Question

Derive the relationship between lowering of vapour pressure and molar mass of solute?

Answer 1

hii there,


Relationship b/w relative lowering of vapour pressure and molality

Relative lowering of vapour pressure is one of the colligative properties.Colligative properties are those properties of solutions which depend only on the number of solute particles and independent of the nature of the solute particles.
Relative lowering of vapour pressure is the ratio of lowering of vapour pressure to the vapour pressure of pure solvent.It is given as--
p0 – p / p0 = xB
where,
p0 = Vapour pressure of pure solvent
p = Vapour pressure of solution containing non-volatile solute
xB = mole fraction of solute
Now this Relative lowering of vapour pressure is related to the molality of solution by the following mathematical expression--
p0 – p / p0 = m * MA / 1000
where,
m = molality of solution and
MA = molar mass of the solvent

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Answer 2

According to Raoult's law

P1=x1p1......... (1)

The decrease in vapour pressure of the solvent (pi) is given by:

=>