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The vapour pressure of pure liquid solve...

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`, the mole fraction of component `B` in the solution is

A

0

B

0.25

C

2. 0

D

3. 0

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(P_(A)^@=0.8 "atm"),(" "P_(sol)=0.6 "atm"):}}=chi_B=(0.8-0.6)/0.8=0.25`
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Knowledge Check

  • The vapour pressure of pure liquid solvent A is 0.80 atm.When a non volatile substances B is added to the solvent, its vapour pressure drops to 0.60 atm. Mole fraction of the components B in the solution is:

    A
    `0.50`
    B
    `0.25`
    C
    `0.75`
    D
    `0.40`

  • The vapour pressure of pure liquid solvent A is 0.80 atm . When a non - volatile substance B is added to the solvent , as vapour pressure drops to 0.60 atm . Mol fraction of the component B in the solution is :

    A
    `0.50`
    B
    `0.25`
    C
    `0.75`
    D
    `0.40`

  • The vapour pressure of pure liquid solvent 0.50 atm. When a non-volatile solute B is added to the solvent, its vapour pressure drops to 0.30 atm. Thus, mole fraction of the component B is

    A
    `0.6`
    B
    `0.25`
    C
    `0.45`
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    `0.75`

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