Q.At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Concept & Intuition

Osmotic pressure ($\Pi$) is a colligative property — it depends only on the number of solute particles, not their identity. For ionic compounds that dissociate completely, the effective particle concentration is the molar concentration multiplied by the van’t Hoff factor $i$ (the number of ions per formula unit).

Here, $\mathrm{A_3B}$ dissociates into 3 A⁺ ions and 1 B³⁻ ion, so $i = 4$. $\mathrm{MgCl_2}$ dissociates into 1 Mg²⁺ and 2 Cl⁻, so $i = 3$.

The problem states: $\Pi_{\mathrm{A_3B}} = 6 \times \Pi_{\mathrm{MgCl_2}}$. Since $\Pi = iCRT$ (with $R$ and $T$ constant), we can cancel $RT$ and solve for the unknown concentration $C_{\mathrm{A_3B}}$.

Step-by-step solution

1. Write the osmotic pressure formula

   For any solution, $\Pi = i C R T$, where $C$ is the molar concentration, $i$ is the van’t Hoff factor, $R$ is the gas constant, and $T$ is the absolute temperature. Since both solutions are at the same $T$, $R$ and $T$ cancel when we compare them.

2. Determine van’t Hoff factors

   • $\mathrm{MgCl_2} \rightarrow \mathrm{Mg^{2+}} + 2\mathrm{Cl^-}$ → $i = 1 + 2 = 3$
   • $\mathrm{A_3B} \rightarrow 3\mathrm{A^+} + \mathrm{B^{3-}}$ → $i = 3 + 1 = 4$

3. Set up the given relationship

   $\Pi_{\mathrm{A_3B}} = 6 \times \Pi_{\mathrm{MgCl_2}}$

   Substitute $\Pi = iCRT$:

4. Cancel $RT$ (same for both sides) …

The problem gives osmotic pressure in bar, volume in mL, mass in grams, and molar mass in g/mol. The van’t Hoff law says $\pi = cRT$, where $c$ is concentration in mol/L. Since $c = \frac{n}{V_{\text{liters}}}$ and $n = \frac{W_B}{M_B}$, we just need to handle the volume conversion from mL to L. The trick is that many options include extra factors like 760 or 76, which come from converting pressure units (e.g., atm to mm Hg), but here pressure is already in bar — and if we use R in appropriate units, no such factor is needed. Let’s derive step by step.

1. Start with the van’t Hoff equation Osmotic pressure $\pi$ (in bar) is related to molar concentration $c$ (in mol/L) by

where $R$ is the gas constant in L·bar/(mol·K) and $T$ is temperature in K.

2. Express concentration in terms of given quantities Molarity $c = \frac{\text{moles of solute}}{\text{volume in liters}} = \frac{W_B / M_B}{V/1000}$ because $V$ mL = $V/1000$ L. So

3. Substitute into the osmotic pressure equation

4. Solve for $W_B$ Multiply both sides by $M_B V$ and divide by $1000 RT$:

5. Check the options …

Step 1 — Volume of the sample.

Density relates mass to volume:

Step 2 — Volume of one FCC unit cell.

Convert the edge length to centimetres (to match the density's units):

Step 3 — Number of unit cells in the sample.

Step 4 — Atoms per FCC unit cell ($Z$).

In a face-centred cubic cell:

• 8 corner atoms, each shared by 8 cells $\Rightarrow 8\times\tfrac18 = 1$ atom;
• 6 face-centred atoms, each shared by 2 cells $\Rightarrow 6\times\tfrac12 = 3$ atoms.

Step 5 — Total atoms. …

Moles of glucose $=\dfrac{3.6}{180}=0.02$ mol.

Moles of urea $=\dfrac{1.2}{60}=0.02$ mol.

Both are non-electrolytes, so total solute particles $n=0.02+0.02=0.04$ mol.

Volume $V=200\text{ mL}=0.2$ L, $T=27^\circ C=300$ K. …

Step 1 — Why the other three fail for macromolecules.

A macromolecule has a huge molar mass ($M \sim 10^4$–$10^6$ g/mol), so even a few grams dissolved give a tiny number of moles ⇒ a very small molality/mole fraction. Now look at the magnitudes each property produces:

With $m \sim 10^{-4}$ mol/kg and $K_f, K_b$ of order $1$ K kg mol$^{-1}$, we get $\Delta T \sim 10^{-4}\,^\circ$C — far below what an ordinary thermometer can resolve. Relative lowering of vapour pressure is equally negligible. Any measurement error is then enormous relative to the signal, so the computed $M$ is unreliable.

Step 2 — Why osmotic pressure works.

The multiplier here is $RT \approx 0.0821 \times 300 \approx 25$ L atm mol$^{-1}$ — orders of magnitude larger in effect than $K_f$ or $K_b$. A concentration of just $10^{-4}$ mol/L still gives …

Step 1 — The governing law.

Osmotic pressure of a dilute solution (van 't Hoff):

This is the colligative property that measures the pressure needed to stop solvent flowing into the solution through a semipermeable membrane. Glucose is a non-electrolyte ($i = 1$), so no van 't Hoff factor correction is needed.

Step 2 — Compute the moles of solute.

Step 3 — Get the volume in litres.

The solution is dilute, so its volume is essentially the volume of the water. Using density of water $= 1$ g/mL:

Step 4 — Convert the temperature to kelvin.

(Never use $^\circ$C in a gas-law-type equation — the $T$ must be absolute.) …

Concept & Intuition

Osmotic pressure $\Pi$ is given by $\Pi = iCRT$, where $C$ is the molar concentration, $R$ is the gas constant, $T$ is the temperature, and $i$ is the van’t Hoff factor — the number of particles each formula unit produces in solution. For strong electrolytes (which all these salts are), $i$ equals the number of ions per formula unit.

Two solutions are isotonic when their osmotic pressures are equal at the same temperature. Since $R$ and $T$ are constant, this reduces to comparing the product $iC$ for each solution. The pair with matching $iC$ values is the answer.

Step-by-step solution

1. Determine the van’t Hoff factor $i$ for each salt

   • $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$ → $i = 3$
   • $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$ → $i = 3$
   • $NaCl \rightarrow Na^+ + Cl^-$ → $i = 2$
   • $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ → $i = 5$

2. Compute the effective particle concentration $iC$ for each solution in the options

   Option	Solution	$C$ (M)	$i$	$iC$ (M)
   (A)	$BaCl_2$	0.01	3	0.03
   (A)	$CaCl_2$	0.001	3	0.003
   (B)	$BaCl_2$	0.01	3	0.03

Step 1 — The definition.

Osmotic pressure $\pi$ is the pressure that must be applied to a solution to just stop the inward flow of solvent through a semipermeable membrane. Two solutions are said to be isotonic when

Placed on either side of a semipermeable membrane, isotonic solutions show no net osmosis — solvent crosses both ways at equal rates.

Step 2 — What that implies about concentration.

From the van't Hoff equation for dilute solutions,

so at the same $T$, isotonic solutions have the same molar concentration of solute particles — e.g. $0.91\%$ NaCl (normal saline) is isotonic with human blood cells, which is why it can be safely infused.

Step 3 — Rule out the others. …