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According to Raoult's law the relative l...

According to Raoult's law the relative lowering of vapour pressure for a solution is equal to

A

Mole fraction of the solvent

B

Mole fraction of the solute

C

Weight percentage of a solute

D

Weight percentage of solvent

Text Solution

Verified by Experts

The correct Answer is:
B
`(p^(@)-p)/(p^(@))=((w)/(m))/((w)/(m)+(W)/(M))` = mole fraction of solute.
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Knowledge Check

  • According to Raoult's law, relative lowering of vopour pressure of a solvent is equal to

    A
    moles of solute
    B
    mole fraction of solute
    C
    moles of solvent
    D
    mole fraction of solvent

  • The relative lowering in vapour pressure is:

    A
    `prop X_("solute")`
    B
    `prop (1)/(X_("solute"))`
    C
    `=X_("solute")`
    D
    `prop m`

  • The law of relative lowering of vapour pressure was given by

    A
    Vant Hoff
    B
    Ostwald
    C
    Lewis
    D
    Raoult

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    The law of relative lowering of vapour pressure was given by

    Equal mass of a soute are dissolved in equal mass of two solvents A and B and formed very dilute solution. The relative lowering of vapour pressure for the solution B has twice the relative lowering of vapour pressure for the solution A. If m_(A) "and" M_(B) are the molecules mass of solventds A and B respectively, then :

    If a non-volatile solute is dissolved in a volatile solvent relative lowering of vapour pressure is equal to

    Relative lowering of vapour pressure is a colligative property because

    Relative lowering of vapour pressure is a colligative property because

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