JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    x g of non-electrolyte compound of molecular mass 200, when dissolved in 1L of 0.05 M NaCl solution, the osmotic pressure of the resulting solution is found to be 4.92 atm at \[27{}^\circ C\]. Assuming the complete dissociation of NaCI and ideal behaviour of solution, the value of x would be

    A)  10 g                      

    B)  20.0g                 

    C)  40.0 g                  

    D)  80.0 g

    Correct Answer: B

    Solution :

     While solving this process, students are advised to use the concept of osmotic pressure as follows \[\therefore \]\[\pi =C\,RT=\left\{ \frac{x}{200}+2\times 0.05 \right\}0.0821\times 300\] \[\Rightarrow \]\[\frac{x}{200}+0.1=\frac{4.92}{0.0821\times 300}\] \[\therefore \]\[\frac{x}{200}+0.1=0.2\Rightarrow x=0.1\times 200=20g\]

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