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Decomposition of non-volatile solute 'A'...

Decomposition of non-volatile solute 'A' into another non-volatile solute B and C , when dissolved in water follows first order kinetics as A `to` 2B + C . Vapour pressure of a solution in which 1 mole of A was dissolved in 180 g of water was found to be 20 mm of Hg after 12 hrs . What will the vapour pressure after 24 hrs ? [Given : Vapour pressure of `H_(2)O` at the given temperature = 24 mm Hg ]

A

19.2 mm of Hg

B

20 mm of Hg

C

10 mm of Hg

D

12 mm of Hg

Text Solution

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The correct Answer is:
C
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Decomposition of non-volatile solute 'A' into another non-volatile solute B and C, when dissolved in water follows first order kinetics as : A to 2B + C when one mole of A is dissolved in 180 gm of water and left for decomposition, the vapour pressure of solution was found to be 20 mm Hg after 12 hrs. Determine the vapour pressure of the solution (in mm of Hg) after 24 hars Assume constant temperature of 25^(@) C, throughout . The vapour pressure of pure water at 25^(@) C is 24mm Hg. [Fill your answer by multiplying it with 100]

How is vapour pressure of a solvent affected when a non-volatile solute is dissolved in it ?

Knowledge Check

  • Vapour pressure of the solution of a non- volatile solute is always __.

    A
    equal to the vapour pressure of pure solvent
    B
    higher than vapour pressure of pure solvent
    C
    lower than vapour pressure of pure solvent
    D
    constant

  • When one mole of non-volatile solute is dissolved in three moles of solvent, the vapour pressure of the solution relative to the vapour pressure of the pure solvent is -

    A
    `1/3`
    B
    `1/4`
    C
    `3/4`
    D
    1

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