Question

derive equation of relative lowering of vapour pressure from Raoult's law!!!


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Answer 1

According to Raoult’s law,

“The relative lowering of vapour pressure of solvent is equal to the mole fraction of the solute.”

Let, No. of moles of solute = n

No. of moles of solvent = N

Total moles = n+N

Vapour pressure of pure solvent= p

Vapour pressure of solution =Ps

lowering of vapour pressure= p-ps

Relative lowering of vapour pressure= p- ps/p

So, According to Raoult’s Law,

p- ps/p =n / n+N

Answer 2

For a solution of volatile liquids Raoult’s law, is given as

P = PA + PB

If solute (component B) is non-volatile then

Thus, relative lowering of vapour pressure is equal to the mole fraction of non-volatile solute.

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