Masterclass: 25 Solved JEE Advanced Numericals on Colligative Properties
Test your mettle with the most complex problem-solving scenarios covering the Ostwald-Walker method, partial dissociation, dimerization, and thermodynamic deviations. Click "View Solution" to reveal the answers.
These numericals combine chemical equilibrium, thermodynamics, and ionic behavior. Before proceeding, ensure you have mastered the core theories from our exhaustive guides on Elevation in Boiling Point, Depression in Freezing Point, and Osmosis.
View Solution
Step 1: Calculate Theoretical $\Delta T_f$
Theoretical $\Delta T_f$ (assuming no dissociation, $i=1$) = $K_f \times m = 1.86 \times 0.01 = 0.0186^{\circ}\text{C}$.
Step 2: Determine Van't Hoff Factor ($i$)
Observed $\Delta T_f = 0.062^{\circ}\text{C}$.
$i = \frac{\text{Observed } \Delta T_f}{\text{Theoretical } \Delta T_f} = \frac{0.062}{0.0186} = 3.33$.
Step 3: Calculate Degree of Dissociation ($\alpha$)
The complex $K_3[Fe(CN)_6]$ dissociates into $3K^+$ and $1 [Fe(CN)_6]^{3-}$, giving $n = 4$ ions.
$i = 1 + \alpha(n - 1) \implies 3.33 = 1 + \alpha(4 - 1)$
$2.33 = 3\alpha \implies \alpha = \frac{2.33}{3} = 0.777$.
View Solution
Step 1: Use the Exact Algebraic Formula
Instead of $\frac{P^{\circ} - P_s}{P^{\circ}} \approx \frac{n_2}{n_1}$, we must use the exact form derived for concentrated solutions: $\frac{P^{\circ} - P_s}{P_s} = \frac{n_2}{n_1}$.
Step 2: Calculate moles of solvent ($n_1$)
$n_1 = \frac{90.0 \text{ g}}{18 \text{ g/mol}} = 5.0 \text{ mol}$.
Step 3: Solve for moles of solute ($n_2$)
$\frac{23.76 - 23.32}{23.32} = \frac{n_2}{5.0}$
$\frac{0.44}{23.32} = \frac{n_2}{5.0} \implies n_2 = \frac{0.44 \times 5.0}{23.32} = 0.0943 \text{ mol}$.
Step 4: Find Molar Mass
$M_2 = \frac{\text{Mass}}{n_2} = \frac{5.40}{0.0943} = 57.2 \text{ g mol}^{-1}$.
View Solution
Step 1: Calculate Van't Hoff Factor ($i$)
$\pi = iCRT$
$2.60 = i \times 0.1 \times 0.0821 \times 300$
$2.60 = i \times 2.463 \implies i = 1.0556$.
Step 2: Calculate Degree of Dissociation ($\alpha$)
For a monobasic acid ($HA \rightleftharpoons H^+ + A^-$), $n = 2$.
$i = 1 + \alpha(2 - 1) \implies 1.0556 = 1 + \alpha \implies \alpha = 0.0556$.
Step 3: Apply Ostwald's Dilution Law
$K_a = \frac{C\alpha^2}{1 - \alpha}$. Since $\alpha$ is small, $K_a \approx C\alpha^2$.
$K_a = 0.1 \times (0.0556)^2 = 0.1 \times 0.00309 = 3.09 \times 10^{-4}$.
View Solution
Step 1: Calculate Molality
Moles = $1.0 / 238 = 0.0042 \text{ mol}$.
Molality ($m$) = $0.0042 / 0.1 \text{ kg} = 0.042 \text{ m}$.
Step 2: Find 'i' for complete ionization
$\Delta T_f = i K_f m$
$0.155 = i \times 1.86 \times 0.042$
$0.155 = i \times 0.07812 \implies i \approx 2$.
Step 3: Deduce Structure
Since $i = 2$ and ionization is complete, the complex yields exactly 2 ions per formula unit. This means only one chloride ion is outside the coordination sphere acting as the counter-ion.
Structure: $[Co(H_2O)_5Cl]Cl_2 \cdot H_2O$ yields 3 ions (Wrong).
Structure: $[Co(H_2O)_4Cl_2]Cl \cdot 2H_2O$ yields 2 ions (Correct).
View Solution
Step 1: Analyze Electrolysis stoichiometry
$1.08 \text{ g}$ of $Ag$ deposited = $1.08 / 108 = 0.01 \text{ mol}$ of $Ag^+$ removed from solution at the cathode.
Simultaneously at the inert anode, water oxidizes: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$.
For every $1 \text{ mol}$ of $e^-$ that deposits $1 \text{ mol}$ of $Ag$, $1 \text{ mol}$ of $H^+$ is generated. Thus, $0.01 \text{ mol}$ of $H^+$ is added to the solution.
Step 2: Total Ion Count
The $Ag^+$ ions are perfectly replaced 1-to-1 by $H^+$ ions. The total number of moles of ions in the solution remains completely unchanged! The original $0.2 \text{ M } AgNO_3$ yields $0.4 \text{ M}$ total ions ($0.2$ of cation, $0.2$ of $NO_3^-$).
Step 3: Calculate $\Delta T_f$
Effective molality $m \approx 0.4 \text{ m}$ (for total particles).
$\Delta T_f = 1.86 \times 0.4 = 0.744^{\circ}\text{C}$.
View Solution
Step 1: Apply Ostwald-Walker Proportionalities
Loss in solution bulbs $\propto P_s \implies P_s \propto 2.50$
Loss in solvent bulbs $\propto (P^{\circ} - P_s) \implies P^{\circ} - P_s \propto 0.04$
Total gain in $CaCl_2$ tube $\propto P^{\circ} \implies P^{\circ} \propto 2.54$
Step 2: Apply Exact Raoult's Law
$\frac{P^{\circ} - P_s}{P_s} = \frac{n_2}{n_1}$
$\frac{0.04}{2.50} = \frac{n_2}{80 / 18}$
$0.016 = \frac{n_2}{4.444} \implies n_2 = 0.0711 \text{ mol}$.
Step 3: Calculate Molar Mass
$M_2 = \frac{5.0 \text{ g}}{0.0711 \text{ mol}} = 70.3 \text{ g mol}^{-1}$.
View Solution
Step 1: Formula and Variables
$K_b = \frac{R M_1 (T_b^{\circ})^2}{1000 \Delta H_{\text{vap}}}$
$R = 8.314$, $T_b^{\circ} = 373.15$, $M_1 = 18 \text{ g mol}^{-1}$.
$\Delta H_{\text{vap}} = 40.66 \text{ kJ mol}^{-1} = 40660 \text{ J mol}^{-1}$.
Step 2: Calculation
$K_b = \frac{8.314 \times 18 \times (373.15)^2}{1000 \times 40660}$
$K_b = \frac{149.652 \times 139240.9}{40660000} = \frac{20837861}{40660000} = 0.512 \text{ K kg mol}^{-1}$.
View Solution
Step 1: Isotonic Condition
$\pi_{NaCl} = \pi_{\text{glucose}}$
$i \times C_{NaCl} \times RT = 1 \times C_{\text{glucose}} \times RT$
$i \times \left(\frac{1.2 / 58.5}{V}\right) = \left(\frac{7.2 / 180}{V}\right)$
Step 2: Solve for 'i'
$i \times 0.02051 = 0.040$
$i = \frac{0.040}{0.02051} = 1.95$.
Step 3: Calculate $\alpha$
$NaCl$ yields 2 ions ($n=2$).
$i = 1 + \alpha(2 - 1) \implies 1.95 = 1 + \alpha \implies \alpha = 0.95$.
View Solution
Step 1: Calculate Partial Pressures (Raoult's Law)
$P_A = x_A \times P^{\circ}_A = 0.40 \times 300 = 120 \text{ torr}$.
$P_B = x_B \times P^{\circ}_B = 0.60 \times 800 = 480 \text{ torr}$.
Step 2: Total Pressure
$P_{\text{total}} = P_A + P_B = 120 + 480 = 600 \text{ torr}$.
Step 3: Vapor Phase Mole Fraction (Dalton's Law)
$y_A = \frac{P_A}{P_{\text{total}}} = \frac{120}{600} = 0.20$.
View Solution
Step 1: Calculate Molality
Molar mass of Benzoic acid = $122 \text{ g/mol}$.
Molality $m = \frac{12.2 / 122}{0.1} = \frac{0.1}{0.1} = 1.0 \text{ m}$.
Step 2: Calculate 'i' for Dimerization
Dimerization means $n = 2$. Association extent $\alpha = 0.80$.
$i = 1 + \alpha\left(\frac{1}{n} - 1\right) = 1 + 0.80\left(\frac{1}{2} - 1\right)$
$i = 1 + 0.80(-0.5) = 1 - 0.40 = 0.60$.
Step 3: Calculate $\Delta T_f$
$\Delta T_f = i \times K_f \times m = 0.60 \times 5.12 \times 1.0 = 3.072 \text{ K}$.
View Solution
Step 1: Identify Moles and 'i'
$AB_2 \rightarrow A^{2+} + 2B^-$. Fully dissociated ($100\%$), so $i = 3$.
$n_2 = 10 / 100 = 0.1 \text{ mol}$. Effective solute moles $= 3 \times 0.1 = 0.3 \text{ mol}$.
$n_1 = 90 / 18 = 5.0 \text{ mol}$.
Step 2: Use Exact RLVP Equation
$\frac{P^{\circ} - P_s}{P_s} = \frac{i \times n_2}{n_1}$
$\frac{25.0 - P_s}{P_s} = \frac{0.3}{5.0} = 0.06$
$25.0 - P_s = 0.06 P_s$
$25.0 = 1.06 P_s \implies P_s = \frac{25.0}{1.06} = 23.58 \text{ torr}$.
View Solution
Step 1: Calculate Effective Moles
$BaCl_2$: Moles = $0.1 \text{ M} \times 0.1 \text{ L} = 0.01 \text{ mol}$. Since $i=3$, effective moles = $0.03 \text{ mol}$.
Urea: Moles = $0.1 \text{ M} \times 0.1 \text{ L} = 0.01 \text{ mol}$. Since $i=1$, effective moles = $0.01 \text{ mol}$.
Total effective moles $= 0.04 \text{ mol}$.
Step 2: Calculate New Volume and Osmotic Pressure
New Volume ($V$) = $100 + 100 = 200 \text{ mL} = 0.2 \text{ L}$.
$\pi = \frac{n_{\text{effective}}}{V} RT = \left(\frac{0.04}{0.2}\right) \times 0.0821 \times 300$
$\pi = 0.2 \times 24.63 = 4.926 \text{ atm}$.
View Solution
Step 1: Find Molality
$Ag_2SO_4 \rightarrow 2Ag^+ + SO_4^{2-}$, so $i = 3$.
$\Delta T_f = i \times K_f \times m \implies 0.00372 = 3 \times 1.86 \times m$
$m = \frac{0.00372}{5.58} = 6.66 \times 10^{-4} \text{ m}$.
For dilute aqueous solutions, molality $\approx$ molarity. So Solubility ($S$) = $6.66 \times 10^{-4} \text{ M}$.
Step 2: Calculate $K_{sp}$
$K_{sp} = [Ag^+]^2[SO_4^{2-}] = (2S)^2(S) = 4S^3$
$K_{sp} = 4(6.66 \times 10^{-4})^3 = 4(2.96 \times 10^{-10}) = 1.18 \times 10^{-9}$.
View Solution
A positive deviation from Raoult's law means the intermolecular forces between the mixed molecules (A-B) are weaker than those in the pure liquids (A-A or B-B). This causes the combined vapor pressure to be significantly higher than expected.
Because the vapor pressure is artificially high, it takes less thermal energy to push the vapor pressure up to match the external atmospheric pressure ($1 \text{ atm}$). Therefore, the mixture will boil at a temperature lower than both pure components. This is called a minimum-boiling azeotrope (e.g., $95\%$ Ethanol-Water).
View Solution
According to the Van't Hoff equation ($\pi = CRT$), if the concentration ($C$) and the universal gas constant ($R$) are held constant, osmotic pressure is directly proportional to absolute temperature ($T$).
$\frac{\pi_1}{T_1} = \frac{\pi_2}{T_2} \implies \frac{2.00}{300} = \frac{\pi_2}{330}$
$\pi_2 = \frac{2.00 \times 330}{300} = \frac{660}{300} = 2.20 \text{ atm}$.
View Solution
Proteins and polymers have massive molar masses (e.g., $60,000 \text{ g/mol}$). Because $M$ is in the denominator of the molality equation, a $1 \text{ gram}$ sample yields an extraordinarily tiny molality.
This tiny molality would result in a $\Delta T_b$ or $\Delta T_f$ on the order of $10^{-4 \circ}\text{C}$, which is impossible to measure accurately. Additionally, boiling or freezing a protein will denature (destroy) it. Conversely, that same tiny concentration produces a very large, easily readable osmotic pressure (like $2 \text{ mmHg}$) on a physical gauge at safe room temperatures.
View Solution
Step 1: Calculate 'i'
$\Delta T_b = i \times K_b \times m \implies 0.65 = i \times 2.6 \times 0.5$
$0.65 = 1.3 i \implies i = 0.5$.
Step 2: Calculate Extent of Dimerization ($\alpha$)
For dimerization, $n = 2$.
$i = 1 + \alpha\left(\frac{1}{2} - 1\right) = 1 - 0.5\alpha$
$0.5 = 1 - 0.5\alpha \implies 0.5\alpha = 0.5 \implies \alpha = 1.0$.
View Solution
Step 1: Calculate the natural Osmotic Pressure
Reverse osmosis begins only when the applied mechanical pressure exceeds the natural osmotic pressure of the seawater.
$NaCl$ dissociates into 2 ions, so $i = 2$.
$\pi = iCRT = 2 \times 0.6 \times 0.0821 \times 300$
$\pi = 1.2 \times 24.63 = 29.56 \text{ atm}$.
View Solution
Step 1: Calculate Effective Moles
$NaCl \rightarrow Na^+ + Cl^-$. $0.1 \text{ mol}$ yields $0.2 \text{ mol}$ of ions.
$HgCl_2$ is famously highly covalent and remains largely undissociated in water ($i \approx 1$). $0.1 \text{ mol}$ yields $0.1 \text{ mol}$ of particles.
Total effective particles $= 0.2 + 0.1 = 0.3 \text{ mol}$.
Step 2: Calculate $\Delta T_f$
$m = 0.3 \text{ mol} / 1 \text{ kg} = 0.3 \text{ m}$.
$\Delta T_f = K_f \times m = 1.86 \times 0.3 = 0.558^{\circ}\text{C}$.
View Solution
Step 1: Convert Equal Masses to Moles
Let mass be $m$. Molar mass of $H_2O = 18$, $D_2O = 20$.
$n_{H_2O} = m / 18$. $n_{D_2O} = m / 20$.
Step 2: Calculate Mole Fractions
$x_{H_2O} = \frac{m/18}{m/18 + m/20} = \frac{1/18}{(20+18)/360} = \frac{1/18}{38/360} = \frac{20}{38} = \frac{10}{19}$.
$x_{D_2O} = 1 - \frac{10}{19} = \frac{9}{19}$.
Step 3: Apply Raoult's Law
$P_{\text{total}} = x_{H_2O}P^{\circ}_{H_2O} + x_{D_2O}P^{\circ}_{D_2O}$
$P_{\text{total}} = \left(\frac{10}{19} \times 24\right) + \left(\frac{9}{19} \times 20\right) = \frac{240 + 180}{19} = \frac{420}{19} = 22.1 \text{ torr}$.
View Solution
Boiling point elevation is directly proportional to the Van't Hoff factor ($i$). The solution with the highest $i$ will have the highest elevation, and thus the highest absolute boiling point.
- Urea: Non-electrolyte, $i = 1$.
- $NaCl$: Yields $Na^+$ and $Cl^-$, $i = 2$.
- $BaCl_2$: Yields $Ba^{2+}$ and $2Cl^-$, $i = 3$.
- $Al_2(SO_4)_3$: Yields $2Al^{3+}$ and $3SO_4^{2-}$, $i = 5$.
View Solution
Step 1: Convert atm to Pascals (SI unit)
$\pi = 0.0821 \times 1.013 \times 10^5 \text{ Pa} = 8316.7 \text{ Pa}$.
Step 2: Equate to Hydrostatic Pressure Formula
$\pi = \rho g h$
$8316.7 = 1000 \times 9.8 \times h$
$8316.7 = 9800 h \implies h = \frac{8316.7}{9800} = 0.848 \text{ meters}$.
View Solution
Trimerization means three separate molecules combine to form a single entity ($n = 3$). Complete association means $\alpha = 1$.
Using the association formula: $i = 1 + \alpha\left(\frac{1}{n} - 1\right)$.
$i = 1 + 1\left(\frac{1}{3} - 1\right) = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$.
View Solution
Hemolysis in $0.1 \text{ M}$ means the $0.1 \text{ M}$ solution is hypotonic (water entered the cell to dilute the higher internal concentration).
Plasmolysis in $0.2 \text{ M}$ means the $0.2 \text{ M}$ solution is hypertonic (water left the cell to dilute the higher external concentration).
Since the internal concentration acts as the pivot point, it must lie exactly between the hypotonic floor and the hypertonic ceiling.
View Solution
For a mixture of two volatile liquids, Raoult's law states the total vapor pressure is $P_s = x_1 P^{\circ}_1 + x_2 P^{\circ}_2$.
If the solute is strictly non-volatile, it cannot enter the gas phase, meaning its pure vapor pressure is exactly zero ($P^{\circ}_2 = 0$). This eliminates the second term, simplifying the equation to $P_s = x_1 P^{\circ}_1$.
Since $x_1 = 1 - x_2$, we get $P_s = (1 - x_2)P^{\circ}_1$, which algebraically rearranges to the classic RLVP equation $\frac{P^{\circ}_1 - P_s}{P^{\circ}_1} = x_2$. If the solute were volatile, $P^{\circ}_2 \neq 0$, and this simplification would be mathematically impossible.
Mastering Solution Thermodynamics
Congratulations on reviewing these 25 grueling, top-tier numericals. Colligative properties are not just abstract math; they are the thermodynamic foundation of biology and chemical engineering. Whether you are correcting abnormal molar masses with the Van't Hoff factor, extracting the exact RLVP without dilution approximations, or deriving constants from enthalpies of vaporization, you now possess the analytical toolkit to conquer any JEE Advanced or NEET problem.
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