Normal boiling point of water is `373 K`. Vapour pressure of water at `298 K` is `23 mm` enthalpy of vaporisation is `40.656 kJ mol^(-1)` if atmopheric pressure becomes `23 mm`, the water will boil at:

Options

A. 250 K
B. 294 K
C. 51.6 K
D. 12.5 K

Answer

B

Explanation

Applying clausius clapeytron equation
`log.(P_(2))/(P_(1))=(Delta H_(V))/(2.303R)[(T_(2)-T_(1))/(T_(1)xxT_(2))]`
`log.(760)/(23)=(40656)/(2.303xx8.314)[(373-T_(1))/(373T_(1))]`
This gives `T_(1)=294.4 K`.