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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Critical Thinking Solutions

  • question_answer
    Normal boiling point of water is 373 K (at 760mm). Vapour pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mole, the boiling point of water at 23 mm pressure will be      [CBSE PMT 1995]

    A)                 250 K    

    B)                 294 K

    C)                 51.6 K   

    D)                 12.5 K

    Correct Answer: B

    Solution :

             Applying clausius clapeytron equation            \[\log \frac{{{P}_{2}}}{{{P}_{1}}}=\frac{\Delta {{H}_{V}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}\times {{T}_{2}}} \right]\]       \[\log \frac{760}{23}=\frac{40656}{2.303\times 8.314}\left[ \frac{373-{{T}_{1}}}{373T} \right]\]                 This gives \[{{T}_{1}}=294.4K\].    


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