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In the case of immiscible liquids, the a...

In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature.
If we mix two immiscible liquids A and B then boiling point of the mixture will be (Given `p_(A)^(@) gt p_(B)^(@))`

A

greater than `T_(B)` and lower than `T_(A)`

B

greater than either of `T_(A)` and `T_(B)`

C

lower than `T_(B)` and higher than `T_(A)`

D

lower than either of `T_(A)` and `T_(B)`

Text Solution

Verified by Experts

The correct Answer is:
D
Liquid will boil when
`p_(A)^(@)+p_(B)^(@)=` Atmospheric pressure.
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In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature. In which of the following mixtures, vapour pressure of each component is independent of the amount of the other components?

Knowledge Check

  • In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature. In question 6, if we decrease the amount of liquid X in the mixture then what will be the effect on the boiling point of the mixture at constant temperature?

    A
    It will remain constant
    B
    Decrease
    C
    Increases
    D
    Can't predict

  • In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature. An immiscible mixture of water and quinoline is prepared and distillation is done. If p_("quinoline")^(@) = 7.9 torr and p_("water")^(@) = 732.04 torr at the boiling point of 98.9^(@)C . Find out the relative masses of water and quinoline in the distillate. If it is given that no of moles of any component in the vapour phase prop its vapour pressure. (Molar mass of water = 18 g/mol and molar mass of quinoline = 129g//mol )

    A
    `W_("water"): W_("quinoline") : : 13:1`
    B
    `W_("water"): W_("quinoline"):: 1:1`
    C
    `W_("water") : W_("quinoline") ::1:14`
    D
    `W_("Water"):W_("quinoline") :: 7.1:1`

  • In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature. If the vapour pressure of two liquids X and Y in their pure state at 300 K are 85 mm of Hg and 120 mm of Hg respectively and if their mole fraction in liquid solution are 0.45 and .55 respectivel. Then find out the total vapour pressure above the mixture containing two immiscible liquids at 300 K.

    A
    `104.25 mm` of Hg
    B
    `205 mm` of Hg
    C
    `27.75 mm` of hg
    D
    35 mm of Hg

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