krit.club logo

Solutions - Colligative properties

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Colligative properties are properties of a solution that depend only on the total number of solute particles (ions or molecules) present in a given amount of solvent, regardless of their chemical identity.

•

Relative Lowering of Vapour Pressure: According to Raoult's Law, for a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute: P0−PsP0=Xsolute\frac{P^0 - P_s}{P^0} = X_{solute}.

•

Elevation of Boiling Point (ΔTb\Delta T_b): The boiling point of a solution containing a non-volatile solute is always higher than that of the pure solvent. ΔTb=Tb−Tb0\Delta T_b = T_b - T_b^0. It is directly proportional to the molality (mm) of the solution.

•

Depression of Freezing Point (ΔTf\Delta T_f): The freezing point of a solution is lower than that of the pure solvent. ΔTf=Tf0−Tf\Delta T_f = T_f^0 - T_f. The lowering is proportional to the molal concentration of the solute.

•

Osmotic Pressure (π\pi): The minimum excess pressure that has to be applied on the solution side to prevent the entry of the solvent into the solution through a semi-permeable membrane. It follows the equation π=CRT\pi = CRT.

•

van't Hoff Factor (ii): This factor accounts for the extent of association or dissociation of solute particles in a solution. i=Observed Colligative PropertyCalculated Colligative Propertyi = \frac{\text{Observed Colligative Property}}{\text{Calculated Colligative Property}}. For dissociation, i>1i > 1; for association, i<1i < 1.

•

Abnormal Molar Mass: When a solute undergoes association or dissociation, the molar mass determined by colligative properties differs from the theoretical value. Mobserved=McalculatediM_{observed} = \frac{M_{calculated}}{i}.

📐Formulae

P0−PsP0=nBnA+nB≈wB⋅MAMB⋅wA\frac{P^0 - P_s}{P^0} = \frac{n_B}{n_A + n_B} \approx \frac{w_B \cdot M_A}{M_B \cdot w_A}

ΔTb=i⋅Kb⋅m=i⋅Kb⋅wB⋅1000MB⋅wA\Delta T_b = i \cdot K_b \cdot m = i \cdot K_b \cdot \frac{w_B \cdot 1000}{M_B \cdot w_A}

ΔTf=i⋅Kf⋅m=i⋅Kf⋅wB⋅1000MB⋅wA\Delta T_f = i \cdot K_f \cdot m = i \cdot K_f \cdot \frac{w_B \cdot 1000}{M_B \cdot w_A}

π=i⋅C⋅R⋅T=i⋅nBV⋅R⋅T\pi = i \cdot C \cdot R \cdot T = i \cdot \frac{n_B}{V} \cdot R \cdot T

i=1+(n−1)α (For Dissociation)i = 1 + (n-1)\alpha \text{ (For Dissociation)}

i=1+(1n−1)α (For Association)i = 1 + (\frac{1}{n}-1)\alpha \text{ (For Association)}

Kb=R⋅MA⋅(Tb0)21000⋅ΔHvapK_b = \frac{R \cdot M_A \cdot (T_b^0)^2}{1000 \cdot \Delta H_{vap}}

💡Examples

Problem 1:

Calculate the osmotic pressure of a 0.05 M0.05 \, M solution of NaClNaCl at 27∘C27^{\circ}C, assuming NaClNaCl is 100%100\% dissociated. (Given R=0.0821 L atm K−1 mol−1R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1})

Solution:

  1. For NaClNaCl, the number of ions n=2n = 2 (Na+Na^+ and Cl−Cl^-). Since it is 100%100\% dissociated, i=2i = 2.
  2. Temperature in Kelvin: T=27+273=300 KT = 27 + 273 = 300 \, K.
  3. Use the formula π=iCRT\pi = iCRT: π=2⋅0.05⋅0.0821⋅300\pi = 2 \cdot 0.05 \cdot 0.0821 \cdot 300 π=0.1⋅24.63=2.463 atm\pi = 0.1 \cdot 24.63 = 2.463 \, atm.

Explanation:

The van't Hoff factor ii is applied because NaClNaCl is an electrolyte that dissociates into two ions, doubling the number of particles compared to a non-electrolyte.

Problem 2:

A solution is prepared by dissolving 18 g18 \, g of glucose (C6H12O6C_6H_{12}O_6) in 1 kg1 \, kg of water (H2OH_2O). At what temperature will this solution boil at 1 atm1 \, atm pressure? (Given KbK_b for H2O=0.52 K kg mol−1H_2O = 0.52 \, K \, kg \, mol^{-1} and Tb0=373.15 KT_b^0 = 373.15 \, K)

Solution:

  1. Molar mass of glucose MB=180 g mol−1M_B = 180 \, g \, mol^{-1}.
  2. Molality m=wBMB⋅WA(kg)=18180⋅1=0.1 mol kg−1m = \frac{w_B}{M_B \cdot W_A(kg)} = \frac{18}{180 \cdot 1} = 0.1 \, mol \, kg^{-1}.
  3. Since glucose is a non-electrolyte, i=1i = 1.
  4. ΔTb=Kb⋅m=0.52⋅0.1=0.052 K\Delta T_b = K_b \cdot m = 0.52 \cdot 0.1 = 0.052 \, K.
  5. Boiling point of solution Tb=Tb0+ΔTb=373.15+0.052=373.202 KT_b = T_b^0 + \Delta T_b = 373.15 + 0.052 = 373.202 \, K.

Explanation:

The addition of a non-volatile solute like glucose increases the boiling point of the solvent. The elevation is calculated using the molal elevation constant.

Colligative properties - Revision Notes & Key Formulas | ICSE Class 12 Chemistry