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The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm.what is the mole fraction of the component B in the solution ?
    Correct answer is '0.25'. Can you explain this answer?
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    The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-v...
    Solution:

    Given,

    Vapour pressure of pure solvent A = 0.80 atm

    Vapour pressure of solution = 0.60 atm

    Let us assume that the mole fraction of component B in the solution is x.

    Then, mole fraction of solvent A in the solution = (1 - x)

    According to Raoult's law,

    P_A = P^0_A * x_A

    where,

    P_A = Vapour pressure of solvent A in the solution

    P^0_A = Vapour pressure of pure solvent A

    x_A = Mole fraction of solvent A in the solution

    Similarly,

    P_B = P^0_B * x_B

    where,

    P_B = Vapour pressure of component B in the solution

    P^0_B = Vapour pressure of pure component B (which is assumed to be zero)

    x_B = Mole fraction of component B in the solution

    Total vapour pressure of the solution can be given as:

    P_total = P_A + P_B

    => P_total = P^0_A * x_A + P^0_B * x_B

    => 0.60 atm = 0.80 atm * (1 - x) + 0 * x

    => 0.60 atm = 0.80 atm - 0.80 atm * x

    => 0.80 atm * x = 0.80 atm - 0.60 atm

    => 0.80 atm * x = 0.20 atm

    => x = 0.25

    Therefore, the mole fraction of component B in the solution is 0.25.
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    The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm.what is the mole fraction of the component B in the solution ?Correct answer is '0.25'. Can you explain this answer?
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    The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm.what is the mole fraction of the component B in the solution ?Correct answer is '0.25'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm.what is the mole fraction of the component B in the solution ?Correct answer is '0.25'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of pure liquid solvent A is 0.80 atm. When a non-volatile substance B is added to the solvent, its vapour pressure drops to 0.60 atm.what is the mole fraction of the component B in the solution ?Correct answer is '0.25'. Can you explain this answer?.
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