You haven't seen pickles being made at home, have you? That's all there is to osmosis' magic. How does osmosis work, though? Do you realise that osmosis is one of the most important factors in the survival of many species? We will learn everything about osmosis and osmotic pressure in this document.
In the process of Osmosis, solvent moves from a region of its higher concentration to a region of its lower concentration through a semipermeable membrane, which allows the free passage of only solvent molecules.
Osmosis is the phenomenon of solvent flow through a semipermeable membrane to equalize the solute concentrations on both sides of the membrane.
Osmotic Pressure can be defined as:
The minimum pressure which should be applied on the solution, so as to prevent the migration of solvent molecules into the solution through a semipermeable membrane.
The osmotic pressure p of a solution is related to the molar concentration of solute M as: π = MRT,
Where R is the gas constant and T is the absolute temperature.
Van’t Hoff equation for a dilute solution is similar to the ideal gas equation,
πV= nRT
Where π = osmotic pressure
R = gas constant
Some typical examples of osmosis and osmotic pressure are:
(i) The climbing of water up a tall tree from the soil.
(ii) Bursting of redblood cells when placed in water.
(ii) Swelling of dry raisins when placed in water.
Difference between Endosmosis and Exosmosis
Key concept: The semipermeable membranes allow a selective passage to only the solvent molecules. Cellophane, parchment paper and animal protein membranes are typical semipermeable membranes.
The hydrostatic pressure set up as a result of osmosis is a measure of the osmotic pressure of the solution.
For instance, if the solution of density d rises to height h, then osmotic pressure is expressed as π = h × d × g, where g is the acceleration due to gravity.
The ability of an extracellular solution to make water move into or out of a cell by osmosis is known as its tonicity.
Three terms—hypertonic, hypotonic, and isotonic—are used to describe whether a solution will cause water to move into or out of a cell:
A pair of solutions having the same osmotic pressure are known as isotonic solutions.
If a cell is placed in a hypertonic solution, there will be a net flow of water out of the cell, and the cell will lose volume. A solution will be hypertonic to a cell if its solute concentration is higher than that inside the cell, and the solutes cannot cross the membrane.
When placed in hypertonic solutions, the fluid from the plant cells comes out and thus the cells contract in size (plasmolysis). When excess fertilizers (like urea) are applied, plasmolysis takes place and plants dry up (wilt).
If a cell is placed in a hypotonic solution, there will be a net flow of water into the cell, and the cell will gain volume. If the solute concentration outside the cell is lower than inside the cell, and the solutes cannot cross the membrane, then that solution is hypotonic to the cell. When placed in water or hypotonic solutions, cell swell and burst (haemolysis).
Example 1. Calculate the osmotic pressure and vapour pressure of 0.6% aqueous solution of a nonvolatile, nonelectrolyte solute, urea (NH_{2}CONH_{2}) at 25˚C. The vapour pressure of pure water at 25˚C is 24 mm Hg. Take densities to be 1 g mL^{1} and assume ideal behaviour of the solution. Gas constant, R = 0.082 L atm. mol^{1} K^{1}.
Key concept: Please refer to difference between osmotic pressure and vapour pressure while solving this illustration.
Solution.
Concentration of urea solution = 0.6%
This means, solution contains 0.6 g of urea per 100 g, and hence 100 mL of the solution,
So,
Mass of urea per litre of solution
= 0.6/100 × 100 = 6 g
Molar mass of urea, M = 60 g/mol
Concentration of urea in solution
= 6/60 mol/L = 0.1 mol/L
Example 2. At 25˚C, the osmotic pressure of human blood due to the pressure of various solutes in the blood is 7.65 atm. Assuming that molarity and molality are almost same, calculate the freezing point of blood. K_{f} = 1.86 K kg/mol.
Solution.
According to the Van't Hoff equation,
p = CRT
c = 7.65/(0.082)(298) = 0.313 mol/L
Since molarity and molality are same,
ΔT_{f} = K_{f}m
ΔT_{f} = 1.86 × 0.313 = 0.582
Freezing point of blood is  0.582˚C.
Example 3. At 27°C, 36 g of glucose per litre has an osmotic pressure of 4.92 atm. If the osmotic pressure of solution is 1.5 atm at the same temperature, what should be its concentration?
Solution.
Given that,
π_{1 }= 4.92 atm, π_{2} = 1.5 atm
π_{1}V_{1} = n_{1}RT_{1}
π_{2}V_{2} = n_{2}RT_{2}
At same temperature,
∴ C_{2} = 0.061 mol/litre
Example 4. At 25°C, a solution containing 0.2 g of polyisobutylene in 100 mL of benzene developed a rise of 2.5 mm at osmotic equilibrium. Calculate the molecular weight of polyisobutylene if the density of the solution is 0.88 g/mL.
Solution.
Height developed = 2.4mm
Osmotic pressure = h.d.g
207.187 dyne cm^{−2}
Now πV = nRT
207.187 × 100
(R in erg; V in mL, using CGS system)
M = 2.39
Try yourself!
RO is a process in which a large pressure is applied to the solution side so as to overcome the osmotic pressure.
This pushes the pure solvent under pressure, out of the solution through the semipermeable membrane. This process finds a number of practical applications. Some of them are the purification of drinking water, removal of salt from water molecules, removal of effluents from water, etc.
RO is considered one of the most important water purification technologies. The contaminants present in water are removed by pushing water under pressure through a semipermeable membrane. Pure water is thus squeezed out which is fit for drinking. RO is also used in the desalinization of seawater. RO is a reliable source for obtaining potable water.
ABNORMAL MOLAR MASS
Molecular masses of the solute can be easily determined with the help of colligative properties; relative lowering in vapour pressure, boiling point elevation, freezing point depression, and osmotic pressure easily. But, we derived the relation between the molecular mass of solute and colligative properties under some assumptions, which are:
So, how do we determine the molecular mass of solute in case the above assumptions are not valid? Basically, the abnormality is due to:
(i) Association of Solute Particles
Image 2: Ethanoic Acid dimer due to hydrogen bonding
(ii) Dissociation of Solute Particles
NaCl dissociate into two ions when dissolved in water
Following cases become possible.
Thus, for
Potassium chloride (KCl) in aqueous solution, i is nearly equal to 2.
Barium chloride (BaCl_{2}) in aqueous solution, i is nearly equal to 3.
Benzoic acid (C_{6}H_{5}COOH) in benzene, i is nearly equal to 0.5.
Acetic acid (CH_{3}COOH) in benzene, i is nearly equal to 0.5.
First, we write an equation of the solute being associated or dissociated
CASE 1: WHEN SOLUTE UNDERGOES DISSOCIATION WHEN DISSOLVED IN SOLVENT
Degree of dissociation: It is defined as the number of moles dissociated over the initial number of moles taken.
Van’t Hoff factor = Total number of particles at equilibrium/Total initial number of particle
Let c moles per litre of an electrolyte A_{x}B_{y} be taken and let it dissociated as
A_{x}B_{y} →xA^{y+} + yB^{x–}
t = 0 C 0 0
c (1 − α) xcα ycα At equilibrium
i = concentration or number of moles after dissociation/concentration or number of moles before dissociation
i =[c  cα + xcα + ycα ]/c
or i = [1  α + xα + yα ]/1
= 1 + [(x + y)–1]α
Let x + y = n = number of particles formed after dissociation
i = 1 + (n – 1)α
Degree of Association may be defined as the number of molecules associated per unit volume over initial concentration
i = Total number of particles or concentration after association/Total number of particles or concentration before association
Let n molecules of an electrolyte A undergo association
then, nA → A_{n}
C 0 t = 0
C – C · α C.α/n At. equilibrium
MODIFIED EQUATIONS FOR COLLIGATIVE PROPERTIES
For solutes which undergo dissociation or association in solutions, the equations for the colligative properties are modified by inserting the van'tHoff's factor in them as follows:
Example 1. Acetic acid (CH_{3}COOH) associates in benzene to form double molecules. 1.65 g of acetic acid when dissolved in 100 g of benzene raised the boiling point by 0.36˚C. Calculate the Van't Hoff factor and the degree of association of acetic acid in benzene. (K_{b} = 2.57 K kg mol^{ 1}).
Solution.
Total moles at equilibrium∆T_{b} = K_{b} × m,
α = 0.981Van't Hoff factor = no. of moles at equilibrium/Initial no. of moles
Example 2. Derive a relationship between K_{b}, T_{b} and ∆H_{vap} for a liquid.
Solution.
Let us consider a pure liquid. At its boiling point, T_{b}, its vapour pressure P˚ would be equal to the external pressure.
P˚ = P_{ext} at temperature T_{b}.
Now let us consider a nonvolatile solute dissolved in the liquid. When the solution reaches temperature T_{b}, the vapour pressure of the system, would be less than the external pressure. When the solution is heated further and it reaches its boiling T_{b}, the vapour pressure of the solution would be equal to P_{ext} which is equal to the P˚ of the pure liquid at its boiling point. Since vapour pressure of a system are the K_{p}'s for the respective equilibrium,
Assuming the solution to be highly dilute, T_{b} would be very close to T_{b}.
Since X_{solute} is very small, we can make the approximation that in (1  x) = x (when x is very small).
Ignoring n in comparison to N in the denominatorMultiplying by 1000 on both the sides, we get,Similarly, K_{f} = Example 3. To 500 cm^{3} of water, 3.0 × 10^{ 3} kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression in freezing point? K_{f} and density of water are 1.86 K kg^{1} mol^{1} and 0.997 g cm^{3} respectively.
Solution. Molality of acetic acid
Total number of moles per kg of solvent = 0.1003 (1 + 0.23)ΔT_{f} = K_{f}m = 1.86 × 0.1003 × 1.23 = 0.23o C
Example. 4. At 300 K, the vapour pressure of an ideal solution containing one mole of A and 3 moles of B is 500 mm of Hg. At the same temperature, if one mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Calculate the V.P. of A and B in their pure state.
Solution.
When, one mole of B is further added to it,
Solving equations (i) and (ii), we getP°_{A}= 440 mm, P°_{B}= 600 mm
Example 5. A decinormal solution of NaCl exerts an osmotic pressure of 4.6 atm. at 300 K. Calculate its degree of dissociation. (R = 0.082 L atm. K^{1}mol^{1}).
Solution.
No. of moles of NaCl per litre of solution = 0.1
Osmotic pressure (p) = 4.6 atm., Temperature (T) = 300 K
Had NaCl not dissociated, then
p normal = CRT = 0.1 × 0.082 × 300 atm. = 2.46 atm.
But, pobs = 4.6 atm.
As per definition,
For the dissociation of an electrolyte producing n ions,
So, Percentage dissociation = 100 × a = 100 × 0.87 = 87%
Example 6. Vapour pressure of C_{6}H_{6} and C_{7}H_{8} mixture at 50˚C are given by: P = 179 X_{B} 92, where X_{B} is mole fraction of C_{6}H_{6}. Calculate (in mm):
(A) Vapour pressure of pure liquids.
(B) Vapour pressure of liquid mixture obtained by mixing 936 g C_{6}H_{6} and 736 g toluene.
(C) If the vapours are removed and condensed into liquid and again brought to the temperature of 50˚C, what would be mole fraction of C_{6}H_{6} in vapour state?
Solution.
(A) Given, P = 179X_{B} + 92
For pure, C_{6}H_{6}, X_{B} = 1
P°_{B}= 179 + 92 = 271 mm
For pure C_{7}H_{8}, X_{B} = 0
P°_{T}= 179 x 0 + 92 = 92 mm
(B) Now PM = P°_{B}X_{B} + P°_{T}X_{T}
= 162.6 +36.8 = 199.4 mmMoles of C_{6}H_{6} = 936/78 = 12
Moles of C_{7}H_{8} = 736/92 = 8
(C) Now mole fraction of C_{6}H_{6} in vapour phase of initial mixture (X'_{T})
Moles fraction of C_{7}H_{8} in vapour phase of initial mixture (X'_{L})
These vapours are taken out and condensed into liquid. The liquid is again brought to 50˚C to get again vapourliquid equilibrium.
Thus, mole fraction of C_{6}H_{6} in vapour phase of initial mixture
= Mole fraction of C_{6}H_{6} in liquid phase on II mixture X'_{B}
Similarly, mole fraction of C_{7}H_{8} in vapour phase of initial mixture
= Mole fraction of C_{7}H_{8} in liquid phase on II mixture X'_{T}
Therefore,
= 271 × 0.815 + 92 × 0.185 mm= 220.865 + 17.02 = 237.885 mm
New mole fraction of C_{6}H_{6} in vapour phase
New mole fraction of C_{7}H_{8} in vapour phase = 0.072.
Example 7. The freezing point depression of an HF (aq) solution with molality 0.10 m is 0.201˚C. What will be the Van’t Hoff factor? Take the value of Kf to be 1.86˚Cm^{1}
Solution.
HF (aq) → H^{+} + F^{}
We have Δ T_{f }= 0.201 C0 , m = 0.10 and K_{f} = 1.86 ˚Cm^{1}
The freezing point depression is Δ T_{f} = i K_{f}m
0.201 = i × 1.86 × 0.10
The value of Van’t Hoff factor will be 1.08
Example 8. A millimolar solution of potassium ferricyanide is 70% dissociated at 27°C. Find out the osmotic pressure of the solution.
Solution.
Given, concentration of solution = C = 1/1000 M = 10^{–3}M
Where α = degree of dissociation = 70% = 70/100 = 0.7
Total moles of after dissociation = (1 – α) + 3α + α = 1+ 3α
∴ i = 1 + 3α/1
i = 1+3α = 1+(3 × 0.7) = 3.1
Since osmotic pressure
π = iCRT
∴ π = 3.1 × 10^{–3} × 0.0821 × 300
= 0.31 × 3 × 0.0821 atm = 0.07635 atm
TRY YOURSELF!
Q.1. A 0.5 percent aqueous solution of Potassium Chloride was found to freeze at 0.24°C. Calculate the Van't Hoff factor and degree of dissociation of the solute at this concentration (K_{f} for water is 1.86).
Ans.
i =
= 1.923; i = 1+(n–1)α
1.923 = 1+ (2–1)α ⇒ 1.923 = 1+ α
α = 1.923 – 1 = 0.923
or α = 0.923 or 92.3%
Q.2. If the apparent degree of ionization of KCl (KCl = 74.5 gm mol^{–1}) in water at 290K is 0.86. Calculate the mass of KCl which must be made up to 1 dm^{3} of aqueous solution so as to produce the same osmotic pressure as the 4.0% solution of glucose at that temperature.,
Ans.
Due to ionization of KCl
From question, degree of ionization = a = 0.86
∴ i = 1+ 0.86 = 1.86
For Osmotic pressure of glucose,
For 4% glucose solution,
Weight of glucose = 40 gm,
Volume of solution = V = 1L = 1dm^{3}
Molecular weight of glucose C_{6}H_{12}O_{6}= m = 180
where π_{glucose} = osmotic pressure of glucose
Similarly,As both solutions are isotonic,∴ π_{KCl} = π_{glucose}
= 8.9 gm
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1. What is osmosis? 
2. What is osmotic pressure? 
3. What are the types of osmosis? 
4. What is tonicity? 
5. What is the difference between osmotic pressure and vapor pressure? 

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