Q.Suggest a way to determine the $\Lambda^0_m$ value of water.

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.KMnO$_4$ oxidizes M$^{2+}$ to M$^{4+}$ in acid medium. 500 mL of 0.02 M M$^{2+}$ solution requires 500 mL of $x$ M MnO$_4^-$ solution for complete oxidation. The value of $x$ is (A) $16 \times 10^{-3}$ (B) $8 \times 10^{-3}$ (C) $4 \times 10^{-3}$ (D) $2 \times 10^{-3}$

   ›Reveal solutionSolution

   This problem involves a redox titration where the gram equivalents of the oxidizing agent (KMnO$_4$) and the reducing agent (M$^{2+}$) are equal at the equivalence point. By determining the n-factors for both species and using the given concentrations and volumes, we find the unknown molarity $x$ to be $\boxed{8 \times 10^{-3} \text{ M}}$.

   When a redox reaction reaches its equivalence point in a titration, it means that the amount of oxidizing agent added is stoichiometrically exactly enough to react with all of the reducing agent present. In terms of chemical quantities, this is most conveniently expressed using gram equivalents.

   The concept of gram equivalents simplifies calculations in redox titrations because it accounts for the number of electrons transferred per mole of reactant. Instead of balancing the full redox equation, we can simply equate the gram equivalents of the oxidant and reductant.

   At the equivalence point of a titration:

   $\text{Gram equivalents of oxidant} = \text{Gram equivalents of reductant}$

   Where $\text{Gram equivalents} = \text{Moles} \times \text{n-factor}$

   And $\text{Moles} = \text{Molarity} \times \text{Volume (in L)}$

   So, $(\text{Molarity} \times \text{n-factor} \times \text{Volume})_\text{oxidant} = (\text{Molarity} \times \text{n-factor} \times \text{Volume})_\text{reductant}$

   The 'n-factor' (or valence factor) for a redox species is the number of electrons gained or lost per mole of the substance in the reaction.

   1. Determine the n-factor for M$^{2+}$:

      The problem states that M$^{2+}$ is oxidized to M$^{4+}$. This means the oxidation state of M changes from +2 to +4.

      The half-reaction for M$^{2+}$ is:

      $\text{M}^{2+} \rightarrow \text{M}^{4+} + 2\text{e}^-$

      Since 2 electrons are lost per mole of M$^{2+}$, its n-factor is 2.

   2. Determine the n-factor for MnO$_4^-$ (from KMnO$_4$) in acid medium:

      In KMnO$_4$, the oxidation state of manganese (Mn) is +7 (since K is +1 and each O is -2, so $1 + \text{Mn} + 4(-2) = 0 \Rightarrow \text{Mn} = +7$).

      In acid medium, MnO$_4^-$ is a strong oxidizing agent and is typically reduced to Mn$^{2+}$ ions.

      The half-reaction for MnO$_4^-$ in acid medium is:

      $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

      Since 5 electrons are gained per mole of MnO$_4^-$, its n-factor is 5.

   3. Calculate the gram equivalents of M$^{2+}$:

      Given:

      Volume of M$^{2+}$ solution ($V_{\text{M}^{2+}}$) = 500 mL = 0.5 L

      Molarity of M$^{2+}$ solution ($M_{\text{M}^{2+}}$) = 0.02 M

      n-factor of M$^{2+}$ ($n_{\text{M}^{2+}}$) = 2

      Gram equivalents of M$^{2+}$ = $M_{\text{M}^{2+}} \times V_{\text{M}^{2+}} \times n_{\text{M}^{2+}}$

      Gram equivalents of M$^{2+}$ = $0.02 \text{ M} \times 0.5 \text{ L} \times 2 = 0.02$

   4. Set up the equivalence condition for MnO$_4^-$:

      At the equivalence point, the gram equivalents of MnO$_4^-$ must be equal to the gram equivalents of M$^{2+}$. …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The solubility of BaCO$_3$ (molar mass 197 g mol$^{-1}$) is $1.4 \times 10^{-3}$ g$|$100mL. The solubility product constant of BaCO$_3$ is $x \times 10^{-9}$ mol$^2$ L$^{-2}$. The value of $x$ is (nearest integer) (A) 4.0 (B) 5.0 (C) 6.0 (D) 4.8

   ›Reveal solutionSolution

   The solubility product is found by converting the given mass solubility to molar solubility, then squaring it (since BaCO₃ dissociates 1:1). The calculated value is $5.0 \times 10^{-9}$, so the nearest integer is 5, corresponding to option (B).

   The key idea is that for a sparingly soluble salt like BaCO₃, the solubility product $K_{sp}$ is directly related to its molar solubility. Because BaCO₃ dissociates into one Ba²⁺ and one CO₃²⁻ ion, $K_{sp} = s^2$, where $s$ is the molar solubility. The trick is to carefully convert the given solubility in grams per 100 mL into moles per liter.

   1. Convert the given solubility to grams per liter. The solubility is $1.4 \times 10^{-3}$ g per 100 mL. Since 1 L = 1000 mL, multiply by 10:

   $\text{Solubility in g/L} = 1.4 \times 10^{-3} \times 10 = 1.4 \times 10^{-2} \text{ g/L}.$

   2. Convert grams per liter to molar solubility (mol/L). Molar mass of BaCO₃ = 197 g/mol. So:

   $s = \frac{1.4 \times 10^{-2} \text{ g/L}}{197 \text{ g/mol}} = \frac{1.4}{197} \times 10^{-2} \text{ mol/L}.$

   Compute: $1.4 / 197 \approx 0.0071066$, so:

   $s \approx 7.1066 \times 10^{-5} \text{ mol/L}.$

   3. Write the dissociation equilibrium and $K_{sp}$ expression.

   $\text{BaCO}_3(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{CO}_3^{2-}(aq)$

   At equilibrium: $[\text{Ba}^{2+}] = s$ and $[\text{CO}_3^{2-}] = s$.

   Hence:

   $K_{sp} = s \times s = s^2.$

   4. Calculate $K_{sp}$. $K_{sp} = (7.1066 \times 10^{-5})^2 = 5.050 \times 10^{-9} \text{ mol}^2 \text{ L}^{-2}.$ …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The wavelength of electron in the third orbit of hydrogen atom is $6\pi a_0$. The kinetic energy of electron (in J) is equal to (Where, $K = \dfrac{h^2}{\pi^2 a_0^2 m_e}$, $a_0$ = radius of first orbit of hydrogen, $m_e$ = mass of electron, $h$ = Planck's constant) (A) $\dfrac{K}{36}$ (B) $\dfrac{K}{108}$ (C) $\dfrac{K}{72}$ (D) $\dfrac{K}{18}$

   ›Reveal solutionSolution

   De Broglie momentum $p=h/\lambda$ gives $KE=\dfrac{p^2}{2m_e}=\dfrac{K}{72}$.

   For the $n=3$ orbit the de Broglie wavelength is $\lambda = 6\pi a_0$ (consistent with the quantisation $2\pi r_3 = 3\lambda$, since $r_3 = 9a_0$).

   Momentum of the electron:

   $p = \frac{h}{\lambda} = \frac{h}{6\pi a_0}.$

   Kinetic energy: …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.For a first order reaction, rate constants at $50^\circ \mathrm{C}$ and $100^\circ \mathrm{C}$ are $1.5 \times 10^7 \ \mathrm{s}^{-1}$ and $4.5 \times 10^7 \ \mathrm{s}^{-1}$ respectively. What is the approximate activation energy of the reaction (in $\mathrm{kJ\,mol}^{-1}$)? \ $(\log 3 = 0.48) \ (R = 8.3 \ \mathrm{J\,mol}^{-1}\mathrm{K}^{-1})$ (A) 22 (B) 28 (C) 38 (D) 44

   ›Reveal solutionSolution

   Use the two-point form of the Arrhenius equation to relate the rate constants at two temperatures. The activation energy comes out to approximately 22 kJ/mol.

   The Arrhenius equation tells us how the rate constant $k$ depends on temperature $T$ and activation energy $E_a$:

   $k = A e^{-E_a / RT}$

   When you have rate constants at two different temperatures, the pre-exponential factor $A$ cancels out if you take the ratio. That gives the two-point form:

   $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

   This is the direct route to $E_a$ without needing $A$.

   1. Convert temperatures to Kelvin.

      $T_1 = 50^\circ\mathrm{C} = 323\ \mathrm{K}$

      $T_2 = 100^\circ\mathrm{C} = 373\ \mathrm{K}$

   2. Write the ratio of rate constants.

      $k_1 = 1.5 \times 10^7\ \mathrm{s}^{-1}$, $k_2 = 4.5 \times 10^7\ \mathrm{s}^{-1}$

   $\frac{k_2}{k_1} = \frac{4.5 \times 10^7}{1.5 \times 10^7} = 3$

   3. Use the two-point Arrhenius equation in log base 10 form. Since the problem gives $\log 3$ (base 10), convert the natural log: $\ln x = 2.303 \log x$.

   $\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

   Plug in $\log 3 = 0.48$:

   $0.48 = \frac{E_a}{2.303 \times 8.3}\left(\frac{1}{323} - \frac{1}{373}\right)$

   4. Compute the temperature difference term.

   $\frac{1}{323} - \frac{1}{373} = \frac{373 - 323}{323 \times 373} = \frac{50}{120479}$

   Approximate $323 \times 373 \approx 120500$ (close enough for an estimate).

   So the term is roughly $\frac{50}{120500} = \frac{1}{2410}$.

   5. Solve for $E_a$.

   $0.48 = \frac{E_a}{2.303 \times 8.3} \times \frac{1}{2410}$

   $E_a = 0.48 \times 2.303 \times 8.3 \times 2410$ …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The molar conductivity of acetic acid solution at infinite dilution is $390~\mathrm{S~cm^2~mol^{-1}}$. What is the molar conductivity of $0.01~\mathrm{M}$ acetic acid solution (in $\mathrm{S~cm^2~mol^{-1}}$)? (Given: $K_a(\mathrm{CH_3COOH}) = 1.8 \times 10^{-5}$, assume $1-\alpha = 1$) (A) 10.64 (B) 16.54 (C) 51.64 (D) 15.64

   ›Reveal solutionSolution

   The key idea is to use the degree of dissociation $\alpha$ from the acid dissociation constant $K_a$, then apply Kohlrausch’s law to find molar conductivity at a given concentration. The final molar conductivity is approximately $16.54~\mathrm{S~cm^2~mol^{-1}}$, which corresponds to option (B).

   We are given the molar conductivity at infinite dilution, $\Lambda_m^\infty = 390~\mathrm{S~cm^2~mol^{-1}}$, for acetic acid. For a weak electrolyte like acetic acid, the molar conductivity at a finite concentration is related to $\Lambda_m^\infty$ by $\Lambda_m = \alpha \, \Lambda_m^\infty$, where $\alpha$ is the degree of dissociation. So the problem reduces to finding $\alpha$ for a $0.01~\mathrm{M}$ solution using the given $K_a$.

   Why this works:

   For weak acids, the conductivity is proportional to the number of ions present. At infinite dilution, the acid is fully dissociated ($\alpha=1$). At any finite concentration, only a fraction $\alpha$ dissociates, so the molar conductivity scales linearly with $\alpha$. The value of $\alpha$ is obtained from the equilibrium expression for $K_a$.

   Step-by-step solution:

   1. Write the dissociation equilibrium For acetic acid:

   $\mathrm{CH_3COOH \rightleftharpoons CH_3COO^- + H^+}$

   Initial concentration: $c = 0.01~\mathrm{M}$.

   At equilibrium:

   $[\mathrm{CH_3COOH}] = c(1-\alpha), \quad [\mathrm{CH_3COO^-}] = c\alpha, \quad [\mathrm{H^+}] = c\alpha$

   2. Apply the acid dissociation constant expression

   $K_a = \frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}$

   Given $K_a = 1.8 \times 10^{-5}$ and $c = 0.01~\mathrm{M}$.

   3. Simplify using the assumption $1-\alpha \approx 1$ The problem explicitly says “assume $1-\alpha = 1$”. This is valid because $\alpha$ is very small for a weak acid at this concentration. Then:

   $K_a \approx c\alpha^2$

   So:

   $\alpha = \sqrt{\frac{K_a}{c}} = \sqrt{\frac{1.8 \times 10^{-5}}{0.01}} = \sqrt{1.8 \times 10^{-3}} = \sqrt{0.0018}$

   4. Calculate $\alpha$ $$$$

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The molar conductivity of acetic acid solution at infinite dilution is $390~\mathrm{S~cm^2~mol^{-1}}$. What is the molar conductivity of $0.01~\mathrm{M}$ acetic acid solution (in $\mathrm{S~cm^2~mol^{-1}}$)? (Given: $K_a~(\mathrm{CH_3COOH}) = 1.8 \times 10^{-5}$, assume $1-\alpha = 1$) (A) 15.64 (B) 51.64 (C) 16.54 (D) 10.64

   ›Reveal solutionSolution

   Using the Ostwald dilution law and the relationship between molar conductivity and degree of dissociation, the molar conductivity of 0.01 M acetic acid is found to be approximately 16.54 S cm² mol⁻¹, corresponding to option (C).

   Concept & Intuition

   Acetic acid is a weak electrolyte — it does not fully dissociate in solution. Its molar conductivity at a given concentration is less than at infinite dilution because fewer ions are present. The degree of dissociation α is related to the acid dissociation constant $K_a$ via Ostwald’s dilution law. Once α is known, the molar conductivity $\Lambda_m$ is simply $\alpha \times \Lambda_m^\infty$, assuming that ionic mobilities are constant (which is reasonable at low concentrations). The problem also tells us to assume $1 - \alpha \approx 1$, which simplifies the calculation.

   Step-by-step solution

   1. Write the relationship between molar conductivity and degree of dissociation For a weak electrolyte, the molar conductivity at a given concentration is:

   $\Lambda_m = \alpha \, \Lambda_m^\infty$

   where $\Lambda_m^\infty = 390~\mathrm{S~cm^2~mol^{-1}}$ is the molar conductivity at infinite dilution.

   2. Express α using the acid dissociation constant For acetic acid:

   $\mathrm{CH_3COOH \rightleftharpoons CH_3COO^- + H^+}$

   Initial concentration: $c = 0.01~\mathrm{M}$.

   At equilibrium: $[H^+] = [CH_3COO^-] = c\alpha$, $[CH_3COOH] = c(1-\alpha)$.

   The dissociation constant is:

   $K_a = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}$

   3. Apply the approximation $1-\alpha \approx 1$ Since acetic acid is weak, α is very small, so:

   $K_a \approx c\alpha^2$

   This gives:

   $\alpha = \sqrt{\frac{K_a}{c}}$

   4. Plug in the numbers $\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.01}} = \sqrt{1.8 \times 10^{-3}} = \sqrt{0.0018}$ …

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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.Identify the correct statements from the following A) At 298 K, the potential of hydrogen electrode placed in a solution of pH = 10, is $-0.59$ V B) The limiting molar conductivity of $\mathrm{Ca^{2+}}$ and $\mathrm{Cl^-}$ is 119 and 76 S cm$^2$ mol$^{-1}$ respectively. The limiting molar conductivity of CaCl$_2$ is 195 S cm$^2$ mol$^{-1}$ C) The correct relationship between $K_c$ and $E^\ominus_{\text{cell}}$ is $E^\ominus_{\text{cell}} = \dfrac{2.303RT}{nF} \log K_c$ (A) A, B, C (B) A, B only (C) A, C only (D) B, C only

   ›Reveal solutionSolution

   The key idea is to check each statement using standard electrochemistry and conductivity relations: statement A uses the Nernst equation for a hydrogen electrode at pH 10, statement B applies Kohlrausch’s law of independent migration of ions, and statement C relates the standard cell potential to the equilibrium constant. Only statements A and C are correct, so the answer is option (C).

   Concept and Intuition

   We need to verify three independent statements.

   • For A: The hydrogen electrode potential depends on $[H^+]$, which is given by pH. The Nernst equation directly gives the potential vs. SHE.
   • For B: Limiting molar conductivity of a salt is the sum of the limiting molar conductivities of its ions, each multiplied by the number of ions per formula unit.
   • For C: The relationship between $E^\ominus_{\text{cell}}$ and $K_c$ is a standard thermodynamic equation derived from $\Delta G^\ominus = -nFE^\ominus_{\text{cell}} = -RT \ln K_c$.

   Let’s examine each step by step.

   1. Statement A: Hydrogen electrode at pH = 10 at 298 K The hydrogen electrode reaction is $2H^+ + 2e^- \rightarrow H_2$. Under standard conditions (1 atm $H_2$, $[H^+] = 1$ M), $E^\ominus = 0$ V. For non-standard conditions, the Nernst equation at 298 K is:

   $E = E^\ominus - \frac{0.059}{n} \log \frac{1}{[H^+]^2}$

   Here $n = 2$, so:

   $E = 0 - \frac{0.059}{2} \log \frac{1}{[H^+]^2} = -0.059 \times \log \frac{1}{[H^+]} = -0.059 \times \text{pH}$

   At pH = 10: $E = -0.059 \times 10 = -0.59$ V.

   Tip

   A common mistake is forgetting that the Nernst equation for the hydrogen electrode simplifies directly to $E = -0.059 \times \text{pH}$ at 298 K.

   So statement A is correct.

   2. Statement B: Limiting molar conductivity of CaCl$_2$ Kohlrausch’s law: $\Lambda_m^\infty(\text{CaCl}_2) = \lambda^\infty(\text{Ca}^{2+}) + 2 \cdot \lambda^\infty(\text{Cl}^-)$ Given: $\lambda^\infty(\text{Ca}^{2+}) = 119$ S cm$^2$ mol$^{-1}$, $\lambda^\infty(\text{Cl}^-) = 76$ S cm$^2$ mol$^{-1}$. Then: $\Lambda_m^\infty(\text{CaCl}_2) = 119 + 2 \times 76 = 119 + 152 = 271$ S cm$^2$ mol$^{-1}$. …

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8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The molar conductivity of 0.02 M solution of an electrolyte is $124 \times 10^{-4}$ S m$^2$ mol$^{-1}$. What is the resistance of the same solution (in ohms), kept in a cell of cell constant 129 m$^{-1}$? (A) 390 (B) 130 (C) 260 (D) 520

   ›Reveal solutionSolution

   The key is to relate molar conductivity, concentration, and cell constant to resistance via conductivity. The resistance is found to be 520 Ω, so the correct option is (D).

   We are given:

   • Molar conductivity, $\Lambda_m = 124 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1}$
   • Concentration, $c = 0.02 \, \text{M} = 0.02 \, \text{mol/L} = 20 \, \text{mol m}^{-3}$ (since 1 M = 1000 mol/m³)
   • Cell constant, $G^* = 129 \, \text{m}^{-1}$

   We need the resistance $R$ in ohms.

   Concept and intuition:

   Molar conductivity $\Lambda_m$ is the conductivity per mole of electrolyte. Conductivity $\kappa$ (in S/m) is related to $\Lambda_m$ by $\kappa = \Lambda_m \times c$. Then, resistance is found from the cell constant: $R = \frac{G^*}{\kappa}$. This is a direct chain: molar conductivity → conductivity → resistance.

   ---

   1. Convert concentration to SI units.

      Molarity is mol/L, but conductivity uses mol/m³.

      $0.02 \, \text{mol/L} = 0.02 \times 1000 = 20 \, \text{mol/m}^3$.

   2. Find conductivity $\kappa$.

   $\kappa = \Lambda_m \times c = (124 \times 10^{-4}) \times 20$

   Compute:

   $124 \times 10^{-4} = 0.0124$

   $0.0124 \times 20 = 0.248 \, \text{S/m}$.

   3. Use the cell constant to find resistance. The cell constant $G^* = \frac{\text{length}}{\text{area}}$ and relates conductivity and resistance:

   $R = \frac{G^*}{\kappa}$

   So,

   $$$$

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9. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.At 300 K, the conductivity of 0.01 mol dm$^{-3}$ aqueous solution of acetic acid is 19.5$\times$10$^{-5}$ mho cm$^{-1}$ and limiting molar conductivity of acetic acid at the same temperature is 390 mho cm$^2$ mol$^{-1}$. The degree of dissociation of acetic acid is (A) $5.0 \times 10^{-5}$ (B) $5.0 \times 10^{-2}$ (C) $2.5 \times 10^{-5}$ (D) $7.5 \times 10^{-2}$

   ›Reveal solutionSolution

   The degree of dissociation $\alpha$ is the ratio of molar conductivity at a given concentration to the limiting molar conductivity. Using $\kappa = 19.5\times10^{-5}\ \text{mho cm}^{-1}$, $c = 0.01\ \text{mol dm}^{-3}$, and $\Lambda_m^\circ = 390\ \text{mho cm}^2\ \text{mol}^{-1}$, we find $\alpha = 5.0\times10^{-2}$, which corresponds to option (B).

   ---

   Concept and intuition:

   For a weak electrolyte like acetic acid, the molar conductivity $\Lambda_m$ increases as the solution is diluted, approaching a limiting value $\Lambda_m^\circ$ at infinite dilution. The degree of dissociation $\alpha$ tells us what fraction of the acid molecules have actually ionized. Kohlrausch’s law and the relation $\alpha = \Lambda_m / \Lambda_m^\circ$ let us extract $\alpha$ directly from conductivity measurements — no equilibrium constants needed, just a ratio.

   ---

   Step-by-step solution:

   1. Convert concentration to consistent units. The concentration is given as $0.01\ \text{mol dm}^{-3}$. Since $1\ \text{dm}^3 = 1000\ \text{cm}^3$, we have

   $c = 0.01\ \text{mol dm}^{-3} = 0.01\ \text{mol per } 1000\ \text{cm}^3 = 1.0 \times 10^{-5}\ \text{mol cm}^{-3}.$

   This conversion is essential because conductivity $\kappa$ is in $\text{mho cm}^{-1}$ and molar conductivity will be in $\text{mho cm}^2\ \text{mol}^{-1}$.

   2. Compute the molar conductivity $\Lambda_m$ of the solution. Molar conductivity is defined as

   $\Lambda_m = \frac{\kappa}{c},$

   where $\kappa$ is the measured conductivity and $c$ is the concentration in $\text{mol cm}^{-3}$.

   Substituting:

   $\Lambda_m = \frac{19.5 \times 10^{-5}\ \text{mho cm}^{-1}}{1.0 \times 10^{-5}\ \text{mol cm}^{-3}} = 19.5\ \text{mho cm}^2\ \text{mol}^{-1}.$

   3. Recall the relation between degree of dissociation and molar conductivities. For a weak electrolyte, the degree of dissociation $\alpha$ is given by

   $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ},$

   where $\Lambda_m^\circ$ is the limiting molar conductivity (at infinite dilution). This holds because at infinite dilution the electrolyte is fully dissociated, so the ratio tells us the fraction dissociated at the given concentration. …

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10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.The conductivity of a solution of concentration $0.1~\mathrm{mol~L^{-1}}$ of a weak monobasic acid (HA) (in $\mathrm{Scm^{-1}}$) is (Given $\wedge_m^0 = 400~\mathrm{Scm^2~mol^{-1}}$ and degree of dissociation $(\alpha)$ of HA $= 0.02$) (A) $32 \times 10^{-4}$ (B) $16 \times 10^{-4}$ (C) $4 \times 10^{-4}$ (D) $8 \times 10^{-4}$

    ›Reveal solutionSolution

    The conductivity $\kappa$ is found from $\kappa = \alpha c \Lambda_m^0$. With $\alpha = 0.02$, $c = 0.1~\mathrm{mol\,L^{-1}}$, and $\Lambda_m^0 = 400~\mathrm{S\,cm^2\,mol^{-1}}$, we get $\kappa = 8 \times 10^{-4}~\mathrm{S\,cm^{-1}}$, so the correct option is (D).

    The key idea is that for a weak electrolyte, the molar conductivity at a given concentration is $\Lambda_m = \alpha \Lambda_m^0$, because only the dissociated fraction contributes to conduction. Conductivity $\kappa$ is then $\kappa = \Lambda_m \times c$ (with careful unit handling). This avoids needing the full Kohlrausch law — just use the degree of dissociation directly.

    1. Understand the relationship

       Molar conductivity $\Lambda_m$ is defined as $\kappa / c$, where $\kappa$ is conductivity (in $\mathrm{S\,cm^{-1}}$) and $c$ is concentration (in $\mathrm{mol\,cm^{-3}}$). For a weak acid, $\Lambda_m = \alpha \Lambda_m^0$, because only the fraction $\alpha$ of molecules has dissociated into ions, each with the limiting molar conductivity $\Lambda_m^0$.

    2. Convert concentration units

       Given $c = 0.1~\mathrm{mol\,L^{-1}}$. Since $1~\mathrm{L} = 1000~\mathrm{cm^3}$, we have

    $c = 0.1~\mathrm{mol\,L^{-1}} = 0.1 \times 10^{-3}~\mathrm{mol\,cm^{-3}} = 1.0 \times 10^{-4}~\mathrm{mol\,cm^{-3}}.$

    This step is crucial because $\Lambda_m^0$ is in $\mathrm{S\,cm^2\,mol^{-1}}$, so $c$ must be in $\mathrm{mol\,cm^{-3}}$ for $\kappa$ to come out in $\mathrm{S\,cm^{-1}}$.

    3. Compute the molar conductivity at this concentration $\Lambda_m = \alpha \Lambda_m^0 = 0.02 \times 400~\mathrm{S\,cm^2\,mol^{-1}} = 8~\mathrm{S\,cm^2\,mol^{-1}}.$ …

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11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.A polymer sample contains 3 molecules of molar mass $10^3$, 3 molecules of molar mass 500 and 4 molecules of molar mass 200. What is its weight average molecular mass? (A) 530 (B) 737.7 (C) 834.4 (D) 821.6

    ›Reveal solutionSolution

    The weight‑average molecular mass weights each molecule by its mass, so larger molecules contribute more heavily. For this sample, the result is approximately 737.7 g/mol, which corresponds to option (B).

    The key idea is that weight‑average molecular mass ($\bar{M}_w$) is not a simple arithmetic mean of the molar masses. Instead, it accounts for the fact that heavier molecules make up a larger fraction of the total mass of the sample. This is crucial in polymer chemistry because properties like viscosity and mechanical strength depend more on the mass of the larger chains.

    The formula is:

    $\bar{M}_w = \frac{\sum n_i M_i^2}{\sum n_i M_i}$

    where $n_i$ is the number of molecules of molar mass $M_i$. The numerator is the sum of (number × mass²), and the denominator is the total mass of the sample.

    Let’s work through it step by step.

    1. List the data clearly

       • 3 molecules of $M_1 = 1000$
       • 3 molecules of $M_2 = 500$
       • 4 molecules of $M_3 = 200$

    2. Compute the total number of molecules (not needed for $\bar{M}_w$, but good for context):

       $N = 3 + 3 + 4 = 10$ molecules.

    3. Compute the total mass of the sample (denominator):

    $\sum n_i M_i = (3 \times 1000) + (3 \times 500) + (4 \times 200)$

    $= 3000 + 1500 + 800 = 5300$

    4. Compute the sum of (number × mass²) (numerator):

    $\sum n_i M_i^2 = (3 \times 1000^2) + (3 \times 500^2) + (4 \times 200^2)$

    $= (3 \times 1\,000\,000) + (3 \times 250\,000) + (4 \times 40\,000)$

    $= 3\,000\,000 + 750\,000 + 160\,000 = 3\,910\,000$ …

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12. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.The molar conductances of sodium acetate, hydrochloric acid and sodium chloride are 91.0, 425.9 and 126.4 S cm$^2$ mol$^{-1}$ respectively at 25 $^\circ$C at infinite dilution. What is the molar conductance of acetic acid at infinite dilution? (A) 390.5 S cm$^2$ mol$^{-1}$ (B) 461.3 S cm$^2$ mol$^{-1}$ (C) 643.3 S cm$^2$ mol$^{-1}$ (D) 43.5 S cm$^2$ mol$^{-1}$

    ›Reveal solutionSolution

    Use Kohlrausch's law of independent migration of ions to combine the given molar conductances algebraically: $\Lambda^\circ_{\text{CH}_3\text{COOH}} = \Lambda^\circ_{\text{CH}_3\text{COONa}} + \Lambda^\circ_{\text{HCl}} - \Lambda^\circ_{\text{NaCl}} = 390.5 \text{ S cm}^2 \text{ mol}^{-1}$.

    The key insight is that at infinite dilution, each ion migrates independently and contributes a fixed amount to the total molar conductance, regardless of which other ion it's paired with. This is Kohlrausch's law.

    For a weak electrolyte like acetic acid, we cannot measure the molar conductance at infinite dilution directly because it never fully dissociates. Instead, we construct it from strong electrolytes whose limiting conductances we can measure.

    Think of it as an algebraic puzzle with ions. We want:

    $\Lambda^\circ_{\text{CH}_3\text{COOH}} = \lambda^\circ_{\text{CH}_3\text{COO}^-} + \lambda^\circ_{\text{H}^+}$

    where $\lambda^\circ$ represents the limiting ionic conductance of each ion. We need to find a combination of the three given electrolytes that gives us exactly these two ions.

    Building the combination:

    1. Identify what each electrolyte contributes:

       • Sodium acetate: $\Lambda^\circ_{\text{CH}_3\text{COONa}} = \lambda^\circ_{\text{CH}_3\text{COO}^-} + \lambda^\circ_{\text{Na}^+} = 91.0$
       • Hydrochloric acid: $\Lambda^\circ_{\text{HCl}} = \lambda^\circ_{\text{H}^+} + \lambda^\circ_{\text{Cl}^-} = 425.9$
       • Sodium chloride: $\Lambda^\circ_{\text{NaCl}} = \lambda^\circ_{\text{Na}^+} + \lambda^\circ_{\text{Cl}^-} = 126.4$

    2. Construct the target combination:

       We need $\text{CH}_3\text{COO}^-$ and $\text{H}^+$, but we want to eliminate $\text{Na}^+$ and $\text{Cl}^-$.

       Start with sodium acetate (gives us $\text{CH}_3\text{COO}^-$) and add HCl (gives us $\text{H}^+$):

       $\Lambda^\circ_{\text{CH}_3\text{COONa}} + \Lambda^\circ_{\text{HCl}} = \lambda^\circ_{\text{CH}_3\text{COO}^-} + \lambda^\circ_{\text{Na}^+} + \lambda^\circ_{\text{H}^+} + \lambda^\circ_{\text{Cl}^-}$ …

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