JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer At room temperature, a dilute soluton of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, lowering of vapour pressure will be (molar mass of urea \[=60g\,mo{{l}^{-1}}\]):- [JEE Main 10-4-2019 Morning]

    A) 0.027 mmHg  

    B) 0.028 mmHg

    C) 0.017 mmHg

    D) 0.031 mmHg

    Correct Answer: C

    Solution :

    Lowering of vapour pressure \[={{p}^{0}}-p={{p}^{0}}.{{x}_{solute}}\] \[\therefore \]\[\Delta p=35\times \frac{0.6/60}{\frac{0.6}{60}+\frac{360}{18}}\] \[=35\times \frac{.01}{.01+20}=35\times \frac{.01}{20.01}\]\[=.017mm\,Hg\]   


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