N2O4=2NO2 the observed molecular weight of N2O4 IS 80g/mol at 350K the

1.	N2O4=2NO2 the observed molecular weight of N2O4 IS 80g/mol at 350K the percentage of dissociation of N2O4 at 350 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 $t=t_{eq}$ $1-\alpha$ $2\alpha$ Van't Hoff factor for dissociation is given by: $i=\frac{\text{Normal Molecular <klux>MASS</klux>}}{\text{Observed Molecular mass}}$ For Association, i < 1 For dissociation, i > 1 Normal molecular mass of $N_2O_5$ = 92 g/mol Observed molecular mass of $N_2O_5$ = 80 g/mol 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: $\text{Percentage dissociation of }N_2O_4=\alpha \times 100$ $\text{Percentage dissociation of }N_2O_4=0.15\times 100=15\%$