Solutions Question 83

Question: The latent heat of vapourisation of water is $ 9700,Cal/mole $ and if the b.p. is $ 100^{o}C $ , ebullioscopic constant of water is [CBSE PMT 1989]

Options:

A) $ {{0.513}^{o}}C $

B) $ {{1.026}^{o}}C $

C) $ {{10.26}^{o}}C $

D) $ {{1.832}^{o}}C $

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Answer:

Correct Answer: A

Solution:

$ K_{b}=\frac{M_1RT_0^{2}}{1000\Delta H_{V}}=\frac{18\times 1.987\times {{(373)}^{2}}}{1000\times 9700} $

$ ={{0.513}^{o}}C $



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