1.

1.0 gram of a monobasic acid HA in 100 gram H_2O lower the freezing point by 0.155 K.0.45 gram of same acid require 15 ml of 1/5M KOH solution for complete neutralisation.If the degree of dissociation of acid is alpha, then value of '20alpha' is : (K_f for H_2O =1.86 K.Kg/mole)

Answer»


SOLUTION :`DeltaT_F=K_Fxxm=K_Fxx[(W_("acid")xx1000)/(M_("acid")xxH_2O)]`
`(M_("acid")_(exp)=120`
Also `N_1V_1=N_2V_2`
`0.45/((M_("acid")_"thor"))=1/5xx15xx1/1000 IMPLIES (M_("acid")_("thor"))=(0.45xx5xx1000)/15=150`
`i=(M_("thor")/M_("exp"))=150/120=1.25 implies thereforealpha=(i-1)=1.25-1=0.25`


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