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

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Q. The latent heat of vaporization of water is $9700\, Cal / mole$ and if the b.p. is $100^{\circ} C$, ebullioscopic constant of water is

Solutions

Solution:

$K_{b}=\frac{M_{1} R T_{0}^{2}}{1000 \Delta H_{V}}$

$=\frac{18 \times 1.987 \times(373)^{2}}{1000 \times 9700}$

$=0.513^{\circ} C$