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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    A solution is prepared by dissolving 0.6 g of urea (molar mass \[=60g\,mo{{l}^{-1}}\]) and 1.8 g of glucose (molar mass \[=180g\,mo{{l}^{-1}}\]) in 100 mL of water at \[27{}^\text{o}C.\]The osmotic pressure of the solution is : \[\left( R=0.08206\text{ }L\text{ atm }{{K}^{1}}mo{{l}^{1}} \right)\] [JEE Main 12-4-2019 Afternoon]

    A) 4.92 atm

    B)              1.64 atm

    C) 2.46 atm                     

    D) 8.2 atm

    Correct Answer: A

    Solution :

    \[\Pi =\frac{\left( \frac{0.6}{60}+\frac{1.8}{180} \right)}{0.1}\times 0.08206\times 300\]           \[\Pi =4.9236\,atm\]


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