At equilibrium N2o4=2no2 the observed molecular weight of n2o4 is 80 g

1.	At equilibrium N2o4=2no2 the observed molecular weight of n2o4 is 80 g at 350 k the %of dissociation of n2o4(g) at350 k is
Answer» $N_2O_5$ Answer: Percentage dissociation of for the given reaction is 15%. Explanation: We are given a chemical EQUATION: $N_2O_4\rightleftharpoons 2NO_2$ at t = 0 1 0 at $t=t_{eq}$ $1-\alpha$ $2\alpha$ Normal molecular mass of $N_2O_5$ = 92 g/mol Observed molecular mass of $N_2O_5$ = 80 g/mol Van't Hoff FACTOR for dissociation is given by: $i=\frac{\text{Normal Molecular mass}}{\text{Observed Molecular mass}}$ For ASSOCIATION, i < 1 For dissociation, i > 1 Putting values in above equation, we get: $i=\frac{92}{80}=1.15$ Expression for Van't Hoff equation is given by: $i=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$ $i=\frac{1-\alpha +2\alpha }{1}$ $i=\frac{1+\alpha }{1}$ where, i = Van't Hoff factor $\alpha$ = DEGREE of dissociation or association. $1.15=1-\alpha \\\alpha = 0.15$ $N_2O_4$ Percentage of dissociation for is given by: $\text{Percentage dissociation of }N_2O_4=\alpha \times 100=0.15\times 100=15\%$