Class 12 Chemistry

# Solution

## Vapor Pressure of Liquid Solutions

Let us consider a binary solution of two volatile liquids with two components 1 and 2. When taken in a closed vessel, both the components would evaporate and eventually and equilibrium would be reached between vapor phase and liquid phase. At this stage, let us assume the total vapor pressure is p_{total} and p_{1} and p_{2} are partial vapor pressures of the two components 1 and 2. These partial pressures are related to mole fractions x_{1} and x_{2} of the two components in that order.

**Roult Law:** For a solution of volatile liquids, the partial vapor pressure of each component of the solution is directly proportional to its mole fraction present in the solution.

Thus, for component 1

`p_1 ∝ x_1`

And `p_1=p_1^0x_1` ……………(3)

Here, `p_1^0` is the vapor pressure of pure component 1 at the same temperature.

Similarly, for component 2

`p_2=p_2^0x_2` ………….(4)

According to *Dalton’s Law of Partial Pressures*, the total pressure (p_{total}) over the solution phase in the container will be the sum of partial pressures of the components of the solution.

Hence, `p_text(total)=x_1p_1^0+x_2p_2^0` …………….(5)

`=(1-x_2)p_1^0+x_2p_2^0`

`=p_1^0+(p_2^0-p_1^0)x_2` ………………(6)

Following conclusions can be drawn from equation (6)

- Total vapor pressure over the solution can be related to the mole fraction of any one component.
- Total vapor pressure over the solution varies linearly with the mole fraction of component 2.
- Depending on the vapor pressures of pure components, total vapor pressure decreases or increases with the increase of mole fraction of component 1.

A graph of p_{1} or p_{2} versus mole fractions x_{1} and x_{2} gives a linear plot. The lines I and II pass through the points for which x_{1} and x_{2} are equal to unity. The minimum value of p_{total} is p_{1}^{0} and the maximum value is p_{2}^{0}.

The composition of vapor phase in equilibrium with the solution is determined by partial pressures of the components. If y_{1} and y_{2} are the mole fractions of components 1 and 2 in vapor phase then, as per Dalton’s law of partial pressures:

`p_1=y_1p_text(total)`

`p_2=y_2p_text(total)1`

In general

`p_1=y_1p_text(total)` ………….(7)

**Example:** Vapor pressure of chloroform (CHCl_{3}) and dichloromethane (CH_{2}Cl_{2}) at 298 K are 200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapor pressure of the solution prepared by mixing 25.5 g of CHCl_{3} and 40 g of CH_{2}Cl_{2} at 298 K and (ii) mole fractions of each component in vapor phase.

**Answer:** Molar mass of CH_{2}Cl_{2} `12xx1+x_1xx2+35.5xx2=85` g mol^{-1}

Molar mass of CHCl_{3} `=12xx1+1xx1+35.5xx3=11.5` g mol^{-1}

Moles of CH_{2}Cl_{2} `=(40g)/(85g\text(mol)^(-1))=0.47` mol

Moles of CHCl_{3} `=(25.5g)/(119.5g\text(mol)^(-1))=0.213` mol

Total number of moles `=0.47+0.213=0.683` mol

`X_(CH_2Cl_2)=(0.47mol)/(0.683mol)=0.689`

`X_(CHCl_3)=1.00-0.688=0.312`

Vapor pressure of solution can be calculated as follows:

`p_text(total)=p_1^0+(p_2^0-p_1^0)x_2`

`=200+(415-200)xx0.688`

`=200+147.9=347.9` mm Hg

Now, mole fraction of components in vapor phase can be calculated as follows:

`p_(CH_2Cl_2)=0.688xx415 = 285.5` mm Hg

`p_(CHCl_3)=0.312xx200=62.4` mm Hg

`y_(CH_2Cl_2)=(285.5)/(347.9)=0.82`

`y_(CHCl_3)=(62.4)/(347.9)=0.18`

#### Raoult’s Law as a special case of Henry’s Law

We know that as per Rault’s law, the vapor pressure of a volatile component in a given solution is given as follows:

`p_1=x_1p_1^0`

In case of a solution of a gas in a liquid, one of the components is so volatile that it exists as a gas. Solubility of this is given by Henry’s law which is expressed as follows:

`p = K_H\X`

In both the equations, partial pressure of the volatile component is directly proportional to its mole fraction in solution. Thus, Rault’s law becomes a special case of Henry’s law in which K_{H} becomes p_{1}^{0}.

Molality

Solubility

Liquid in Liquid

Solid in Liquid

Boiling Point

Osmosis

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