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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    In Kjeldahl’s method, ammonia from 5 g of food neutralizes \[30\,c{{m}^{3}}\] of 0.1 N acid. The percentage of nitrogen in the food is

    A) 8.4

    B)   8.4

    C)   16.8                                       

    D)   1.68

    Correct Answer: A

    Solution :

      From Kjeldahl’s method, percentage of nitrogen                 \[=\frac{1.4\times N\times V}{w}=\frac{1.4\times 0.1\times 30}{5}\]                 \[=0.84\,\,%\]


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