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Molar mass of a solute using colligative properties of solution
Colligative properties
Those properties of solutions which depend on the
concentration of the solute components while that are not depend on the identity of
the solute
...

Relative lowering of vapor pressure
If we dissolved the non-volatile solute into
the solution then the vapor pressure of the solvent becomes lowered that is known
as the lowering of the vapor pressure
...

P1 is called as vapor pressure of the solution PO1 is called as the vapor pressure of
the solvent, x1 is the mole fraction of the solvent then in the result of Raoult’s law;
P1 = PO1x1
Lowering in the vapor pressure of solvent is the ∆P1
ΔP1 = P1o – P1
Changing the P 1 = P 1 o x1 of the equation;
ΔP1 = P1o – P1o x1
ΔP1 = P1o(1 – x1)
So, total of the mole fraction of solute is equal to ‘x 2’ while the mole fraction
of solvent ‘x 1’ 1that is;
1 – x1 = x2
Δ P1 = P10 x2
Hence that lower in vapor pressure that depends on the mole fraction of the
solute x2 as like via the Raoult’s law
...
When the n 1 , n 2 are the known as the moles of
solute and solvent so the equation become:

When the solution is little diluted, that can be disrepair the moles of solute is
n1 in moles of solvent n 2
...

The mole of solvent ‘n1’ is equivalent to;
n1 = m1 / M1
Changing the values of n 1 , n2 in the dilute solution of vapor pressure, the
equation become;

Hence by knowing the other quantities, we easily measured the molar mass of
solute M 2 in the above equation
...
850 bars on room temperature
...
6g in benzene solution’ that is equal to 39g,
then vapour pressure of benzene become lowered to 0
...
Calculate the
molar mass of solute?

So, that the
P1o = 0
...
845 ba,
m1 = 39 g
M1 = 78 g
m2= 0
...

By using the equation of lowering of vapor pressure;

Change in the vapor pressure is equal to;
ΔP1 = P10 – P1, that are equivalent to 0
...
845 = 0
...
By
changing the values;

= 20
...

So, by using these values the molar mass of solute is equal to 20
...

Elevation of Boiling Point:
We know that when we dissolved the non-volatile solute into the solution then the
vapor pressure of the solvent becomes lowered
...
P of the solution is
higher than the initial solvent that is due the vapor pressure is directly
proportional to the temperature
...
This increasing of
temperature is known as the elevation of boiling point such as the lowering of
vapor pressure that is too dependent on the solute components in the solution
...

ΔT b ∝ m
ΔT b = K b m
In the equation m is reported as the no
...
In the equation K b is known as the molal elevation

constant or also called as the ebullioscopic constant
...

Suppose that m1 and m 2 are the masses of solvent and solute while the molar
masses of solute is equal to M2 and the solvent is equal to M1, then the
molality is equivalent to;

By put the value of molality in the boiling of elevation equation then;

M2 = K b × 1000 × m2 / ∆T b × m1
So, by knowing the others quantities we can calculate the molar mass of solute
of the elevation of boiling point equation
...
23 K
...
70 g of the non-volatile solute in to the liquid (90g), then solution’ boiling
point increased to 354
...
Calculate the molar mass of the solute? By using
the value of K b of liquid that is equal to 2
...

Since that;
T bo = 353
...
11 K
m2 = 2
...
53 K kg mol -1
and m1 = 90 g
Elevation of boiling point is equivalent to;
ΔT b = T b – T b0
= 354
...
23K = 0
...
53 ×1000 ×2
...
88 ×90

= 86
...


So, by using that equation we calculate the molar mass of solute that is equal
to 86
...

Depression of Freezing Point:
If the vapor pressure of the solution is lowered then the freezing point of the
solution can be lowered
...
When the solutions’ vapor pressure is equivalent
to the initial solvent vapor pressure, in the result solution become cold or
frozen
...

That is known as the depression of freezing point
...

So, the molality ‘m’ is equivalent to;

In that equation the ‘m2’ is known as the mass of solute while the m1 is the
mass of solvent and the M2 is called as the molar mass of added non-volatile
solute
...


❖ For example;
Nonelectrolyte solution is equal to 38
...

5
...
Calculate the solute’ molar mass?
Given data;
Δ Tf = −5
...
218 kg
mass solute = 38
...
86o C/m
Calculate molar mass solute =? g/mol
Solutions;
m = ∆𝑇f / Kf

= -5
...
86 Cm-1 = 2
...
97m × 0
...
648mol
38
...
648 g = 59
...


Example;
The substance with 0
...
2g benzene that
decreased the boiling point of benzene that is 0
...
Calculate the substance’
molecular weight?
m = 1000 × K f × w / ∆T × W
w = 0
...
567 oC
W = 22
...
12 OC / mol
m = 1000 × 5
...
440 / 0
...
2
m = 178
...

Osmotic pressure calculations are less precise and are tough to perform
...

If we know about the osmotic pressure of the solution then we can
measure the molecular weight of the solute by using the van’t Hoff eq;
π V = nRT
𝑤

= RT
𝑀

M = wRT / πV
In these equations the M is equivalent to the molecular mass of the solute,
w is known as the amount of the solute, R is equal to 0
...

Example
A glucose solution which consists of 18g per liter with osmotic pressure
of 2
...
0821l-atm
T = 23 + 273 = 296 k
Π = 2
...
0821 × 296 / 2
...
0g



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