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The normal boiling point of water is 373 K (at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vapourisation is 40.656 KJ/mol, the boiling point of water at 23 mm atmospheric pressure will be

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294 K

51.6 K

12.5 K

250 K

answer is C.

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Detailed Solution

$\ln \frac{760}{23} = \frac{40656}{8.314} [\frac{1}{T_{1}} - \frac{1}{373}]$

T₁ = boiling point = 294 K

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