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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Lowering of vapour pressure

  • question_answer
    The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be [DPMT 2001]

    A)                 18.0

    B)                 342

    C)                 60          

    D)             180

    Correct Answer: D

    Solution :

             \[\frac{{{P}^{0}}-{{P}_{s}}}{{{P}^{0}}}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}}\] or \[0.00713=\frac{{71.5}/{m}\;}{\frac{71.5}{m}+\frac{1000}{18}}\]                             \[m=180\]


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