For a 5% solution of urea (Molar mass = 60 g/mol), calculate the osmotic pressure at 300 K. [R = 0·0821 L atm

Question

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(OR)

Visha took two aqueous solutions — one containing 7\cdot5 g of urea (Molar mass = 60 g/mol) and the other containing 42\cdot75 g of substance Z in 100 g of water, respectively. It was observed that both the solutions froze at the same temperature. Calculate the molar mass of Z.

CBSECBSE Class XII Board 2020Subjective· 2mImportance★★★★★

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Concept understanding — Osmotic Pressure

Osmotic Pressure: The Push of Pure Solvent

Imagine you have a U-shaped tube with a special membrane at the bottom that only lets water molecules pass through — not sugar molecules. On one side you put pure water, on the other side you put a sugar solution. What happens?

Water moves from the pure side into the solution side. The solution level rises. This is osmosis — the spontaneous net movement of solvent across a semipermeable membrane from a region of lower solute concentration to higher solute concentration.

But here's the key question: what if you don't want that level to rise? What if you want to keep the solution side exactly where it is?

You would have to push down on the solution side with extra pressure — just enough to stop the water from coming in. That extra pressure is osmotic pressure.

Note

Osmotic pressure is not a pressure the solution "has" inside it. It is the external pressure you must apply to prevent osmosis. Think of it as the "resistance pressure" that exactly balances the tendency of solvent to dilute the solution.

The Precise Definition

Osmotic pressure ( $\Pi$ ) is the minimum excess pressure that must be applied to a solution to prevent the inward flow of solvent across a semipermeable membrane.

The membrane must be permeable only to solvent molecules, not to solute particles. This is the defining condition — if the membrane leaks solute, you don't get true osmotic pressure.

The van't Hoff Equation

For dilute solutions, osmotic pressure follows a beautifully simple law:

\Pi = CRT

Where:

$\Pi$ = osmotic pressure (in atm or Pa)
$C$ = molar concentration of solute (mol/L or mol/m³)
$R$ = universal gas constant
$T$ = absolute temperature (K)

This is van't Hoff's law of osmotic pressure. It looks exactly like the ideal gas law ( $PV = nRT$ rearranged as $P = (n/V)RT$ ), and that's no coincidence — van't Hoff noticed that solute particles in dilute solution behave like gas molecules bouncing around, creating a "pressure" against the membrane.

Watch out

This equation works only for non-electrolyte solutions at low concentrations. For electrolytes, you must include the van't Hoff factor $i$ : $\Pi = iCRT$ . A 0.1 M NaCl solution gives nearly twice the osmotic pressure of a 0.1 M glucose solution because NaCl dissociates into two ions.

Why It Matters for Macromolecules

Here's where osmotic pressure becomes a powerful tool. Suppose you have a protein — say, hemoglobin — and you want to know its molar mass. You can't easily measure its concentration in mol/L because you don't know the molar mass yet. But you can measure:

The mass of protein dissolved (say, $w$ grams in $V$ liters)
The osmotic pressure $\Pi$ of that solution

Since $C = n/V = (w/M)/V$ , where $M$ is the molar mass:

\Pi = \frac{wRT}{MV}

Rearrange:

M = \frac{wRT}{\Pi V}

Tip

| Property | Why osmotic pressure wins |

|----------|---------------------------|

| Boiling point elevation | Very small for macromolecules — hard to measure |

| Freezing point depression | Very small — same problem | …

(a) $\Pi = CRT = 0.833\times0.0821\times300 \approx 20.5$ atm.

(b) Equal $\Delta T_f$ ⇒ equal molality ⇒ $M_Z = 42.75/(7.5/60) = 342$ g mol $^{-1}$ .

Osmotic pressure of 5% urea

A 5% (w/V) urea solution contains 5 g urea in 100 mL of solution, i.e. 50 g in 1 L.

Moles of urea per litre:

C = \frac{50\ \text{g}}{60\ \text{g mol}^{-1}} = 0.833\ \text{mol L}^{-1}

Applying the van't Hoff equation ( $i = 1$ for urea):

\Pi = CRT = 0.833 \times 0.0821 \times 300 = 20.5\ \text{atm}

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Concept understanding — Depression of Freezing Point

Depression of Freezing Point

Imagine a cold winter morning. You see water on the road turning to ice at 0°C. But if you sprinkle salt on that ice, it melts — even though the temperature is still below zero. That’s the same phenomenon that keeps roads safe in snowy countries and makes ice cream freeze in a churn. The salt lowers the freezing point of water.

That is the core idea: when you dissolve a non-volatile solute (like salt, sugar, or urea) in a solvent (like water), the freezing point of the solution becomes lower than that of the pure solvent. This drop is called the depression of freezing point, denoted by $\Delta T_f$ .

Why does this happen? The intuition

In a pure liquid, molecules at the surface escape into the solid (freeze) when the temperature is low enough — the solid and liquid are in equilibrium at the freezing point. Now add a solute. The solute particles sit between solvent molecules, getting in the way. For the solvent to freeze, its molecules must arrange themselves into an orderly crystal lattice. The solute particles disrupt this order — they make it harder for the solvent to solidify.

Think of it like trying to pack a suitcase full of neatly stacked blocks. If you throw in a few marbles, the blocks can’t settle as tightly. You’d need to cool the system further (lower the temperature) to force the blocks into place. That extra cooling is the depression.

Note

The solute must be non-volatile (it doesn’t evaporate) and non-electrolyte (it doesn’t break into ions) for the simplest formula to work. If the solute dissociates (like NaCl → Na⁺ + Cl⁻), the effect is larger — but that’s a refinement you’ll meet later.

The precise statement

For a dilute solution, the depression in freezing point is directly proportional to the molality of the solution (moles of solute per kilogram of solvent).

\Delta T_f = K_f \cdot m

Where:

$\Delta T_f = T_f^\circ - T_f$ (pure solvent freezing point minus solution freezing point)
$K_f$ = cryoscopic constant or molal freezing point depression constant — a property of the solvent alone (units: K kg mol⁻¹)
$m$ = molality of the solution

\Delta T_f = K_f \cdot m

Each solvent has its own $K_f$ . For water, $K_f = 1.86 \text{ K kg mol}^{-1}$ . That means: a 1 molal aqueous solution freezes at $-1.86^\circ\text{C}$ (instead of $0^\circ\text{C}$ ).

How it helps find molar mass

If you dissolve a known mass $w_2$ of an unknown solute in a known mass $w_1$ of solvent, measure the freezing point depression $\Delta T_f$ , you can calculate the molar mass $M_2$ of the solute.

Start from the definition of molality:

m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{w_2 / M_2}{w_1 / 1000}

Substitute into $\Delta T_f = K_f \cdot m$ :

\Delta T_f = K_f \cdot \frac{w_2 \times 1000}{M_2 \times w_1}

Rearrange for $M_2$ :

M_2 = \frac{K_f \cdot w_2 \cdot 1000}{\Delta T_f \cdot w_1}

Tip

This is the most common exam formula. Remember: $w_1$ is in grams, $w_2$ in grams, and the factor 1000 converts grams of solvent to kilograms.

A quick example

You dissolve 5.00 g of a non-electrolyte in 100 g of water. The freezing point drops to $-0.93^\circ\text{C}$ . Find the molar mass.

Given: $K_f = 1.86$ , $\Delta T_f = 0.93$ , $w_2 = 5.00$ , $w_1 = 100$ . …

(a) $\Pi = CRT = 0.833\times0.0821\times300 \approx 20.5$ atm.

(b) Equal $\Delta T_f$ ⇒ equal molality ⇒ $M_Z = 42.75/(7.5/60) = 342$ g mol $^{-1}$ .

Molar mass of Z

Freezing-point depression is colligative: $\Delta T_f = K_f \, m$ . Both solutions use 100 g water and freeze at the same temperature, so they have equal $\Delta T_f$ and hence equal molality (urea and Z are both non-electrolytes, $i=1$ ). …