2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main - Mock Test - 16

  • question_answer
    At \[80{}^\circ C,\] the vapour pressure of pure liquid 'A' is \[520\text{ }mm\text{ }Hg\]and that of pure liquid 'B' is\[1000\text{ }mm\text{ }Hg\]. If a mixture solution of 'A' and 'B' boils at \[80{}^\circ C\] and 1 atm pressure, the amount of 'A' in the mixture is \[(1\,atm=760\,mm\,Hg)\]

    A) \[52\text{ }mol\]percent   

    B)   \[34\text{ }mol\]percent

    C) \[48\text{ }mol\]percent     

    D) \[50\text{ }mol\]percent

    Correct Answer: D

    Solution :

    At 1 atmospheric pressure the boiling point of mixture is\[80{}^\circ C\]. At boiling point the vapour pressure of mixture, \[{{P}_{T}}=1\]atmosphere \[=760\text{ }mmHg.\] Using the relation, \[{{P}_{T}}=p_{A}^{o}{{x}_{A}}+p_{B}^{o}{{x}_{B}},\]'we get \[{{P}_{T}}=520{{x}_{A}}+1000\,(1-{{x}_{A}})\] \[\{\because \,\,p_{A}^{o}=520\,mm\,\,Hg,\,p_{B}^{o}=1000mm\,Hg,{{x}_{A}}+{{x}_{B}}=1\}\]\[760=520{{x}_{A}}+1000-1000{{x}_{A}}\] \[{{x}_{A}}=\frac{240}{480}=\frac{1}{2}\] or \[50\,mol\] percent


You need to login to perform this action.
You will be redirected in 3 sec spinner