Question

The freezing point of 0.08 molal aqueous soln of NaHSO4 is -. 372°C.The dissociation constant for the following reaction is (Kf for H2O is 1.86)
HSO4^- ---> H^+ + SO4^2- ​

Answer 1

Answer:

Explanation:

ΔTf = -0.372oC =  0.372K

Kf = 1.86 K Kg/mol

 Molality of NaHSO4 = 0.08 m

 ΔTf = iKf x m

0.372.  = i x1.86x 0.08

i=2.5

HSO4 →→  H+ + SO4-

c(1-α)        cα      cα

Degree of dissociation ,α

i=1+α

​2.5 = 1+ α

α =1.5