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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 27

  • question_answer
    \[526.3\text{ }mL\]of \[0.5\text{ }mL\]\[HCl\] is shaken with \[0.5\text{ }g\]of activated charcoal and filtered. The concentration of the nitrate is reduced to \[0.4\text{ }m\]. The amount of adsorption \[(x/m)\] is

    A) 3          

    B)                    4                   

    C) 6                        

    D)        8

    Correct Answer: B

    Solution :

    [b] Mass of \[HCl\]acid adsorbed by \[10\text{ }g\]charcoal \[=526.3\times {{10}^{-3}}(0.5-0.4)\times 38\approx 2\] (Mw of \[HCl=38g\,mo{{l}^{-1}}\]) The amount of adsorption \[\frac{x}{m}=\frac{2}{0.5}=4\]   


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