JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    In an adsorption experiment, a graph between \[\log \frac{x}{m}\] versus log p was found to be linear with a slope of \[{{45}^{o}}\]. The intercept on the \[\log \frac{x}{m}\] axis was found to be 0.30103. Calculate the amount of the gas adsorbed per gram of charcoal under a pressure 1 atm. \[[{{\log }_{10}}2=0.301032]\]

    A)  1                                

    B)  2

    C)  4                                

    D)  0

    Correct Answer: B

    Solution :

     \[\because \]    \[\log \,\frac{x}{m}=\log \,k+\frac{1}{n}\,\log \,p\] \[(y=c+mx\] form ) \[\frac{1}{n}=\tan \,{{45}^{o}}=1\] \[\log \,k=0.30103\] \[\therefore \]    \[k=2\] \[\therefore \]    \[\frac{x}{m}=k{{p}^{1/n}}=2\times {{1}^{1/1}}=2\]

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