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A mixture of two immiscible liquids at a constant pressure of 1.0 atm boils at temperature
A.Equal to the normal boiling point of more volatile liquid
B.Equal to the mean of the normal boiling of two liquids
C.Greater than normal boiling point of either of liquids
D.Smaller than the normal boiling point of either of liquids.

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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: Immiscible liquids are nothing but those liquids which won't mix to give a single phase. Water and oil are the examples for immiscible liquids. One liquid floats on top of the other liquid when mixed together.

Complete answer:
We know that the vapour pressure of a mixture of two immiscible liquids is equal to the sum of their vapour pressures of liquids in their pure states.
Vapour pressure is independent of their relative amounts present in the mixture.
So, the boiling point of the mixture of the liquids is less than that of either of the boiling points of the liquids.
Therefore a mixture of two immiscible liquids at a constant pressure of 1.0 atm boils at a temperature lower than the normal boiling point of either of liquids.

So, the correct option is D.

Note:
As we know that a solute (nonvolatile substance) lowers the vapour pressure of a liquid, and hence leads to a higher boiling point of the solution.
So here, the mixture of two immiscible liquids does not contain any solute. Therefore the mixture will boil at lower temperature of the boiling point of either of liquids.
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