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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

  • question_answer
    Specific rotation of sugar solution is 0.5 deg m2/kg. 200 kgm-3 of impure sugar solution is taken in a sample polarimeter tube of length 20 cm and optical rotation is found to be \[19{}^\circ \].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, \[\theta ={{19}^{o}},l=20cm=0.20m,\] \[S=0.5\deg \,{{m}^{2}}/kg\] \[\therefore \] \[c=\frac{19}{0.20\times 0.5}=190kg-{{m}^{-3}}\] 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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