# RAOULT'S LAW - RELATIVE LOWERING OF VAPOUR PRESSURE

## 1) For dilute solution, Raoult's law of lowering of vapour pressure states that:

(IIT JEE 1985)

(a) the relative lowering of vapour pressure is equal to mole fraction of solute

(b) the vapor pressure of the solution is equal to mole fraction of the solvent

(c) the relative lowering of vapour pressure is proportional to amount of solute

(d) the lowering of vapor pressure is equal to the mole fraction of solute  in solution

Logic and Solution:

According to Raoult's law, for dilute solutions containing non volatile solute, the relative lowering of vapor pressure, Δp/po is equal to the mole fraction of solute, X2

i.e.,

Δp/po = X2

It is a colligative property.

Conclusion:

The correct option is "a".

Extra information:

* The lowering of vapor pressure is proportional to the mole fraction of solute or inversely related to mole fraction of solvent.

* Colligative properties depend only on the number of particles and not on the nature of substances.

## Related questions

### 2) The vapor pressure of ethanol and methanol are 44.5 and 88.7 mm Hg, respectively. An ideal solution is formed at the same temperature by mixing 60 g ethanol and 40 g methanol. Calculate the total vapor pressure of the solution and mole fraction of methanol in the vapour.

(IIT JEE 1986)

Logic:

Let

pT = total vapor pressure of the solution,

then

According to Raoult's law:

pT = pe + pm = peoXe + pmoXm

Where:

pe & pm are partial vapor pressures of ethanol and methanol, respectively.

peo & pmo are pure vapor pressures of ethanol and methanol, respectively.

Xe & Xm are the mole fractions of ethanol and methanol in the liquid phase, respectively.

However, according to Dalton's law:

pT = pe + pm = pTYe + pTYm

Where:

Ye & Ym are the mole fractions of ethanol and methanol in the vapour phase, respectively.

Solution:

Calculation of vapor pressure of solution:

First of all,  let's calculate the number of moles of ethanol and methanol.

no. of moles of ethanol = ne = mass of ethanol / molar mass of ethanol = 60 g / 46 g mol-1 = 1.3 mol

no. of moles of methanol = nm = mass of methanol / molar mass of methanol = 40 g / 32 g mol-1 = 1.25 mol

Total number of moles of mixture = 1.3 + 1.25 = 2.55 mol

Now calculate the mole fractions.

mole fraction of ethanol = Xe = ne / ne + nm = 1.3 / 2.55 = 0.51

mole fraction of methanol = Xm = 1- Xe = 1 - 0.51 = 0.49

Now calculate the partial vapor pressures of ethanol and methanol.

pe = peoXe = 44.5 x 0.51 = 22.7 mm Hg

pm =  pmoXm = 88.7 x 0.49= 43.5 mm Hg

Therefore:

pT =  pe + pm = 22.7 + 43.5 = 66.2 mm Hg

Calculation of mole fraction of methanol in vapor phase:

According to Dalton's law:

pm =  pTYm

or

Ym = pm /  pT = 43.5 / 66.2 = 0.66

Note:

pe = peoXe = pTYe

pm =  pmoXm = pTYm

### 3) Which of the following liquid pairs shows a positive deviation from Raoult’s law?

(AIEEE 2004)

1) Water – hydrochloric acid

2) Acetone – chloroform

3) Water – nitric acid

4) Benzene – methanol

Logic:

Positive deviation indicates weaker intermolecular forces of attractions between different components in the mixture than attractions in the pure individual components. Due to weakened forces of attractions, the escaping tendency of the liquids in the mixture will be more than expected i.e., the vapor pressure of the mixture is greater than expected.

Solution:

In all of the above mixtures, except Benzene - methanol, the total vapor pressure of the mixture is less than expected i.e., shows negative deviation, since the attraction forces are stronger when compared to those in pure liquids. The weaker forces of attraction are replaced by stronger forces of attraction.

However, in Benzene - methanol mixture the intermolecular forces of attractions are weaker than those in pure liquids. Hence this mixture shows positive deviation.

### 4) A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statement is correct regarding the behavior of the solution?

(AIEEE 2009)

1) The solution formed is an ideal solution

2) The solution is non-ideal, showing +ve deviation from Raoult’s Law

3) The solution is non-ideal, showing –ve deviation from Raoult’s Law

4) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s Law.

Logic:

n-Heptane is non polar and ethanol is polar. Hence in the mixture formed by these two liquids is non ideal. The stronger forces of attraction between n-heptane-n-heptane and ethanol-ethanol are replaced by weaker forces of attraction between n-heptane-ethanol in the solution. Therefore the escaping tendency of the liquid molecules into the vapor phase increases. This will increase the vapor pressure more than expected from Raoult's law. This is positive deviation.

Solution:

The correct option is '2'.

## Raoult's law - Followup questions for Practice - IIT JEE NEET

1) What is vapor pressure? Explain the phenomenon of lowering of vapor pressure.

2) State Raoult's law of lowering of vapor pressure for dilute solutions.

3) Discuss Raoult's law of relative lowering of vapour pressure.

3) What is relative lowering of vapor pressure?

4) What is a colligative property?

5) The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm, respectively. What will happen when about 5 mL of methanol is added to 100 mL of ethanol?

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Author: Aditya vardhan Vutturi