Dear Student ,
Hypo solution( Na2S2O3 ) reacts with I2 according to the following rxn :
2 Na2S2O3 + I2 --------> 2NaI + Na2S4O6
xf = 1
N= 1/10, 15.7 mL
using , N= M * xf
0.1 = M *1
M= 0.1
Now, molarity (M) = no.of moles of solute / vol. of solution in L
0.1 = n * 15.7 * 10-3
n = 6.36 moles of Na2S2O3
According to the reaction moles of I2 required = 6.36 / 2 = 3.18 moles
now consider the eqbm given :
2H ----> H2 + I2
t=0 n = 0.96/ 128 – –
=0.0075
t= eqbm 0.0075 – 2x x x moles
and we know the value of x = 3.18 moles
Now, degree of dissociation = ( final moles / initial moles ) * 100
Hope this will help you.