H-2ユcalculate the percentage degree of dissociation of an electrolyt

1.	H-2ユcalculate the percentage degree of dissociation of an electrolyte XY2 (Normal molar mass164) inwater if the observed molar mass by measuring elevation in boiling point is 65.6
Answer» The compound XY2 will dissociate as XY2>>>>>X+2Y Thus, the Van't Hoff factor will be given by 1+2d where d is the degree of dissociation. But by definition, Van't Hoff factor is given by Normal mass/Observed mass. In this case, 164/65.6=2.5 Thus, by equating above two equations: 1+2d=2.5 which implies that d=0.75 Thus, percentage dissociation is 0.75×100=75% The answer is 75%

H-2ユcalculate the percentage degree of dissociation of an electrolyte XY2 (Normal molar mass164) inwater if the observed molar mass by measuring elevation in boiling point is 65.6

Answer»

The compound XY2 will dissociate as

XY2>>>>>X+2Y

Thus, the Van't Hoff factor will be given by 1+2d where d is the degree of dissociation.

But by definition, Van't Hoff factor is given by Normal mass/Observed mass. In this case, 164/65.6=2.5

Thus, by equating above two equations:

1+2d=2.5 which implies that d=0.75

Thus, percentage dissociation is 0.75×100=75%

The answer is 75%