5 g of a nonvolatile solute in 100 g of water has a vapour pressure of 2980 Pa. If the vapour pressure of pure water is 3000 Pa, the molar mass of solute is

5 g of a nonvolatile solute in 100 g of water has a vapour pressure of 2980 Pa. If the vapour pressure of pure water is 3000 Pa, the molar mass of solute is

  1. A

    120 g mol-1

  2. B

    125 g mol-1

  3. C

    130 g mol-1

  4. D

    134 g mol-1

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    Solution:

    Mole fraction of water in solution is

    x=(100/18)mol(5 g/M)+(100/18)mol

    Hence, p=x1p1*

    2980 Pa=(100/18)mol(5 g/M)+(100/18)mol(3000 Pa) This gives M=134.1 g mol-1

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