Why is osmotic pressure the best method for polymer molar mass determination?

Polymers have very large molar masses giving tiny molality even at 1 percent concentration. BP elevation and FP depression produce unmeasurably small temperature changes. Osmotic pressure can still produce measurable pressure differences for dilute polymer solutions.

Colligative Properties

Q: Why are colligative properties called colligative?

The word comes from Latin colligatus meaning bound together. The four properties share a common cause: dissolved solute particles reduce the escape tendency of solvent molecules. Because the effect depends only on the number of solute particles and not their identity, all four are tied together by this mechanism.

Q: Why is freezing point depression preferred over boiling point elevation for molar mass determination?

The cryoscopic constant K f is generally larger than the ebullioscopic constant K b. This gives a bigger temperature shift per mole of solute, making measurement more precise. Freezing point measurements also avoid decomposition of heat-sensitive solutes.

Q: What happens if a plant is watered with sea water?

Sea water is hypertonic to plant cell sap. Water flows out of root cells by osmosis instead of into them, dehydrating the plant. The plant wilts and eventually dies. This is why irrigation with brackish water damages crops.

Q: Why does adding salt to icy roads melt the ice?

Salt dissolves in the thin liquid film on ice, depressing the freezing point well below 0 degrees C. If ambient temperature is above this depressed value, the salt solution stays liquid and continues to dissolve more ice, melting it. Calcium chloride works even better than sodium chloride because it dissociates into 3 ions instead of 2.

Q: Why does ethylene glycol work as antifreeze in car radiators?

Ethylene glycol is highly soluble in water and non-volatile. It gives freezing point depression to prevent coolant freezing in winter and boiling point elevation to prevent coolant loss on hot engines. Typical 50 percent solution lowers freezing point to about minus 37 degrees C.

Q: What is the difference between osmosis and diffusion?

Diffusion is the general movement of any particles from high to low concentration, requiring no membrane. Osmosis is specifically the movement of solvent through a semipermeable membrane from lower to higher solute concentration. The membrane blocks the solute; only solvent moves. Osmosis is a special case of diffusion restricted by a semipermeable membrane.

Q: Why is Van't Hoff equation for osmotic pressure similar to the ideal gas equation?

In dilute solution, solute particles are spread far apart with little interaction, behaving somewhat like gas molecules in a container. The pressure they exert on a semipermeable barrier obeys the same functional form as gas pressure. The analogy holds only for dilute solutions where solute-solute interactions are negligible.

Q: Why does BP elevation depend on molality but osmotic pressure on molarity?

BP and FP shifts occur over a range of temperatures, and molality is temperature-independent. Osmotic pressure is measured at a fixed temperature so molarity is the natural choice, and it also connects directly to the ideal gas-like form pi V equals n R T.

ChemistrySolutionsFor JEE aspirants

Colligative properties depend only on the number of solute particles in solution, not on their chemical identity. The four colligative properties are: (i) relative lowering of vapour pressure, (ii) elevation of boiling point ( $Δ T_{b} = K_{b} m$ ), (iii) depression of freezing point ( $Δ T_{f} = K_{f} m$ ), and (iv) osmotic pressure ( $π = M R T$ ). All four can be used to determine unknown molar masses of non-volatile solutes. Everyday applications include ethylene glycol as car radiator antifreeze, salting icy roads in winter, and reverse osmosis desalination of sea water.

Key Formulas - Quick Reference

Relative lowering of vapour pressure: $\frac{P _{1}^{\circ} - P _{1}}{P _{1}^{\circ}} = x_{2} = \frac{W _{2} M _{1}}{W _{1} M _{2}}$
Elevation of boiling point: $Δ T_{b} = K_{b} \cdot m$ ; molar mass $M_{2} = \frac{1000 \cdot K _{b} \cdot W _{2}}{Δ T _{b} \cdot W _{1}}$
Depression of freezing point: $Δ T_{f} = K_{f} \cdot m$ ; molar mass $M_{2} = \frac{1000 \cdot K _{f} \cdot W _{2}}{Δ T _{f} \cdot W _{1}}$
Ebullioscopic constant: $K_{b} = \frac{R T _{b}^{2} M _{1}}{1000 \cdot Δ H _{v a p}}$ ; Cryoscopic constant: $K_{f} = \frac{R T _{f}^{2} M _{1}}{1000 \cdot Δ H _{f u s}}$
Osmotic pressure: $π = M R T$ (Van't Hoff equation) or $π V = n R T$
Molar mass from osmotic pressure: $M_{2} = \frac{W _{2} R T}{π V}$
Isotonic condition: $π_{1} = π_{2} \Rightarrow C_{1} = C_{2}$ (same molar concentration at same T)
For water: $K_{b} = 0.52$ K kg mol $^{- 1}$ , $K_{f} = 1.86$ K kg mol $^{- 1}$

1. What Are Colligative Properties

Colligative properties are properties of a solution that depend only on the number of solute particles dissolved, not on their nature (mass, size, or chemistry).

The Latin root colligatus means "bound together" - colligative properties tie together several apparently different observations under one common cause: dissolved solute particles reduce the escape tendency of solvent molecules from the liquid phase.

There are four colligative properties:

Relative lowering of vapour pressure - the drop in vapour pressure caused by dissolving a non-volatile solute
Elevation of boiling point - the increase in boiling point of the solvent when solute is added
Depression of freezing point - the decrease in freezing point of the solvent
Osmotic pressure - the pressure needed to stop osmosis of pure solvent into the solution

Figure 1: The four colligative properties. Each is measured in a different way, but all four count exactly the same thing: the number of dissolved solute particles per unit of solvent. That shared cause is why any one of them can be used to find an unknown molar mass.

All four are useful for finding the molar mass of an unknown non-volatile solute, because the "number of particles" is directly related to $\frac{W}{M}$ .

Assumption: the standard colligative-property formulas assume the solute is non-volatile and does not dissociate or associate. When a solute dissociates (like NaCl into Na $^{+}$ and Cl $^{-}$ ) or associates (like benzoic acid dimerising in benzene), the actual number of particles differs from the assumed count. This is handled by the Van't Hoff factor $i$ , covered in the next concept (Abnormal Molar Masses).

2. Relative Lowering of Vapour Pressure

When a non-volatile solute is added to a solvent, the vapour pressure of the solvent decreases. If $P_{1}^{\circ}$ is the vapour pressure of the pure solvent and $P_{1}$ is that of the solution, applying Raoult's law:

$P_{1} = x_{1} \cdot P_{1}^{\circ} = (1 - x_{2}) \cdot P_{1}^{\circ}$

Rearranging gives the relative lowering of vapour pressure:

$\frac{P _{1}^{\circ} - P _{1}}{P _{1}^{\circ}} = x_{2} = \frac{n _{2}}{n _{1} + n _{2}}$

The relative lowering equals the mole fraction of the solute - it is independent of the nature of the solute, so it is a colligative property.

For dilute solutions, $n_{2} ≪ n_{1}$ , so $x_{2} \approx \frac{n _{2}}{n _{1}} = \frac{W _{2} / M _{2}}{W _{1} / M _{1}} = \frac{W _{2} M _{1}}{W _{1} M _{2}}$

In terms of molality: $x_{2} \approx \frac{m \cdot M _{1}}{1000}$ (where $m$ is molality and $M_{1}$ is molar mass of solvent in g/mol).

Solved Example 1

The vapour pressure of a solvent decreases by 10 mmHg when a non-volatile solute is added, giving a solute mole fraction of 0.2. What is the mole fraction of the solvent if the decrease in vapour pressure is 20 mmHg?

Solution:

$P^{\circ} - P = P^{\circ} \cdot x_{solute}$

From first case: $10 = P^{\circ} \times 0.2 \Rightarrow P^{\circ} = 50$ mmHg.

From second case: $20 = 50 \times x_{solute}^{'} \Rightarrow x_{solute}^{'} = 0.4$ .

Mole fraction of solvent $= 1 - 0.4 = 0.6$ .

Solved Example 2

The density of a 0.438 M solution of

K_{2} CrO_{4}

at 298 K is 1.063 g/cm

^{3}

. Calculate the vapour pressure of water above this solution. Given:

P^{\circ}

(water) = 23.79 mmHg. Assume complete dissociation of

K_{2} CrO_{4}

Solution:

Consider 1 L of solution: mass of $K_{2} CrO_{4}$ dissolved $= 0.438 \times 194 = 84.97$ g.

Mass of 1 L solution $= 1000 \times 1.063 = 1063$ g.

Mass of water in 1 L $= 1063 - 84.97 = 978.03$ g. Moles of water $= 978.03/18 = 54.34$ mol.

$K_{2} CrO_{4}$ dissociates as $K_{2} CrO_{4} \to 2 K^{+} + CrO_{4}^{2 -}$ (3 particles per formula unit).

Total moles of solute particles $= 3 \times 0.438 = 1.314$ mol.

Mole fraction of water $= \frac{54.34}{54.34 + 1.314} = 0.976$ .

$P_{water} = x_{water} \times P^{\circ} = 0.976 \times 23.79 = 23.22$ mmHg.

3. Elevation of Boiling Point

The boiling point of a liquid is the temperature at which its vapour pressure equals the external (usually atmospheric) pressure. When a non-volatile solute is added, the vapour pressure decreases, so the solution must be heated to a higher temperature than the pure solvent to bring its vapour pressure up to atmospheric. This upward shift is called elevation of boiling point.

Figure 2: Vapour pressure against temperature. The solution curve lies below the pure solvent curve at every temperature. Boiling point is where a liquid curve crosses

P = 1

atm, so the solution boils higher (

Δ T_{b}

). Freezing point is where a liquid curve meets the solid curve, and the lower solution curve meets it at a lower temperature (

Δ T_{f}

$Δ T_{b} = T_{b} - T_{b}^{\circ} = K_{b} \cdot m$

where $Δ T_{b}$ is elevation, $m$ is molality (mol/kg of solvent), and $K_{b}$ is the molal elevation constant (also called the ebullioscopic constant) of the solvent. Unit of $K_{b}$ : K kg mol $^{- 1}$ .

$K_{b}$ can be derived from solvent properties:

$K_{b} = \frac{R \cdot T _{b}^{2} \cdot M _{1}}{1000 \cdot Δ H _{v a p}}$

Physical meaning of $K_{b}$ : the boiling point elevation produced (theoretically) by a 1 molal solution of a non-volatile, non-electrolyte solute.

Determining molar mass: if $W_{2}$ g of solute of molar mass $M_{2}$ is dissolved in $W_{1}$ g of solvent, then

$M_{2} = \frac{1000 \cdot K _{b} \cdot W _{2}}{Δ T _{b} \cdot W _{1}}$

For water, $K_{b} = 0.512$ K kg mol $^{- 1}$ - meaning 1 mole of any non-electrolyte solute in 1 kg of water raises the boiling point by 0.512 °C.

Solved Example 3

A non-volatile, non-electrolyte compound has composition 40.0% C, 6.71% H, and 53.3% O by mass. An aqueous solution containing 5% of this solute by mass boils at 100.15 °C. Given

K_{b}

(water)

= 0.512

K kg mol

^{- 1}

, determine the molecular formula.

Solution:

Empirical formula: mole ratios C : H : O $= \frac{40}{12} : \frac{6.71}{1} : \frac{53.3}{16} = 3.33 : 6.71 : 3.33 = 1 : 2 : 1$ .

Empirical formula = $CH_{2} O$ ; empirical mass = 30 g/mol.

Molar mass from BP elevation: $Δ T_{b} = 100.15 - 100 = 0.15$ K.

Mass of solute = 5 g in 100 g total, so mass of water = 95 g.

$M_{2} = \frac{1000 \cdot K _{b} \cdot W _{2}}{Δ T _{b} \cdot W _{1}} = \frac{1000 \times 0.512 \times 5}{0.15 \times 95} = 180$ g/mol.

Molecular formula: $x = \frac{180}{30} = 6 \Rightarrow (CH_{2} O)_{6} = C_{6} H_{12} O_{6}$ (glucose).

Solved Example 4

K_{b}

values for the following alcohols follow the order:

n

-butanol

>

isobutanol

>

tert-butanol. Explain briefly.

Solution:

All three are isomeric $C_{4} H_{10} O$ alcohols with the same molar mass, but their boiling points differ because of shape (branching).

Going from $n$ -butanol to tert-butanol, branching increases, which decreases the surface area available for intermolecular van der Waals interactions. Weaker intermolecular attraction means the liquid boils at a lower temperature - so $T_{b}$ decreases in the same order.

From the formula $K_{b} = \frac{R T _{b}^{2} M _{1}}{1000Δ H _{v a p}}$ and Trouton's rule that $\frac{Δ H _{v a p}}{T _{b}}$ is nearly constant for similar liquids, we get $K_{b} \propto T_{b}$ . So $K_{b}$ follows the same order as $T_{b}$ : $n$ -butanol $>$ isobutanol $>$ tert-butanol.

4. Depression of Freezing Point

The freezing point of a solvent is the temperature at which its solid and liquid phases have the same vapour pressure. When a non-volatile solute is added, the vapour pressure of the liquid solution falls below the vapour pressure of the pure solid at the original freezing point. To re-establish solid-liquid equilibrium, the temperature must be lowered - giving a depression of the freezing point.

$Δ T_{f} = T_{f}^{\circ} - T_{f} = K_{f} \cdot m$

where $Δ T_{f}$ is depression, $m$ is molality, and $K_{f}$ is the molal depression constant (also called cryoscopic constant) of the solvent.

Analogous to $K_{b}$ :

$K_{f} = \frac{R \cdot T _{f}^{2} \cdot M _{1}}{1000 \cdot Δ H _{f u s}}$

Determining molar mass:

$M_{2} = \frac{1000 \cdot K _{f} \cdot W _{2}}{Δ T _{f} \cdot W _{1}}$

For water, $K_{f} = 1.86$ K kg mol $^{- 1}$ . Because $K_{f}$ is generally larger than $K_{b}$ (freezing produces bigger temperature shifts per mole than boiling), FP depression is more sensitive for measuring small solute amounts and is preferred for molar mass determination.

Solved Example 5

The amount of ice that will separate on cooling a solution containing 50 g of ethylene glycol (M = 62) in 200 g of water to

- 9.3

°C is: [

K_{f} = 1.86

K kg mol

^{- 1}

]

Solution:

At $- 9.3$ °C, the remaining liquid solution has $Δ T_{f} = 9.3$ K.

$Δ T_{f} = \frac{1000 \times K _{f} \times W _{solute}}{M _{solute} \times W _{water,left}}$

$9.3 = \frac{1000 \times 1.86 \times 50}{62 \times W _{water,left}} \Rightarrow W_{water,left} = 161.29$ g.

Ice separated $= 200 - 161.29 = 38.71$ g.

Solved Example 6

Calculate the freezing point of an aqueous solution containing 5% urea, 1% KCl, and 10% glucose (all by mass).

K_{f}

(water) = 1.86 K kg mol

^{- 1}

. Molar masses: urea = 60, KCl = 74.5, glucose = 180.

Solution:

Consider 100 g of solution: 5 g urea, 1 g KCl, 10 g glucose, 84 g water.

Total $Δ T_{f}$ = sum of contributions from each solute:

$Δ T_{f} = Δ T_{f} (urea) + Δ T_{f} (KCl) + Δ T_{f} (glucose)$

$= \frac{1000 \times 1.86 \times 5}{60 \times 84} + \frac{1000 \times 1.86 \times 1 \times 2}{74.5 \times 84} + \frac{1000 \times 1.86 \times 10}{180 \times 84}$

(For KCl, $i = 2$ since it dissociates into K $^{+}$ and Cl $^{-}$ .)

$= 1.845 + 0.594 + 1.230 = 3.669$ K

Freezing point $= 273 - 3.67 = 269.33$ K $= - 3.67$ °C.

Solved Example 7

Rank the following aqueous solutions of equal molality in order of increasing boiling point: (a) glucose, (b) NaCl, (c)

BaCl_{2}

, (d)

K_{3} [Fe(CN)_{6}]

. Assume 100% dissociation for electrolytes.

Solution:

Boiling point elevation $Δ T_{b} = i \cdot K_{b} \cdot m$ . For the same $m$ and same solvent, $Δ T_{b} \propto i$ .

Glucose: non-electrolyte, $i = 1$
NaCl $\to$ Na $^{+}$ + Cl $^{-}$ : $i = 2$
$BaCl_{2} \to Ba^{2 +}$ + 2Cl $^{-}$ : $i = 3$
$K_{3} [Fe(CN)_{6}] \to 3 K^{+} + [Fe(CN)_{6}]^{3 -}$ : $i = 4$

Order of increasing $T_{b}$ : glucose $<$ NaCl $<$ $BaCl_{2}$ $<$ $K_{3} [Fe(CN)_{6}]$ .

(The FP depression follows the reverse order for temperatures below 0 °C.)

5. Osmosis and Osmotic Pressure

Certain membranes allow small solvent molecules to pass through but block larger solute molecules. Such membranes are called semipermeable membranes (SPM) - examples are animal bladders, vegetable tissues, cellophane, and cellulose acetate.

Osmosis is the spontaneous flow of solvent molecules through a semipermeable membrane from a region of lower solute concentration (or pure solvent) to a region of higher solute concentration.

Figure 3: Osmosis. Solvent passes freely through the semipermeable membrane into the solution, while the larger solute particles cannot pass back. The solution column rises until its extra hydrostatic pressure just balances the flow, and that balancing pressure is the osmotic pressure

π

of the solution.

Osmotic pressure ( $π$ ) is the external pressure that must be applied to the solution side to just stop osmosis. It is a property of the solution and depends on the molar concentration of solute particles.

$π = M R T$ or equivalently $π V = n R T$

This is the Van't Hoff equation for dilute solutions. Notice its similarity to the ideal gas law - a coincidence that reflects the "gas-like" behaviour of dispersed solute particles.

Determining molar mass by osmotic pressure:

$M_{2} = \frac{W _{2} R T}{π V}$

Osmotic pressure is especially useful for measuring molar masses of polymers and biomolecules because $π$ can be significant even at very low concentrations - unlike BP elevation or FP depression, which give tiny shifts for large molecules.

Solved Example 8

Calculate the osmotic pressure of a sucrose solution containing 1.75 g of sucrose in 150 mL of solution at 17 °C. (Molar mass of sucrose = 342 g/mol;

R

= 0.0821 L atm mol

^{- 1}

^{- 1}

Solution:

$π V = n R T = \frac{W}{M} R T$

$π \times 0.150 = \frac{1.75}{342} \times 0.0821 \times 290$

$π = 0.812$ atm.

Isotonic, Hypertonic, and Hypotonic Solutions

Isotonic solutions have the same osmotic pressure ( $π_{1} = π_{2}$ ). At the same temperature this implies equal molar concentrations of solute particles ( $C_{1} = C_{2}$ ). No net osmosis occurs between two isotonic solutions.
Hypertonic solution: higher osmotic pressure than a reference. Water flows out of a cell placed in a hypertonic solution (cell shrinks - crenation).
Hypotonic solution: lower osmotic pressure than a reference. Water flows into a cell placed in a hypotonic solution (cell swells and may burst - haemolysis for red blood cells).

Figure 4: Tonicity and red blood cells. In a hypotonic solution water flows in and the cell may burst (haemolysis). In an isotonic solution there is no net flow, which is why IV drips are 0.9% w/v saline. In a hypertonic solution water flows out and the cell shrivels (crenation).

Medical application: intravenous drips must be isotonic with blood (osmotic pressure ~7.65 atm at body temperature 310 K), otherwise red blood cells rupture or shrivel. A 0.9% w/v saline solution (or a 5.4% w/v glucose solution) is isotonic with blood plasma and is used for IV drips and eye drops.

Solved Example 9

The osmotic pressure of blood is 7.65 atm at 310 K. Calculate the concentration (in % w/v) of an aqueous glucose solution that will be isotonic with blood. Molar mass of glucose = 180 g/mol.

Solution:

For isotonic solutions: $π_{glucose} = π_{blood}$ .

$π V = \frac{W}{M} R T$

$7.65 \times V = \frac{W}{180} \times 0.0821 \times 310$

$\frac{W}{V} = \frac{7.65 \times 180}{0.0821 \times 310} = 54.1$ g/L $= 5.41$ g/100 mL $= 5.41%$ w/v.

6. Reverse Osmosis and Water Purification

The direction of osmosis can be reversed by applying an external pressure larger than the osmotic pressure on the solution side. Pure solvent then flows out of the solution through the semipermeable membrane. This is called reverse osmosis.

Figure 5: Reverse osmosis. Applying

P > π

on the sea-water side drives solvent through the membrane against the natural direction of osmosis, leaving the dissolved ions behind. At

P = π

exactly, osmosis merely stops; only

P > π

reverses it.

Practical use: reverse osmosis is the primary technology for desalination of sea water. A pressure greater than the osmotic pressure of sea water (~30 atm) is applied on the sea-water side; a suitable membrane (typically cellulose acetate on a porous support) lets pure water through but blocks dissolved salts. Countries such as Israel, Saudi Arabia, and coastal parts of India depend on RO plants for drinking water.

Domestic RO water purifiers work on the same principle at much lower pressures, using thin-film composite membranes to remove dissolved ions, bacteria, and small organic molecules.

7. Everyday Applications of Colligative Properties

Antifreeze in car radiators: ethylene glycol added to radiator water lowers its freezing point (via $Δ T_{f} = K_{f} m$ ), preventing radiator damage in freezing weather. It simultaneously raises the boiling point, protecting against overheating.
Salting icy roads: NaCl and CaCl $_{2}$ spread on winter roads dissolve in the surface layer of water and depress its freezing point to well below 0 °C, melting the ice. CaCl $_{2}$ is more effective because it dissociates into three ions ( $i \approx 3$ ) vs NaCl's two.
Food preservation: salting meats (bacon, fish) and adding sugar to jams create a hypertonic environment. Water is drawn out of bacterial cells by osmosis, killing them and preventing spoilage.
Osmoregulation in living cells: plants take up water through roots by osmosis. Kidneys use osmotic pressure gradients to filter blood.
Reverse osmosis desalination and purification as above.
Cryoscopy and ebullioscopy - laboratory determination of molar masses of unknown solids.

8. Determination of Molar Mass - Summary

Property	Formula	Molar Mass Expression	Best For
Relative VP lowering	$\frac{P ^{\circ} - P}{P ^{\circ}} = x_{2}$	$M_{2} = \frac{W _{2} M _{1} P ^{\circ}}{W _{1} ( P ^{\circ} - P )}$	Non-volatile solutes
BP elevation	$Δ T_{b} = K_{b} m$	$M_{2} = \frac{1000 K _{b} W _{2}}{Δ T _{b} W _{1}}$	Small molecules
FP depression	$Δ T_{f} = K_{f} m$	$M_{2} = \frac{1000 K _{f} W _{2}}{Δ T _{f} W _{1}}$	Small-medium molecules (bigger effect)
Osmotic pressure	$π = M R T$	$M_{2} = \frac{W _{2} R T}{π V}$	Polymers, proteins (large molecules)

Common Mistakes to Avoid

Watch out

Colligative property formulas use molality (mass of solvent), not molarity. The exception is osmotic pressure $π = M R T$ , which uses molarity. Do not mix them up.
For electrolyte solutes, the standard formulas underestimate the effect. Multiply by the Van't Hoff factor $i$ : $Δ T_{f} = i \cdot K_{f} \cdot m$ , $Δ T_{b} = i \cdot K_{b} \cdot m$ , $π = i \cdot M R T$ .
$K_{b}$ and $K_{f}$ are solvent properties, not solute properties. They do not change when you change the solute. Water has $K_{b} = 0.512$ , $K_{f} = 1.86$ ; benzene has $K_{b} = 2.53$ , $K_{f} = 5.12$ .
When calculating amount of ice separated on cooling, use $Δ T_{f} = K_{f} \cdot m_{remaining}$ - the molality is based on the water that stays liquid, not the original water.
Boiling point elevates and freezing point depresses - both are shifts caused by adding solute, but in opposite directions on the temperature axis.
Osmotic pressure formulas use $T$ in kelvin, not Celsius.
Reverse osmosis requires applied pressure greater than $π$ , not equal to $π$ . At $P = π$ exactly, osmosis just stops - it does not reverse.

Frequently Asked Questions

Q1. Why are colligative properties called "colligative"?

The word comes from Latin colligatus meaning "bound together". These four properties share a common cause: dissolved solute particles reduce the escape tendency of solvent molecules. Because the effect depends only on the number of solute particles and not their identity, all four properties are "tied together" by this single mechanism.

Q2. Why is freezing point depression preferred over boiling point elevation for molar mass determination?

The cryoscopic constant $K_{f}$ is generally larger than the ebullioscopic constant $K_{b}$ (for water 1.86 vs 0.512). This means FP depression gives a bigger temperature shift per mole of solute, making the measurement more precise. Also, freezing point can be measured accurately without worry about decomposition of thermolabile solutes, which is a concern at boiling temperatures.

Q3. Why is osmotic pressure the best method for determining molar masses of polymers?

Polymers have very large molar masses (10,000 g/mol and up), so even at 1% concentration the molality is tiny. BP elevation and FP depression give unmeasurably small temperature changes. Osmotic pressure, being expressed in atmospheres and depending on molar concentration, can still produce measurable pressure differences (millimetres of solvent column) for dilute polymer solutions.

Q4. What happens if a plant is watered with sea water?

Sea water is hypertonic to plant cell sap. Water flows out of the root cells by osmosis instead of into them, dehydrating the plant. The plant wilts and eventually dies. This is why irrigation with brackish water damages crops - a real problem in coastal agriculture.

Q5. Why does adding salt to icy roads melt the ice?

Salt dissolves in the thin film of liquid water always present on the ice surface. This saltwater solution has a freezing point well below 0 °C (because $Δ T_{f} = i K_{f} m$ ; for NaCl, $i = 2$ ). If the ambient temperature is above this new freezing point, the salt solution stays liquid and continues to dissolve more ice, gradually melting it away. CaCl $_{2}$ works even better because it dissociates into 3 ions instead of 2.

Q6. Why does ethylene glycol work as antifreeze in car radiators?

Ethylene glycol is highly soluble in water and non-volatile. When added to radiator water it produces two protective effects at once: freezing point depression (prevents the coolant from freezing and cracking the radiator in winter) and boiling point elevation (raises the boiling point above 100 °C, preventing coolant loss on hot engines). Typical 50% ethylene glycol lowers the freezing point to about $- 37$ °C.

Q7. What is the difference between osmosis and diffusion?

Diffusion is the general movement of any particles (solute or solvent) from high to low concentration, requiring no membrane. Osmosis is specifically the movement of solvent through a semipermeable membrane from lower to higher solute concentration. The membrane selectively blocks the solute; only solvent can move. Osmosis is a special case of diffusion restricted by the membrane.

Q8. Why is Van't Hoff's equation $π V = n R T$ similar to the ideal gas equation?

In a dilute solution, solute particles are spread far apart with little interaction, behaving somewhat like gas molecules in a container. Van't Hoff showed that the pressure they exert on a semipermeable barrier obeys the same formula as gas pressure. The similarity is real but limited to dilute solutions where solute-solute interactions are negligible - just as ideal gas behaviour requires low density.

Q9. Why does BP elevation depend on molality but osmotic pressure on molarity?

BP and FP shifts are measured over a range of temperatures, and molality is preferred because it is temperature-independent (mass does not change with T). Osmotic pressure, however, is measured at a fixed temperature and uses molar concentration (mol/L). At the fixed measurement temperature, no temperature-independence issue arises, so molarity is the natural choice - and it also connects directly to the ideal gas-like form $π V = n R T$ .

Previous year questions on Colligative Properties

43 questions from past papers, each with a step-by-step solution.

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1. What Are Colligative Properties

2. Relative Lowering of Vapour Pressure

3. Elevation of Boiling Point

4. Depression of Freezing Point

5. Osmosis and Osmotic Pressure

Isotonic, Hypertonic, and Hypotonic Solutions

6. Reverse Osmosis and Water Purification

7. Everyday Applications of Colligative Properties

8. Determination of Molar Mass - Summary

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why are colligative properties called "colligative"?

Q2. Why is freezing point depression preferred over boiling point elevation for molar mass determination?

Q3. Why is osmotic pressure the best method for determining molar masses of polymers?

Q4. What happens if a plant is watered with sea water?

Q5. Why does adding salt to icy roads melt the ice?

Q6. Why does ethylene glycol work as antifreeze in car radiators?

Q7. What is the difference between osmosis and diffusion?

Q8. Why is Van't Hoff's equation πV=nRT similar to the ideal gas equation?

Q9. Why does BP elevation depend on molality but osmotic pressure on molarity?

Previous year questions on Colligative Properties

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Q8. Why is Van't Hoff's equation $π V = n R T$ similar to the ideal gas equation?