Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The conductivity of 0.001 M acetic acid is $5 \times 10^{-5}\ S\ cm^{-1}$. If the molar conductivity of acetic acid solution at infinite dilution is $390\ S\ cm^2\ mol^{-1}$. What is its degree of dissociation? (A) 0.218 (B) 0.128 (C) 0.138 (D) 0.238

   ›Reveal solutionSolution

   Molar conductivity at 0.001 M works out to 50 S cm² mol⁻¹; dividing by the limiting value 390 gives the degree of dissociation, 0.128.

   Concept and Intuition

   For a weak electrolyte, the degree of dissociation can be estimated (Arrhenius) as the ratio of the molar conductivity at the given concentration to the molar conductivity at infinite dilution: $\alpha = \Lambda_m/\Lambda_m^\circ$. This works because $\Lambda_m^\circ$ represents complete dissociation, so the ratio directly measures what fraction has actually dissociated.

   Step-by-Step Solution

   1. Convert $\kappa$ to $\Lambda_m$: $\Lambda_m = \dfrac{\kappa\times1000}{C(\text{mol/L})}$.
   2. $\Lambda_m = \dfrac{5\times10^{-5}\times1000}{0.001} = \dfrac{5\times10^{-2}}{10^{-3}} = 50\ \mathrm{S\ cm^2\ mol^{-1}}$.
   3. $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ} = \dfrac{50}{390} \approx 0.128$.

   Common Mistakes

   • Forgetting the ×1000 conversion factor between $\kappa$ (per cm³) and $\Lambda_m$ (per litre-based concentration).
   • Dividing the concentration incorrectly (using 0.001 mol/cm³ instead of mol/L).

   ✓Final answer

   The correct option is (B) — 0.128.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The molar conductivity of three electrolytes, NaCl, KCl and CsCl, was plotted against $\sqrt{c}$ (c = concentration) and the obtained figure is shown below. The $\lambda^0$ of $Na^+, K^+$ and $Cs^+$ is 50, 73 and 77 $S\ cm^2mol^{-1}$ respectively. Identify the electrolytes y-axis = $\Lambda_m$ (in $S^2cm^2mol^{-1}$); x-axis = $\sqrt{c}$ [FIGURE] (A) A = KCl ; B = CsCl ; C = NaCl (B) A = KCl ; B = NaCl ; C = CsCl (C) A = CsCl ; B = NaCl ; C = KCl (D) A = NaCl ; B = KCl ; C = CsCl

   ›Reveal solutionSolution

   KCl and CsCl are the two close top curves (A and C), NaCl is the lower separate curve (B): option (B).

   Concept and Intuition

   For a strong electrolyte, $\Lambda_m = \Lambda^0_m - b\sqrt{c}$, so the intercept at $\sqrt{c}\to0$ is the limiting molar conductivity $\Lambda^0_m = \lambda^0_{+} + \lambda^0_{-}$. Because $Cl^-$ is shared, the curves are ordered purely by cation conductivity.

   Step-by-Step Solution

   1. Take $\lambda^0(Cl^-)\approx 76\ S\,cm^2\,mol^{-1}$.
   2. $\Lambda^0_m(NaCl)=50+76=126$; $\Lambda^0_m(KCl)=73+76=149$; $\Lambda^0_m(CsCl)=77+76=153$.
   3. KCl (149) and CsCl (153) differ by only $\sim4$, so their lines nearly coincide — these are the overlapping pair ending at the adjacent points A and C.
   4. NaCl (126) is far lower/apart from the pair — it is the distinct line ending at B.
   5. Within the close pair, the slightly higher curve (C, above) is the larger $\Lambda^0_m$ = CsCl; the slightly lower (A) = KCl.

   Common Mistakes

   • Forgetting that $Cl^-$ is common, and trying to order by anion.
   • Pairing NaCl with a high-conductivity salt: NaCl's low $Na^+$ value makes it the clearly separated line.

   ✓Final answer

   The correct option is (B) — A = KCl ; B = NaCl ; C = CsCl.

   ANSWER: B

   Note

   This solution was worked out by our team and cross-checked by a second independent solve. The official answer key for this question could not be confirmed, so please cross-verify with the official paper where possible.

3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The resistance of a conductivity cell filled with 0.02 M KCl solution is 85 $\Omega$ at 25°C. Conductivity of this solution is 0.3 S m$^{-1}$. Resistance of 0.0025 M $K_2SO_4$ solution taken in the same cell is 300 $\Omega$. The molar conductivity of 0.0025 M $K_2SO_4$ solution (in S m$^2$ mol$^{-1}$) is (A) $6.8 \times 10^{-3}$ (B) $2.4 \times 10^{-2}$ (C) $3.4 \times 10^{-2}$ (D) $3.4 \times 10^{-3}$

   ›Reveal solutionSolution

   Use the KCl data to find the cell constant, then use that fixed cell constant to get the conductivity — and hence molar conductivity — of the $K_2SO_4$ solution. The answer is $3.4\times10^{-2}$ S m² mol⁻¹.

   Concept and Intuition

   A conductivity cell's geometry fixes a constant, the cell constant $G^* = \kappa \times R$, which is the same for any solution measured in that same cell. Once $G^*$ is known from one calibrating solution (here KCl), it can be used to convert any other measured resistance directly into that solution's conductivity, from which molar conductivity follows by dividing by molar concentration.

   Step-by-Step Solution

   1. From the KCl data: $G^* = \kappa \times R = 0.3\ \text{S/m} \times 85\ \Omega = 25.5\ \text{m}^{-1}$.
   2. This same $G^*$ applies to the $K_2SO_4$ measurement in the same cell: $\kappa_{K_2SO_4} = \dfrac{G^*}{R} = \dfrac{25.5}{300} = 0.085\ \text{S/m}$.
   3. Convert concentration to SI (mol/m³): $0.0025\ \text{mol/L} = 2.5\ \text{mol/m}^3$.
   4. Molar conductivity: $\Lambda_m = \dfrac{\kappa}{C} = \dfrac{0.085}{2.5} = 0.034\ \text{S m}^2\ \text{mol}^{-1} = 3.4\times10^{-2}\ \text{S m}^2\ \text{mol}^{-1}$.

   Common Mistakes

   • Forgetting to convert mol/L to mol/m³ (a factor of 1000), which throws off the final power of ten.
   • Re-using the KCl conductivity value directly for $K_2SO_4$ instead of recomputing it via the cell constant.

   ✓Final answer

   The correct option is (C) — $3.4 \times 10^{-2}$.

   ANSWER: C

4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The $\Lambda_m$ (on $y$-axis) of $NaCl$ and $CsCl$ was plotted against $\sqrt{c}$ ($c$ = concentration on $x$-axis). Identify the correct figure for these electrolytes ($\lambda^0$ of $Na^+$ and $Cs^+$ is 50 and 77 S $cm^2\ mol^{-1}$ respectively) (A) [FIGURE] (two straight lines rising from near the origin toward the upper right; the CsCl line lies above and is offset from the NaCl line, both with positive slope) (B) [FIGURE] (two straight lines falling from upper left to lower right with negative slope; the CsCl line lies above the NaCl line, both roughly parallel) (C) [FIGURE] (two horizontal straight lines, constant with $\sqrt{c}$; the NaCl line lies above the CsCl line) (D) [FIGURE] (two straight lines falling with negative slope, starting close together at the y-axis with NaCl above CsCl, and converging/crossing as $\sqrt{c}$ increases)

   ›Reveal solutionSolution

   Strong electrolytes like $NaCl$ and $CsCl$ show $\Lambda_m$ falling linearly with $\sqrt{c}$ (Kohlrausch's law); since $Cs^+$ has a higher limiting ionic conductivity than $Na^+$, the $CsCl$ line sits above the $NaCl$ line, and the two lines run roughly parallel rather than crossing.

   Concept and Intuition

   Kohlrausch found that for a strong electrolyte, the molar conductivity varies with concentration as $\Lambda_m = \Lambda^0_m - A\sqrt{c}$ — a straight line with a negative slope when plotted against $\sqrt{c}$, extrapolating to the limiting molar conductivity $\Lambda^0_m$ at $c\to0$ (i.e. at $\sqrt{c}=0$). This is very different from weak electrolytes, whose $\Lambda_m$ rises steeply (near-vertically) close to $c=0$ and cannot be extrapolated linearly.

   The limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its ions (Kohlrausch's law of independent migration): $\Lambda^0_m(NaCl) = \lambda^0(Na^+) + \lambda^0(Cl^-)$ and $\Lambda^0_m(CsCl) = \lambda^0(Cs^+) + \lambda^0(Cl^-)$. Since both share the same $Cl^-$ contribution, and $\lambda^0(Cs^+) = 77 > \lambda^0(Na^+) = 50$, we must have $\Lambda^0_m(CsCl) > \Lambda^0_m(NaCl)$ — the $CsCl$ line intercepts the y-axis higher than the $NaCl$ line.

   Step-by-Step Solution

   1. Both $NaCl$ and $CsCl$ are strong electrolytes → both lines must be straight with negative slope (rules out options with positive slope or horizontal lines).
   2. $\Lambda^0_m(CsCl) > \Lambda^0_m(NaCl)$ since $\lambda^0(Cs^+) > \lambda^0(Na^+)$ → the $CsCl$ line must lie above the $NaCl$ line, especially at (and near) $\sqrt{c}=0$.
   3. Since both are 1:1 electrolytes, the Onsager slope (which depends mainly on ionic charge and solvent properties, not strongly on which specific monovalent cation is present) is similar for both, so the two lines should run essentially parallel, not converge or cross within the plotted range.
   4. This matches the description of option (B): two falling straight lines, roughly parallel, with $CsCl$ consistently above $NaCl$.

   Common Mistakes

   • Treating $NaCl$/$CsCl$ like weak electrolytes and expecting a steep upward curve near the origin.
   • Assuming the higher-conductivity line must cross the lower one at some point, rather than recognising that for the same electrolyte type (1:1) the lines stay roughly parallel.

   ✓Final answer

   The correct option is (B) — two straight lines falling from upper left to lower right with negative slope; the CsCl line lies above the NaCl line, both roughly parallel.

   ANSWER: B

5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The specific conductance of 0.05 M NaOH solution is $0.0115\ \mathrm{S\, cm^{-1}}$. What is its molar conductance ($\Lambda_m$) in $\mathrm{S\, cm^2\, mol^{-1}}$? (A) 23 (B) $5.75 \times 10^{-7}$ (C) 2300 (D) 230

   ›Reveal solutionSolution

   Converting specific conductance to molar conductance using $\Lambda_m = 1000\kappa/C$ gives 230 S cm² mol⁻¹. Answer: (D).

   Concept and Intuition

   Specific conductance ($\kappa$) is the conductance of a 1 cm cube of solution, while molar conductance ($\Lambda_m$) normalizes this to "per mole of electrolyte," accounting for the fact that the amount of electrolyte (and hence total ions) present per unit volume depends on the molar concentration. The conversion factor of 1000 arises because $\kappa$ is defined per cm³ but concentration is given per litre (1000 cm³).

   Step-by-Step Solution

   1. Formula: $\Lambda_m = \dfrac{\kappa \times 1000}{C}$, where $\kappa$ is in S cm⁻¹ and $C$ is molar concentration in mol L⁻¹.
   2. Given $\kappa = 0.0115$ S cm⁻¹ and $C = 0.05$ mol L⁻¹.
   3. $\Lambda_m = \dfrac{0.0115 \times 1000}{0.05} = \dfrac{11.5}{0.05}$.
   4. $\dfrac{11.5}{0.05} = 230$.
   5. So $\Lambda_m = 230\ \mathrm{S\,cm^2\,mol^{-1}}$, matching option (D).

   Common Mistakes

   • Forgetting the factor of 1000 (mixing up cm³ and litre units), which would give an answer 1000× too small.
   • Misplacing a decimal in the division (e.g., getting 2300 instead of 230).

   ✓Final answer

   The correct option is (D) — 230.

   ANSWER: D

6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The resistance of a conductivity cell filled with 0.1 M $\mathrm{KCl}$ solution is $100\,\Omega$. If the resistance of the same cell when filled with 0.2 M $\mathrm{KCl}$ solution is $520\,\Omega$, the molar conductivity of 0.02 M solution (in S $\mathrm{cm^2\,mol^{-1}}$) is (Given : conductivity of 0.1 M $\mathrm{KCl}$ solution = $1.29$ $\mathrm{Sm^{-1}}$) (A) 124 (B) 186 (C) 248 (D) 104

   ›Reveal solutionSolution

   First get the (fixed) cell constant from the 0.1 M KCl calibration measurement, then use it with the second solution's resistance to get its conductivity, then convert to molar conductivity. The answer is (A) 124 S cm² mol⁻¹.

   Concept and Intuition

   A conductivity cell has a fixed geometric cell constant $G^* = \ell/A$ (length between electrodes / cross-sectional area), which relates measured resistance to the solution's conductivity: $\kappa = G^*/R$, i.e. $G^* = \kappa R$. Since $G^*$ is a property of the cell, not the solution, once we calibrate it with one known solution (0.1 M KCl of known conductivity), we can use the same $G^*$ for any other solution measured in that same cell — here, the 0.02 M KCl solution. Molar conductivity then normalises conductivity by concentration: $\Lambda_m = \dfrac{\kappa \times 1000}{C}$ (with $\kappa$ in S/cm and $C$ in mol/L, giving $\Lambda_m$ in S cm² mol⁻¹).

   Step-by-Step Solution

   1. Calibrate the cell constant using the 0.1 M KCl data: $\kappa_1 = 1.29\ \mathrm{S\,m^{-1}}$, $R_1 = 100\ \Omega$.

   $G^* = \kappa_1 R_1 = 1.29 \times 100 = 129\ \mathrm{m^{-1}}$

   2. Use the same cell (same $G^*$) with the second solution: $R_2 = 520\ \Omega$.

   $\kappa_2 = \frac{G^*}{R_2} = \frac{129}{520} = 0.2481\ \mathrm{S\,m^{-1}}$

   3. Convert to S cm⁻¹ (divide by 100, since $1\ \mathrm{S\,m^{-1}} = 0.01\ \mathrm{S\,cm^{-1}}$):

   $\kappa_2 = 2.481\times10^{-3}\ \mathrm{S\,cm^{-1}}$

   4. Molar conductivity at $C = 0.02$ mol/L:

   $\Lambda_m = \frac{\kappa_2 \times 1000}{C} = \frac{2.481\times10^{-3}\times1000}{0.02} = \frac{2.481}{0.02} \approx 124\ \mathrm{S\,cm^2\,mol^{-1}}$

   Common Mistakes

   • Forgetting to convert $\kappa$ from S/m to S/cm before applying the $\Lambda_m = 1000\kappa/C$ formula (which requires $\kappa$ in S/cm).
   • Re-deriving a new cell constant instead of reusing the one found from the calibration measurement.

   ✓Final answer

   The correct option is (A) — 124.

   ANSWER: A

7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.Match the following. List-I (symbol of electrical property): A) $\Lambda_m$ B) G C) $\kappa$ D) $G^{*}$. List-II (units): I) $S\,cm^{-1}$ II) $m^{-1}$ III) $S\,cm^2\,mol^{-1}$ IV) S. The correct answer is (A) A-IV, B-III, C-I, D-II (B) A-III, B-IV, C-I, D-II (C) A-III, B-IV, C-II, D-I (D) A-II, B-I, C-IV, D-III

   ›Reveal solutionSolution

   Matching electrical/conductance quantities to their SI/CGS units: $\Lambda_m\to S\,cm^2\,mol^{-1}$, $G\to S$, $\kappa\to S\,cm^{-1}$, $G^{*}\to m^{-1}$ — option (B).

   Concept and Intuition

   Electrolytic conductance quantities form a small family that is easy to confuse: conductance $G$ is simply the reciprocal of resistance and carries the base unit siemens; specific conductance (conductivity) $\kappa$ is conductance normalised per unit cell geometry, giving $S\,cm^{-1}$; molar conductivity $\Lambda_m$ further normalises by concentration, giving $S\,cm^2\,mol^{-1}$; and the cell constant $G^{*}=l/A$ is a pure geometric factor with units of reciprocal length.

   Step-by-Step Solution

   1. $\Lambda_m$ (molar conductivity) $= \dfrac{\kappa \times 1000}{C}$, units work out to $S\,cm^2\,mol^{-1}$ → matches III.
   2. $G$ (conductance) $=1/R$, base SI unit siemens (S) → matches IV.
   3. $\kappa$ (specific conductance/conductivity) has units $S\,cm^{-1}$ → matches I.
   4. $G^{*}$ (cell constant, $= l/A$) has units of reciprocal length, i.e. $m^{-1}$ (or cm$^{-1}$) → matches II.
   5. So: A(Λm)-III, B(G)-IV, C(κ)-I, D(G*)-II — option (B).

   Common Mistakes

   • Swapping $\kappa$ and $G^{*}$ since both look like 'per length' quantities — $\kappa$ carries siemens, $G^{*}$ is purely geometric with no siemens.
   • Confusing $G$ (plain conductance, unit S) with $\Lambda_m$ (needs the extra $cm^2\,mol^{-1}$ normalisation).

   ✓Final answer

   The correct option is (B) — A-III, B-IV, C-I, D-II.

   ANSWER: B

8. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.A conductivity cell is filled with solution of KCl of concentration 0.2 mol $\mathrm{dm^{-3}}$ and its conductivity is 0.28 $\mathrm{S m^{-1}}$. The resistance of this solution is 82.2 $\Omega$. The same cell filled with solution of 0.0025 mol $\mathrm{dm^{-3}}$ $\mathrm{K_2SO_4}$ showed a resistance of 325 $\Omega$. The molar conductivity of $\mathrm{K_2SO_4}$ solution (in $\mathrm{S m^2 mol^{-1}}$) is (A) $1.4 \times 10^{-2}$ (B) $2.8 \times 10^{-2}$ (C) $4.2 \times 10^{-3}$ (D) $5.6 \times 10^{-4}$

   ›Reveal solutionSolution

   Tests the cell-constant method for molar conductivity: use the KCl calibration run to find the

   cell constant, then apply it to the K$_2$SO$_4$ run. Answer: $2.8\times10^{-2}\ \mathrm{S\,m^2\,mol^{-1}}$.

   Concept and Intuition

   A conductivity cell has a fixed geometric cell constant $G^* = l/A$ (in $\mathrm{m^{-1}}$),

   which relates the measured resistance to the solution's conductivity via

   $\kappa = G^*/R$ (equivalently $G^* = \kappa R$). Once $G^*$ is found from a solution of known

   conductivity (here, KCl), the same cell constant applies to any other solution placed in that

   cell, letting us find the unknown solution's conductivity, and from that its molar conductivity

   $\Lambda_m = \kappa/c$ (with $c$ expressed in $\mathrm{mol\,m^{-3}}$ to get SI units of $\Lambda_m$).

   Step-by-Step Solution

   1. Find the cell constant from the KCl calibration: $G^* = \kappa_{KCl} \times R_{KCl} = 0.28 \times 82.2 = 23.016\ \mathrm{m^{-1}}$.
   2. Use $G^*$ to find $\kappa$ of the K$_2$SO$_4$ solution: $\kappa_{K_2SO_4} = G^*/R_{K_2SO_4} = 23.016/325 = 0.0708\ \mathrm{S\,m^{-1}}$.
   3. Convert concentration to SI (mol per m$^3$): $0.0025\ \mathrm{mol\,dm^{-3}} \times 1000 = 2.5\ \mathrm{mol\,m^{-3}}$.
   4. Compute molar conductivity: $\Lambda_m = \kappa/c = 0.0708/2.5 \approx 2.83\times10^{-2}\ \mathrm{S\,m^2\,mol^{-1}}$.
   5. This rounds to $2.8\times10^{-2}\ \mathrm{S\,m^2\,mol^{-1}}$.

   Common Mistakes

   • Forgetting to convert $\mathrm{mol\,dm^{-3}}$ to $\mathrm{mol\,m^{-3}}$ (a factor of 1000) before dividing, which throws the answer off by three orders of magnitude.
   • Re-deriving the cell constant from the K$_2$SO$_4$ data instead of reusing the one found from the KCl (known-conductivity) run.

   ✓Final answer

   The correct option is (B) — $2.8 \times 10^{-2}$.

   ANSWER: B

9. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.The conductivity of a solution containing 2.08 g of anhydrous barium chloride in 200 mL solution is $6\times10^{-3}\ \mathrm{ohm^{-1}cm^{-1}}$. The molar conductivity of the solution (in $\mathrm{ohm^{-1}cm^2mol^{-1}}$) is $x\times10^2$. The value of x is (Atomic mass of Ba = 137, Cl = 35.5) (A) 1.2 (B) 2.4 (C) 3.6 (D) 3.0

   ›Reveal solutionSolution

   Compute molarity of the BaCl$_2$ solution, then use $\Lambda_m = 1000\kappa/M$ to find molar conductivity. Answer: $x=1.2$.

   Concept and Intuition

   Molar conductivity $\Lambda_m$ relates a solution's specific conductivity ($\kappa$) to its molar concentration: $\Lambda_m = \dfrac{\kappa \times 1000}{M}$ (with $\kappa$ in $\mathrm{ohm^{-1}cm^{-1}}$, $M$ in mol/L, giving $\Lambda_m$ in $\mathrm{ohm^{-1}cm^2mol^{-1}}$). We first need the molarity from the given mass and volume.

   Step-by-Step Solution

   1. Molar mass of $\mathrm{BaCl_2} = 137 + 2(35.5) = 208\ \mathrm{g/mol}$.
   2. Moles $= \dfrac{2.08}{208} = 0.01$ mol.
   3. Volume $= 200\ \mathrm{mL} = 0.2$ L, so Molarity $= \dfrac{0.01}{0.2} = 0.05\ \mathrm{mol/L}$.
   4. $\Lambda_m = \dfrac{\kappa \times 1000}{M} = \dfrac{6\times10^{-3}\times1000}{0.05} = \dfrac{6}{0.05} = 120\ \mathrm{ohm^{-1}cm^2mol^{-1}}$.
   5. Given $\Lambda_m = x\times10^2$, so $x = 120/100 = 1.2$.

   Common Mistakes

   • Forgetting the factor of 1000 (unit conversion between cm$^3$ and L) in the $\Lambda_m$ formula.
   • Using the wrong molar mass for BaCl$_2$.

   ✓Final answer

   The correct option is (A) — 1.2.

   ANSWER: A

10. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.Molar conductivities at infinite dilution $\Lambda_m^\circ$ for $\mathrm{Ba(OH)_2}$, $\mathrm{BaCl_2}$ and $\mathrm{NH_4Cl}$ are 457.0, 240.6 and 2130 Scm$^2$mol$^{-1}$ respectively. The $\Lambda_m^\circ$ for ammonium hydroxide (in Scm$^2$mol$^{-1}$) is (A) 1683.2 (B) 1080.2 (C) 2130.0 (D) 2238.2

    ›Reveal solutionSolution

    Combine the three given limiting molar conductivities via Kohlrausch's law so that the $Ba^{2+}$ and $Cl^-$ contributions cancel, leaving $\Lambda_m^\circ(\mathrm{NH_4OH})$.

    Concept and Intuition

    Kohlrausch's law states each ion contributes independently to the limiting molar conductivity, so ionic contributions can be added/subtracted across compounds that share ions. Here we want $\lambda^\circ(NH_4^+)+\lambda^\circ(OH^-)$; we can build it from $NH_4Cl$ (gives $NH_4^+ + Cl^-$) plus half of $Ba(OH)_2$ (gives $\tfrac12 Ba^{2+}+OH^-$) minus half of $BaCl_2$ (gives $\tfrac12 Ba^{2+}+Cl^-$), which cancels the $Ba^{2+}$ and $Cl^-$ contributions, leaving exactly $NH_4^++OH^-$.

    Step-by-Step Solution

    1. Write ionic contributions: $\Lambda^\circ(Ba(OH)_2)=\lambda(Ba^{2+})+2\lambda(OH^-)=457.0$; $\Lambda^\circ(BaCl_2)=\lambda(Ba^{2+})+2\lambda(Cl^-)=240.6$; $\Lambda^\circ(NH_4Cl)=\lambda(NH_4^+)+\lambda(Cl^-)=2130$.
    2. Subtract the first two and halve: $\lambda(OH^-)-\lambda(Cl^-) = \dfrac{457.0-240.6}{2}=108.2$.
    3. $\Lambda^\circ(NH_4OH)=\lambda(NH_4^+)+\lambda(OH^-) = [\lambda(NH_4^+)+\lambda(Cl^-)] + [\lambda(OH^-)-\lambda(Cl^-)] = 2130+108.2 = 2238.2$.

    Common Mistakes

    • Forgetting to divide by 2 when isolating a single ionic contribution from a compound with 2 like ions (e.g. $Ba(OH)_2$ has 2 $OH^-$).
    • Adding instead of subtracting the $BaCl_2$ term, which would fail to cancel $Cl^-$/$Ba^{2+}$ properly.

    ✓Final answer

    The correct option is (D) — 2238.2.

    ANSWER: D

11. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.The molar conductivity of 0.027 M methanoic acid is 40.42 $\mathrm{S\,cm^2\,mol^{-1}}$. The value of dissociation constant of this acid is (Given $\lambda^{\circ}_{H^+} = 349.6\ \mathrm{S\,cm^2\,mol^{-1}}$ and $\lambda^{\circ}_{HCOO^-} = 54.6\ \mathrm{S\,cm^2\,mol^{-1}}$) (A) $1.5\times10^{-5}$ (B) $6.0\times10^{-5}$ (C) $4.5\times10^{-4}$ (D) $3.0\times10^{-4}$

    ›Reveal solutionSolution

    Using $\alpha=\Lambda_m/\Lambda_m^\circ$ and Ostwald's dilution law $K_a=\dfrac{c\alpha^2}{1-\alpha}$ gives $K_a=3.0\times10^{-4}$ for this weak acid.

    Concept and Intuition

    For a weak electrolyte, the fraction of molar conductivity actually achieved at a finite concentration compared to its limiting (infinite-dilution) value directly gives the degree of dissociation $\alpha$ (this is Arrhenius's conductance-based definition of $\alpha$). Once $\alpha$ is known, Ostwald's dilution law connects it to the acid's dissociation constant.

    Step-by-Step Solution

    1. Limiting molar conductivity: $\Lambda_m^\circ = \lambda^\circ_{H^+} + \lambda^\circ_{HCOO^-} = 349.6 + 54.6 = 404.2\,\text{S cm}^2\text{mol}^{-1}$.
    2. Degree of dissociation: $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ} = \dfrac{40.42}{404.2} = 0.1$.
    3. Ostwald's dilution law: $K_a = \dfrac{c\,\alpha^2}{1-\alpha}$.
    4. Substitute $c=0.027\,\text{M}$, $\alpha=0.1$: $K_a = \dfrac{0.027\times(0.1)^2}{1-0.1} = \dfrac{0.027\times0.01}{0.9} = \dfrac{2.7\times10^{-4}}{0.9} = 3.0\times10^{-4}$.

    Common Mistakes

    • Forgetting to add the anion's ionic conductivity to the cation's when finding $\Lambda_m^\circ$ (both are needed via Kohlrausch's law).
    • Omitting the $(1-\alpha)$ correction factor in the denominator and just using $K_a=c\alpha^2$, which is only a valid approximation for very small $\alpha$ (here $\alpha=0.1$ is not negligible enough to skip it cleanly, though the numbers still work out to the same option).

    ✓Final answer

    The correct option is (D) — $3.0\times10^{-4}$.

    ANSWER: D

12. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.What is the molar conductivity of $CH_3CO_2H$ at infinite dilution? Given that, $\Lambda_m^o((CH_3CO_2)_2Ba) = x_1$ S $cm^2$ $mol^{-1}$ $\Lambda_m^o(BaCl_2) = x_2$ S $cm^2$ $mol^{-1}$ $\Lambda_m^o(HCl) = x_3$ S $cm^2$ $mol^{-1}$ (A) $\dfrac{x_1 - x_2}{2} + x_3$ (B) $\dfrac{x_1 - x_3}{2} + x_2$ (C) $\dfrac{x_2 - x_3}{2} + x_1$ (D) $x_1 + x_3 - x_2$

    ›Reveal solutionSolution

    Combining the given limiting molar conductivities of $(CH_3COO)_2Ba$, $BaCl_2$, and $HCl$ via Kohlrausch's law of independent ion migration reconstructs $\Lambda^0_m(CH_3COOH)$ as $(x_1-x_2)/2+x_3$.

    Concept and Intuition

    Kohlrausch's law says the limiting molar conductivity of any electrolyte is the sum of the limiting ionic conductivities of its constituent ions, each independent of the other ion present. This lets us build the conductivity of a weak electrolyte (like acetic acid) from strong electrolytes sharing common ions.

    Step-by-Step Solution

    1. Write each given compound in terms of ionic conductivities: $x_1 = \Lambda^0_m((CH_3COO)_2Ba) = \lambda^0(Ba^{2+}) + 2\lambda^0(CH_3COO^-)$ $x_2 = \Lambda^0_m(BaCl_2) = \lambda^0(Ba^{2+}) + 2\lambda^0(Cl^-)$ $x_3 = \Lambda^0_m(HCl) = \lambda^0(H^+) + \lambda^0(Cl^-)$
    2. Subtract: $x_1 - x_2 = 2\lambda^0(CH_3COO^-) - 2\lambda^0(Cl^-)$, so $\dfrac{x_1-x_2}{2} = \lambda^0(CH_3COO^-) - \lambda^0(Cl^-)$.
    3. Add $x_3$: $\dfrac{x_1-x_2}{2} + x_3 = \lambda^0(CH_3COO^-) - \lambda^0(Cl^-) + \lambda^0(H^+) + \lambda^0(Cl^-) = \lambda^0(CH_3COO^-) + \lambda^0(H^+)$.
    4. This equals exactly $\Lambda^0_m(CH_3COOH)$ by Kohlrausch's law.

    Common Mistakes

    • Trying to directly add/subtract the compounds without tracking which ions cancel — the $Ba^{2+}$ must cancel via subtraction of $x_1,x_2$ (both contain it once), and $Cl^-$ must cancel between the $(x_1-x_2)/2$ term and $x_3$.

    ✓Final answer

    The correct option is (A) — $\dfrac{x_1 - x_2}{2} + x_3$.

    ANSWER: A