The normal boiling point of watr is 373 K (at 760 mm), vapour pressure of water at 298 K is 23 mm. If entyalpy of vaporization is `40.656` kj//mol.the boiling point of water at 23 mm pressure will be :

Options

A. 250 K
B. `294.4 K`
C. `51.6 K`
D. `12.5 K`

Answer

B

Explanation

From Clausius- clapeyron equation.
`"log" (P_(2))/(P_(1))=(DeltaH_(v))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`rArr"log" (760)/(23)=(40656)/(2.303xx8.314)[(373-T_(1))/(373xxT_(1))]`
`:.T_(1)=294.4K`.
Hence, (B) is the correct answer.