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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    Dry air is passed through a solution containing 10 g of a solute in 90 g of water and then through pure water. The loss in weight of solution is 2.5 g and that of pure solvent is 0.05 g. Calculate the molecular weight of the solute.

    A) 50                          

    B)        180

    C) 102                        

    D)        25

    E) 51

    Correct Answer: C

    Solution :

    According to Raoults law \[\frac{p-{{p}_{s}}}{p}=\frac{n}{n+N}=\frac{0.05}{2.5+0.05}\] \[=\frac{0.05}{2.55}=\frac{1}{51}\]           Weight of solute\[=\frac{w}{W}\times M\times \frac{p}{p-{{p}_{s}}}\]                                 \[=\frac{10\times 18}{90}\times 51\]                                 \[=102g\]


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