The vapour pressure of a 0.01m solution of a weak base BOH in water at 20 C is 17.536 mm. Calculate Kb for the base. Aq tension at 20 C=17.54 mm. Ans=9.7*10^-4

As you know ,

relative lowering of vapour pressure is proportional to mole fraction of the solute.

Since it is a weak base, in aquoeus solution at equilibrium 

BOH  --> B+   +   OH-

initial:   1  0 0

FInal:  1-x x x

Therefore the vanthoff factor i = 1-x+2x = 1+x.

Now, substitute in formula

(P-p)/P = i * molefraction

(17.54 - 17.536)/17.4 = (1+x) * (0.01/55.5)                {No of moles of water in 1L=1000/18=55.55 and approximate 55.55+0.01 ~ 55.55 } 

From this we get x=0.2668

Now Kb = Cx^2/(1-x)

substitute values and you will get Kb = 9.7 * 10^-4.

Please approve the answer!!!!!

How do u get Kb=Cx^2/(1-x)???