation , the Van't Hoff factor = Total no. of ions Now in K4 [ Fe(CN)6 ] there are 4 K+ ions and one [ Fe(CN)6 ]4- ion , thus total no. of ions = 5 and i = 5 Regards