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Relationship b/w relative lowering of vapour pressure and molality

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5 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 = xBwhere, p0 = Vapour pressure of pure solvent p = Vapour pressure of solution containing non-volatile solute xB = mole fraction of soluteNow this Relative lowering of vapour pressure is related to the molality of solution by the following mathematical expression-- p0 – p / p0 = m * MA / 1000where, m = molality of solution and MA = molar mass of the solvent
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5 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).
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3 years ago
```							1000/M(P-p/p)=mm=molalityM=molar massP=vapour pressure of solutionp= vapour pressure of pure solvent
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2 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 = xBwhere, p0 = Vapour pressure of pure solventp = Vapour pressure of solution containing non-volatile solutexB = mole fraction of soluteNow this Relative lowering of vapour pressure is related to the molality of solution by the following mathematical expression--(p0 – p) / p0 = m * MA / 1000where, m = molality of solutionand MA = molar mass of the solventThanks and Regards
```
6 months ago
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