The latent heat of vaporization of water is 9700cal/mole and if the b.p. is 100°C, the ebullioscopic constant of water is

The equiliscopic constant is given by-

${\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{RT}}^2}{1000\times L_v}$

Here,

R = Gas constant
T = Boiling point of water on water scale =$100+273=373 K$
L_v = Latent heat of vapourisation (in cal /gm) = $\frac{9700}{18}\mathrm{cal}/\mathrm{gm}$

$therefore {\mathrm{K}}_{\mathrm{b}}=\frac{2\times (373{)}^2\times 18}{1000\times 9700}$

$\Rightarrow {\mathrm{K}}_{\mathrm{b}}=0.516$
 

Therefore, Option A is the correct answer