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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
      On mixing, heptane and octane form an ideal solution At 373 K, the vapour pressures of the two liquid components (Heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane\[=100\text{ }g\text{ }mo{{l}^{1}}\]and of octane \[=114\text{ }g\text{ }mo{{l}^{-1}}\])       AIEEE  Solved  Paper-2010

    A) 144.5 kPa            

    B)        72.0 kPa              

    C)        36.1 kPa              

    D)        96.2 kPa  

    Correct Answer: B

    Solution :

    \[{{P}_{T}}={{P}_{0}}{{x}_{0}}+{{P}_{hep}}{{x}_{hep}}\] \[=45\times \frac{0.3}{0.55}+105\times 25\frac{0.25}{0.55}\] \[=45\times 0.545+105\times 0.454\] \[=72.25\text{ }kPa.\]


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