Question

The latent heat of vaporisation of water is 9700 cal/mole and if the boiling point is 100, the ebullioscopic constant of water is:

• A. 0.516
• B. 1.026
• C. 10.26
• D. 1.832

Answer 1

The correct option is A 0.516

Solution:- (A) $0.513$

The equiliscopic constant is given by-

$K_b=\frac{R{T}^2}{1000\times l}$

Here,

$R=$ Gas constant

$T=$ Boiling point of water in water scale $=100+273=373K$

$l=$ Latent heat of vapourisation $\left(incal/gm\right)=\frac{9700}{18}cal/gm$

$\therefore K_b=\frac{2\times {\left(373\right)}^2\times 18}{1000\times 9700}$

$\Rightarrow K_b=0.516$