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The Mistake Bank

Class 12 - Chapter 1: Solutions

Colligative properties depend strictly on the number of particles. Did you count the ions correctly, or did Van't Hoff trick you again?

1. The "i" Factor (Van't Hoff)

Colligative Properties

Scenario: Compare the Osmotic Pressure ($\pi$) of a $0.1\text{M}$ Glucose solution and a $0.1\text{M}$ $NaCl$ solution at the same temperature.

✖ What Students Do

Student sees the identical molar concentration ($0.1\text{M}$) and assumes colligative properties depend only on the written concentration.

Answer given: "Both have the same osmotic pressure."

(They completely ignored the dissociation of the ionic salt!)

✔ The Correct Way

Check if the Solute is an Electrolyte!

Colligative properties depend on the total number of particles in solution, represented by the Van't Hoff factor ($i$).
- Glucose is a non-electrolyte (does not split): $i = 1$.
- $NaCl$ is a strong electrolyte ($NaCl \to Na^+ + Cl^-$): $i = 2$.
Since $\pi = i \cdot C \cdot R \cdot T$, the $NaCl$ solution will exert double the osmotic pressure of the glucose solution!

2. Henry's Law Constant ($K_H$)

Solubility of Gases

Scenario: At a given temperature, Gas A has a higher Henry's Law Constant ($K_H$) than Gas B. Which gas is more soluble in water?

✖ What Students Do

Student assumes a direct relationship based on terminology: "A larger constant must mean a larger solubility."

Answer given: "Gas A is more soluble."

✔ The Correct Way

Higher $K_H$ = Lower Solubility!

According to Henry's Law: $P_{gas} = K_H \times \chi_{gas}$
Rearranging for solubility (mole fraction): $\chi_{gas} = \frac{P_{gas}}{K_H}$
At a constant partial pressure, the solubility is inversely proportional to $K_H$. Because Gas A has a higher $K_H$, it is less soluble.
Answer: Gas B is more soluble.

3. Units of R in Osmosis

Calculation Errors

Scenario: Calculate the osmotic pressure ($\pi = CRT$). The volume is in Liters and the required pressure unit is atmospheres (atm).

✖ What Students Do

Student habitually uses the universal gas constant from Thermodynamics/Physics:

$$ R = 8.314 \text{ J K}^{-1}\text{mol}^{-1} $$

(This will yield an answer in Pascals, not atmospheres! Your numerical answer will be off by a factor of 101325.)

✔ The Correct Way

Match the Pressure Units!

For solution chemistry, the pressure is almost always requested in $atm$ or $bar$, and volume in Liters.
- If pressure is in atm, use: $\mathbf{R = 0.0821 \text{ L atm K}^{-1}\text{mol}^{-1}}$
- If pressure is in bar, use: $\mathbf{R = 0.083 \text{ L bar K}^{-1}\text{mol}^{-1}}$

4. Azeotropes vs Deviation

Non-Ideal Solutions

Scenario: A binary liquid solution shows a large Positive Deviation from Raoult's Law. What kind of Azeotrope does it form?

✖ What Students Do

Student intuitively connects the word "Positive" with "Maximum".

Answer given: "Maximum Boiling Azeotrope."

(The logic is inverted!)

✔ The Correct Way

Positive Deviation = Minimum Boiling!

"Positive deviation" means the A-B interactions are weaker than A-A or B-B. The molecules escape into the vapor phase more easily.
Because the vapor pressure is unusually HIGH, the solution will boil at a much LOWER temperature than either pure component.
Answer: Minimum Boiling Azeotrope (e.g., Ethanol + Water).

5. Relative Lowering Formula

Raoult's Law

Scenario: Calculate the molar mass of an unknown non-volatile solute using the Relative Lowering of Vapor Pressure.

✖ What Students Do

Student confuses the colligative property formulas and uses Molality ($m$) instead of Mole Fraction ($\chi$).

$$ \frac{P^\circ - P}{P^\circ} = \text{Molality} \times K $$

(Incorrect formula. Relative lowering has no Molality or constant $K$!)

✔ The Correct Way

It equals the Mole Fraction of the Solute!

$$ \frac{P^\circ - P_s}{P^\circ} = \chi_{solute} = \frac{n_2}{n_1 + n_2} $$
For very dilute solutions, you can safely approximate the denominator as just $n_1$ (moles of solvent) to make the math easier: $\frac{P^\circ - P_s}{P^\circ} \approx \frac{n_2}{n_1}$. But be careful with concentrated solutions!

6. Mixing Volumes (Non-Ideal)

Solution Properties

Scenario: You mix exactly 10 mL of Ethanol with 10 mL of pure Water at room temperature. What is the total volume of the resulting solution?

✖ What Students Do

Student assumes simple arithmetic addition applies universally.

Answer given: "$10 + 10 = \mathbf{20 \text{ mL}}$".

(Volumes are only perfectly additive in strictly Ideal Solutions!)

✔ The Correct Way

$\Delta V_{mix} \neq 0$ for non-ideal mixtures!

Ethanol and Water form extensive new hydrogen bonds when mixed. These strong intermolecular forces pull the molecules slightly closer together than they were in their pure states ($\Delta V_{mix} < 0$).
The total volume will actually be slightly less than 20 mL (around 19.2 mL)!

7. Boiling Point Elevation Trick

Colligative Properties

Scenario: You calculate the boiling point elevation ($\Delta T_b$) of an aqueous solution to be $0.52\text{K}$. What is the boiling point of the solution?

✖ What Students Do

Student calculates $\Delta T_b = K_b \cdot m$ successfully, gets $0.52$, and immediately boxes it as the final answer.

Answer given: "$0.52^\circ\text{C}$ or $0.52\text{K}$".

($\Delta T_b$ is the CHANGE in boiling point, not the actual boiling point!)

✔ The Correct Way

Add the elevation to the pure solvent's boiling point!

The boiling point of pure water ($T_b^\circ$) is $100^\circ\text{C}$ (or $373.15\text{K}$).
$T_b = T_b^\circ + \Delta T_b$
$T_b = 100^\circ\text{C} + 0.52^\circ\text{C} = \mathbf{100.52^\circ\text{C}}$
(Note: A temperature difference of $0.52\text{K}$ is exactly equal to a difference of $0.52^\circ\text{C}$).

8. Vapor Phase Mole Fraction ($Y_A$)

Raoult's Law

Scenario: A solution contains liquid A and liquid B. Calculate the mole fraction of component A in the vapor phase above the solution.

✖ What Students Do

Student uses the standard liquid phase mole fraction formula:

$$ X_A = \frac{n_A}{n_A + n_B} $$

(This tells you the composition of the LIQUID, not the vapor above it!)

✔ The Correct Way

Use Dalton's Law of Partial Pressures!

To find the mole fraction in the vapor phase (usually denoted as $Y_A$), you must relate the partial pressure of the gas to the total vapor pressure.
$$ Y_A = \frac{P_A}{P_{total}} $$
Where $P_A = P_A^\circ \cdot X_A$ (from Raoult's Law) and $P_{total} = P_A + P_B$.

9. Isotonic Solution Trap

Osmosis

Scenario: A $0.1\text{M}$ solution of Urea is found to be isotonic with an unknown concentration of $NaCl$. Find the concentration of the $NaCl$ solution.

✖ What Students Do

Student remembers "Isotonic means equal concentration."

They immediately conclude $C_1 = C_2$.

Answer given: "$0.1\text{M} \text{ } NaCl$".

✔ The Correct Way

Isotonic means equal OSMOTIC PRESSURE, not molarity!

$\pi_1 = \pi_2 \implies i_1 C_1 R T = i_2 C_2 R T \implies \mathbf{i_1 C_1 = i_2 C_2}$
- Urea is non-electrolyte: $i_1 = 1$.
- $NaCl$ dissociates into 2 ions: $i_2 = 2$.
$1 \times (0.1\text{M}) = 2 \times C_2$
$C_2 = 0.1 / 2 = \mathbf{0.05\text{M}}$ (The $NaCl$ must be half the concentration to exert the same pressure).

10. Dimerization (i < 1)

Van't Hoff Factor

Scenario: Ethanoic acid ($CH_3COOH$) is dissolved in Benzene. What is the value of the Van't Hoff factor ($i$)?

✖ What Students Do

Student remembers that Acetic acid is a weak acid that dissociates in water. They assume it does the same here.

Answer given: "$1 < i < 2$".

✔ The Correct Way

In non-polar solvents, it Associates!

Benzene is a non-polar solvent. Instead of dissociating, two molecules of Ethanoic acid hydrogen-bond to each other to form a Dimer.
Because 2 molecules become 1 particle, the total number of particles decreases.
For 100% association, $i = 1/2 = \mathbf{0.5}$. (Therefore, $i < 1$).

11. Ebullioscopic Constant ($K_b$)

Constants

Scenario: Does the boiling point elevation constant ($K_b$) of water change if you dissolve 1 mole of Glucose versus 1 mole of Sucrose?

✖ What Students Do

Student thinks the constant depends on the nature of the solute added.

Answer given: "Yes, $K_b$ changes because the solutes are different molecules with different masses."

✔ The Correct Way

$K_b$ depends ONLY on the SOLVENT!

$K_b$ (and $K_f$) are thermodynamic constants specific to the solvent itself (e.g., Water has $K_b = 0.52 \text{ K kg mol}^{-1}$).
It does not matter if you dissolve glucose, sucrose, or urea; as long as the solvent is water, the constant remains exactly the same.

12. The Negative Freezing Point

Depression in FP

Scenario: You calculate the depression in freezing point ($\Delta T_f$) of an aqueous solution to be $2^\circ\text{C}$. What is the freezing point of the solution?

✖ What Students Do

Student boxes the $\Delta T_f$ value as the final answer, or assumes the freezing point is positive $2^\circ\text{C}$.

Or worse, they think the formula is $\Delta T_f = T_f - T_f^\circ$ (like boiling point) and get confused by the signs.

✔ The Correct Way

Subtract from the pure solvent's freezing point!

The formula is $\Delta T_f = T_f^\circ - T_f$ (Pure minus Solution) because the solution freezes at a lower temperature.
For water, $T_f^\circ = 0^\circ\text{C}$.
$T_f = 0^\circ\text{C} - 2^\circ\text{C} = \mathbf{-2^\circ\text{C}}$.
The freezing point is negative!

13. Reverse Osmosis Condition

Osmosis

Scenario: To purify seawater via Reverse Osmosis (RO), how much pressure must be applied to the solution side?

✖ What Students Do

Student writes: "A pressure exactly equal to the osmotic pressure ($\pi$) must be applied."

(Applying pressure equal to $\pi$ only STOPS regular osmosis. It doesn't reverse it!)

✔ The Correct Way

Applied Pressure must be GREATER than $\pi$!

To force solvent molecules (pure water) to flow backwards against their natural concentration gradient (from the salty solution side to the pure water side), the external pressure applied ($P_{ext}$) must exceed the osmotic pressure.
Condition for RO: $\mathbf{P_{ext} > \pi}$.

14. Total Pressure Calculation

Raoult's Law

Scenario: Calculate the total vapor pressure of a solution containing two volatile liquids, A and B.

✖ What Students Do

Student calculates the partial pressure of the main solvent ($P_A = P_A^\circ X_A$) and presents it as the total pressure.

(This only works if solute B is non-volatile!)

✔ The Correct Way

Add BOTH partial pressures!

Because both components are volatile (they both evaporate), the total pressure above the solution is the sum of their individual partial pressures (Dalton's Law).
$P_{total} = P_A + P_B$
$\mathbf{P_{total} = (P_A^\circ \cdot X_A) + (P_B^\circ \cdot X_B)}$

15. Fractional Distillation Limits

Azeotropes

Scenario: Can an Azeotropic mixture (like 95% Ethanol + 5% Water) be separated into 100% pure components using fractional distillation?

✖ What Students Do

Student thinks: "Fractional distillation separates liquids based on boiling point differences. Since it's a liquid mixture, yes, it can be separated."

✔ The Correct Way

NO! Azeotropes boil at a constant temperature.

An Azeotrope is a "constant boiling mixture". At this specific composition, the mole fraction of the components in the liquid phase is exactly equal to their mole fraction in the vapor phase ($X_A = Y_A$).
Because the vapor has the exact same composition as the boiling liquid, boiling it further does not change the concentration. Fractional distillation fails completely at this point!

Confess Your Sins!

"A solution is a homogeneous mixture... until you make a heterogeneous mistake."

Did one of these catch you? Or do you have a different horror story from your last exam?

Scroll down to the comments section below and tell us:

"Which Solutions trap cost you the most marks?"

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