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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    Adsorption of a gas follows Freundlich adsorption isotherm x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of \[\log \frac{x}{m}\]versus log p is shown in the given graph. \[\frac{x}{m}\]is proportional to :                                                                               [JEE Main 8-4-2019 Morning]

    A) \[{{p}^{{}^{3}/{}_{2}}}\]                                  

    B) \[{{p}^{3}}\]

    C) \[{{p}^{{}^{2}/{}_{3}}}\]          

    D) \[{{p}^{2}}\]

    Correct Answer: C

    Solution :

    \[\frac{x}{m}=K.{{p}^{1/n}}\] \[\therefore \]\[\log \frac{x}{m}=\log K+\frac{1}{n}.\operatorname{logP}\] slope \[=\frac{1}{n}=\frac{2}{3}\] \[\therefore \]\[\frac{x}{m}=K.{{p}^{2/3}}\] Correct option : (c)                 


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