It discusses in brief about vapor pressure lowering and simple numerical on it.

1. 2. SOLUTIONS
Part -2
colligative properties of nonelectrolyte
solutions.
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2. 2.6 colligative properties of nonelectrolyte solutions
Definition-
The physical properties of solutions that depend on the number of solute
particles in solutions and not on their nature are called colligative
properties.
Colligative properties-
1. Vapour pressure lowering
2. Boiling point elevation
3. Freezing point depression
4. Osmotic pressure
(For colligative properties of nonelectrolyte solutions, the relatively
dilute solutions with concentrations 0.2 M or less are considered)
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3. 2.7 Vapour pressure lowering
Vapour Pressure-
When a liquid in a closed container is in equilibrium with its vapours, the
pressure exerted by the vapour on the liquid is its vapour pressure.
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4. Vapour pressure lowering-
When a nonvolatile, non-ionizable solid
is dissolved in a liquid solvent, the
vapour pressure of the solution is lower
than that of pure solvent.
i.e. the vapour pressure of a solvent is
lowered by dissolving a nonvolatile
solute into it.
A nonvolatile solute does not contribute
to the vapour pressure above the solution.
V.P. of solution = V.P. of solvent above
the solution.
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5. If P0
1 = V. P. of pure solvent and
P1 = V. P. of solvent above the solution
Then P1 < P0
1
The vapour pressure lowering is,
∆ P = P0
1 - P1 ------- (1)
Reason: The vapour pressure of solution P1 < P01 vapour of pure
solvent.
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Reason: The vapour pressure of solution P1 < P0
1 vapour of pure
solvent.
Pure solvent
Molecules of pure solvent
at the surface of pure
solvent

7. After addition of
non volatile solute
As solute is nonvolatile it does not contribute to the vapour
pressure above the solution.
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When nonvolatile solute is dissolved in a solvent, some of the surface
molecules of solvent are replaced by nonvolatile solute molecules.
These solute molecules do not contribute to vapour above the solution.
Surface of pure
solvent after
addition of non
volatile solute

8. Thus, the number of solvent
molecules available for vaporization
per unit surface area of solution is
less than the number at the surface
of pure solvent. The solvent
molecules at the surface of solution
vaporize at a slower rate than
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2.7.1 Raoult’s law for solutions of non-volatile solutes:
According to this law,
P1= P0
1 x1
A plot of P1 versus is a straight line as shown below,

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For a binary solution containing one solute,
x1 = 1- x2
P1= P0
1 x1
P1= P0
1 (1- x2)
= P0
1 - P0
1 x2
OR
P0
1 - P1 = P0
1 x2
Then according to eq. ∆ P = P0
1 - P1 ------- (1)
∆ P = P0
1 x2 ------- (2)
∆ P depends on x2 i.e. number of solute particles.
Thus ∆ P, the lowering of vapour pressure is a colligative
property.

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2.7.2Relative lowering of vapour pressure:
The ratio of vapour pressure lowering of solvent divided by the vapour
pressure of pure solvent is called relative lowering of vapour pressure.
Relative vapour pressure lowering
= ∆P / P0
1 = P0
1 − P1 / P0
1 ----- (3)
But ∆ P = P0
1 x2 ------- (2)
∆P / P0
1 = x2 = P0
1 − P1 / P0
1 ----- (4)
The relative lowering of vapour pressure is equal to the mole fraction of solute in the
solution. Therefore, relative vapour pressure lowering is a colligative property.

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2.7.3 Molar mass of solute from vapour pressure lowering:
Lowering of vapour pressure is equal to the mole fraction x2 of solute
in the solution. ∆P / P0
1 = x2
The mole fraction of a component of solution is equal to its moles
divided by the total moles in the solution. Thus,
x2 = 𝑛2 /𝑛1+𝑛2
where and 𝑛1 , 𝑛2 are the moles of solvent and solute respectively
We are concerned only with dilute solutions
hence 𝑛1 >> 𝑛2 and 𝑛1+𝑛2 ≈ 𝑛1
The mole fraction
x2 = 𝑛2 /𝑛1 and
∆P / P0
1 = 𝑛2 /𝑛1 ------ (5)

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When we prepare a solution by dissolving W2 g of a solute in W1 g of
solvent. Then,
n2 = 𝑊2 /𝑀2
and
n1 = 𝑊1/ 𝑀1
where M1 & M2 molar masses of solvent and solute respectively
∆P = x2 = 𝑊2 /𝑀2
P0
1 𝑊1/ 𝑀1
P0
1 − P1 = ∆P = 𝑊2 /𝑀2
P0
1 P0
1 𝑊1/ 𝑀1
knowing all other quantities molar mass of solute M2 can be calculated.

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Solve numerical on lowering of vapour pressure.
1)The vapour pressure of water at 200c is 17 mm Hg. What is the vapour
pressure of solution containing 2.8 g urea in 50 gram of water.
Given :
1) Vapour pressure of pure solvent i. e. water P1
0=17mm Hg.
2) Mass of urea (solute) W2= 2.8 g. 3)Molar mass of urea 𝑀2 =60 g.
4) Mass of water(pure solvent)W1 = 50 g. 5) Molar mass of water 𝑀1 =18 g
Vapour pressure of solution P1 =?
P0
1 − P1 = 𝑊2 /𝑀2
P0
1 𝑊1/ 𝑀1
17mmHg- P1 = 2.8 g × 18 g/mol = ?
17mm Hg 50 g × 60 g/mol
𝑀2

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