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Relationship b/w relative lowering of vapour pressure and molality

Relationship b/w relative lowering of vapour pressure and molality

Grade:12

4 Answers

Aarti Gupta
askIITians Faculty 300 Points
8 years ago
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
MOHAMMED AFROZ ALU
13 Points
7 years ago
LVP: This difference between the vapour pressure of pute solvent (p°) and the vapour pressure of the solution (ps) is known as lowering of vapour pressure (LVP)Mathematically,delta p=p°-psRLVP: The ratio of lowering of vapour pressure (delta p) to the vapour pressure of pure solvent (p°)is called relative lowering of vapour pressure (RLVP)According to raoult`s law, RLVP=XB (mole fraction of solute) =》p°-ps a W --------- = --- × ---- P° M bFrom this equation, we can find molar weight of solute (M).
Avinash
13 Points
5 years ago
1000/M(P-p/p)=m
m=molality
M=molar mass
P=vapour pressure of solution
p= vapour pressure of pure solvent
 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago

Dear Student,
Please find below the solution to your problem.

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
Thanks and Regards

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