Let the molality of NaHSO4 solution be m, then assuming molarity = molality.
HSO−4 ⟶H+ + SO−24
m(1−α) mα mα
i=1+1+α=2+α
Effective molality =[Na+]+i[HSO−4]
=m+m(1+α)=m(2+α)
△Tf=Kfim
⇒0.372=1.86×m(2+α)
⇒0.2=m(2+α)
Also, Ka=mα21−α=0.4
⇒0.22+α×α21−α=0.4
⇒α2=0.2(2+α)(1−α)
So,
1.2α2+0.2α−0.4=0
⇒6α2+α−2=0
⇒α=0.5
So, m=0.22.5=0.08 mol/kg