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Normal boiling point of water is 373 K. ...

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:

A

250 K

B

294 K

C

51.6 K

D

12.5 K

Text Solution

Verified by Experts

The correct Answer is:
B
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`.
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Knowledge Check

  • 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 vaporisation is 40.656 kJ/mol, the boiling point of water at 23 mm atmospheric pressure will be

    A
    250 K
    B
    51.6 K
    C
    298 K
    D
    12.5 K

  • 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 :

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

  • The boiling point of water at 1 atmospheric pressure is

    A
    `100^(@)`C
    B
    `0^(@)`C
    C
    `100.5^(@)` C
    D
    `-5^(@)` C

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