12th Class Chemistry Solutions / विलयन Question Bank Case Based - Solutions

  • question_answer
    Directions: (Q. 1 to 5)
    Case I: Read the passage given below and answer the following questions from 1 to 5.
    The solubility of gases increases with increase of pressure. William Henry made a systematic investigation of the solubility of a gas in a liquid. According to Henry's law "the mass of a gas dissolved per unit volume of the solvent at constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution".
    Dalton during the same period also concluded independently that the solubility of a gas in a liquid solution depends upon the partial pressure of the gas. If we use the mole fraction of gas in the solution as a measure of its solubility, then Henry's law can be modified as "the partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution".
    Henry's law constant for the solubility of methane in benzene at 298 K is \[4.27\times {{10}^{5}}\] mm Hg. The solubility of methane in benzene at 298 K under 760 mm Hg is

    A) \[4.27\times {{10}^{-5}}\]

    B) \[1.78\times {{10}^{-3}}\]

    C) \[4.27\times {{10}^{-3}}\]

    D) \[1.78\text{ }\times {{10}^{-5}}\]

    Correct Answer: B

    Solution :

    \[{{K}_{H}}=4.27\times {{10}^{5}}\]mm Hg p=760 mm Hg According to Henry's law, \[p={{K}_{H}}\times \,\,{{x}_{C{{H}_{4}}}}\] \[{{X}_{C{{H}_{4}}}}=\frac{p}{{{K}_{H}}}=\frac{760}{4.27\times {{10}^{5}}}=1.78\times {{10}^{-3}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner