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JEE Main & Advanced Sample Paper JEE Main Sample Paper-23

  • question_answer
    The mass of \[{{C}_{6}}{{H}_{12}}{{O}_{6}}\] that should be dissolved in 100 gram of water in order to produce same lowering of vapour pressure as is produced by dissolving 1 gram of urea in same quantity of water is

    A)  1 gram                        

    B)  3 gram

    C)  6 gram                        

    D)  18 gram

    Correct Answer: B

    Solution :

    \[(\Delta P)\]Glucose = \[(\Delta P)\]urea \[({{X}_{B}})\]Glucose = \[({{X}_{B}})\]urea \[\frac{{{\omega }_{B}}}{100}\times \frac{18}{180}\,=\frac{1}{100}\,\times \frac{18}{60}\] \[{{\omega }_{B}}=3\,\,gram\].


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