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Question
Explain two colligative properties.
Explain
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Solution
- Relative lowering of vapour pressure: When a non-volatile solute is added to a solvent, the vapour pressure of the solvent decreases. This is because some surface area is now occupied by solute particles, reducing the number of solvent molecules escaping into the vapour phase.
`(p^circ - p)/p^circ = X_"solute"`
Where:
`p^circ` = vapour pressure of pure solvent
p = vapour pressure of solution
`X_"solute"` = mole fraction of solute
- Elevation of boiling point: Addition of a non-volatile solute raises the boiling point of the solvent. This is because the solution has lower vapour pressure, so a higher temperature is needed for it to equal atmospheric pressure and boil.
ΔTb = Kb ⋅ m
Where:
ΔTb = elevation in boiling point
Kb = ebullioscopic constant
m = molality of the solution
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