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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    Specific rotation of sugar solution is \[0.5\text{ }deg\text{ }{{m}^{2}}/kg.\text{ }200\text{ }kg{{m}^{-3}}\] of impure sugar solution is taken in a sample polarimeter tube of length 20 cm and optical rotation is found to be \[{{19}^{o}}\]. The percentage of purity of sugar is:

    A)  20 %                                     

    B)  80 %

    C)  95 %                                     

    D)  89 %

    Correct Answer: C

    Solution :

    The strength of solution is given by \[c=\frac{\theta }{l\times s}\]where the symbols have their usual meanings. Here,                 \[\theta ={{190}^{o}},\,\,\,l=20\,\,cm=0.20\,\,m\],                                 \[S=0.5\,\,\deg \,{{m}^{2}}/kg\] \[\therefore \]  \[c=\frac{19}{0.20\times 0.5}=190\,\,kg-{{m}^{-3}}\] The sugar sample dissolved in a \[{{m}^{3}}\] of water is 200 kg in which 190 kg is pure sugar. Therefore, purity is                 \[=\frac{190}{200}\times 100=95%\]


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