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Dear Student , Hypo solution( Na2S2O3 ) reacts with I2 according to the following rxn : 2 Na2S2O3 + I2 --------> 2NaI + Na2S4O6 xf = 1N= 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.
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