ADV-2026-P2-CHE-Q01 · Molar Conductivity & Solubility of AgCl

01Original question — verbatim

At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:

Salt	Concentration	Molar conductivity
Salt	(M)	(S cm² mol⁻¹)
NaNO₃	0.01	111
NaNO₃	0.04	101
NaCl	0.01	117
NaCl	0.04	107
AgNO₃	0.01	125
AgNO₃	0.04	116

The conductivity of a saturated aqueous solution of AgCl is 1.40×10⁻⁶ S cm⁻¹ at 300 K. If the solubility of AgCl in water at 300 K is X mol L⁻¹, then log₁₀(X⁻¹) is

(Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)

(A) 3
(B) 4
(C) 5
(D) 6

Click an option to check it.

02Interactive conductivity cell

A pair of platinised-Pt electrodes dips into the electrolyte. The meter drives a small AC signal and reads the conductance G; the conductivity is κ = G × G_cell and the molar conductivity is Λ_m = 1000κ/c. Change the salt and the concentration, then record points: the solution model is the Debye–Hückel–Onsager law Λ_m = Λ°_m − A√c, with Λ° and A taken from the paper's own data.

Cell

Ion count scales with concentration; drift speed is drawn proportional to the solution's molar conductivity (schematic — the paper's data fix the salt totals, not the individual ionic values).

Controls

Electrolyte

Concentration c 0.0100 M

1×10⁻⁴ M0.05 M

Cell constant G_cell 1.00 cm⁻¹

Temperature 300 K

≈2 % per K applied to Λ (illustrative); the question's data are at 300 K.

Live readings

Conductivity κ

—

S cm⁻¹

Molar cond. Λ_m

—

S cm² mol⁻¹

Conductance G

—

S (=Ω⁻¹)

Resistance R

—

√c

—

M^1/2

Λ° − Λ_m

—

S cm² mol⁻¹

No measurements recorded yet.

03Λ_m versus √c — reading Λ° off the intercept

For a strong electrolyte Λ_m falls linearly in √c, so the intercept at √c = 0 is the limiting molar conductivity Λ°_m. Solid markers are the two tabulated points per salt; the dashed violet line is AgCl, which is never measured directly — it is constructed in the next panel. Hollow markers are your own recorded points, and their least-squares fit is drawn when you have two or more for one salt.

NaNO₃ · Λ° = 121 NaCl · Λ° = 127 AgNO₃ · Λ° = 134 AgCl (derived) · Λ° = 140 ○ your recorded points

Table view of the plotted data

Salt	Λ_m at 0.01 M	Λ_m at 0.04 M	Slope A	Λ°_m (intercept)
NaNO₃	111	101	100	121
NaCl	117	107	100	127
AgNO₃	125	116	90	134
AgCl	not measurable (insoluble)		—	140

04Kohlrausch builder — assembling Λ°(AgCl)

At infinite dilution each ion migrates independently, so Λ° of a salt is just the sum of its two ionic contributions. Combine the three measured salts so that Na⁺ and NO₃⁻ cancel and only Ag⁺ + Cl⁻ survives.

+ AgNO₃ Ag⁺+ NO₃⁻ 134

+ NaCl Na⁺+ Cl⁻ 127

− NaNO₃ Na⁺+ NO₃⁻ 121

= AgCl Ag⁺+ Cl⁻ 140

Λ°(AgCl) = 134 + 127 − 121 = 140 S cm² mol⁻¹

05From κ of the saturated solution to the solubility

A saturated AgCl solution is so dilute that its molar conductivity is already the limiting value. Rearranging Λ_m = 1000κ/c gives the solubility directly. Drag κ to see how the answer responds — the paper's value is 1.40×10⁻⁶ S cm⁻¹.

Saturated AgCl

Calculation

κ of saturated solution 1.40×10⁻⁶ S cm⁻¹

Λ°_m(AgCl) used 140 S cm² mol⁻¹

Locked to the Kohlrausch result by default; move it to see the sensitivity.

κ (measured)1.40×10⁻⁶ S cm⁻¹

Λ°_m(AgCl) (Kohlrausch)140 S cm² mol⁻¹

X = 1000κ / Λ°_m1.00×10⁻⁵ mol L⁻¹

K_sp = X²1.00×10⁻¹⁰

log₁₀(X⁻¹)5.00

06Step-by-step solution

Use the linear √c law for strong electrolytes

NaNO₃, NaCl and AgNO₃ are strong electrolytes, so over this dilute range the Debye–Hückel–Onsager relation holds:

Λ_m = Λ°_m − A√c

Two concentrations per salt are given, which is exactly what a straight line needs.
Extrapolate each salt to infinite dilution

With c₁ = 0.01 M (√c = 0.1) and c₂ = 0.04 M (√c = 0.2), the √c values differ by exactly 0.1, so the extrapolation collapses to a one-line formula:

Λ° = Λ(0.01) + A(0.1), A = [Λ(0.01) − Λ(0.04)] / 0.1 ⟹ Λ° = 2Λ(0.01) − Λ(0.04)

NaNO₃: 2(111) − 101 = 121 NaCl: 2(117) − 107 = 127 AgNO₃: 2(125) − 116 = 134
Build Λ°(AgCl) with Kohlrausch's law of independent migration

Λ°(AgCl) = λ°(Ag⁺) + λ°(Cl⁻)

= [λ°(Ag⁺)+λ°(NO₃⁻)] + [λ°(Na⁺)+λ°(Cl⁻)] − [λ°(Na⁺)+λ°(NO₃⁻)]

= Λ°(AgNO₃) + Λ°(NaCl) − Λ°(NaNO₃) = 134 + 127 − 121 = 140 S cm² mol⁻¹

Na⁺ and NO₃⁻ appear once with a plus sign and once with a minus sign, so they cancel identically — no individual ionic conductivity is ever needed.
Convert the measured κ into a concentration

The problem states that for the saturated solution Λ_m = Λ°_m = 140 S cm² mol⁻¹ (legitimate: at ~10⁻⁵ M the A√c correction is only about 0.3 S cm² mol⁻¹, i.e. 0.2 %).

Λ_m = 1000 κ / c ⟹ X = c = 1000 κ / Λ°_m

X = (1000 cm³ L⁻¹)(1.40×10⁻⁶ S cm⁻¹) / (140 S cm² mol⁻¹) = 1.00×10⁻⁵ mol L⁻¹
Take the logarithm

log₁₀(X⁻¹) = −log₁₀(1.00×10⁻⁵) = 5

Answer: (C) 5. As a by-product, K_sp(AgCl) = X² = 1.0×10⁻¹⁰ at 300 K, which matches the accepted value closely.

✓ Independent verification

Slopes recomputed from scratch: A = Δ Λ / Δ√c gives 100, 100 and 90 S cm² mol⁻¹ M^−1/2; intercepts 121, 127, 134 — identical to the shortcut 2Λ(0.01) − Λ(0.04).
Dimensional check: (cm³ L⁻¹)(S cm⁻¹)/(S cm² mol⁻¹) = mol L⁻¹ ✓.
Numerical check (independent script): 1000 × 1.40×10⁻⁶ / 140 = 1.000000×10⁻⁵; −log₁₀ = 5.000000 exactly.
Physical sanity: K_sp = 1.0×10⁻¹⁰ against the literature value ≈1.8×10⁻¹⁰ at 298 K — right order of magnitude for a designed problem.
Self-consistency of the slope pattern: both sodium salts share A = 100 and only AgNO₃ differs (90), consistent with the same 1:1 charge type but a different ion pair — no data point is an outlier.
Distractor audit: 3, 4 and 6 correspond to κ read at the wrong power of ten or to forgetting the 1000 cm³ L⁻¹ factor; none arises from a legitimate route.

07Concepts, formulas and classification

Formulas used

κ = G · G_cell = (1/R)(l/A)

Λ_m = 1000 κ / c (c in mol L⁻¹)

Λ_m = Λ°_m − A√c (Debye–Hückel–Onsager)

Λ°_m = ν₊λ°₊ + ν₋λ°₋ (Kohlrausch)

K_sp(AgCl) = [Ag⁺][Cl⁻] = X²

Classification

Simulation ID	`ADV-2026-P2-CHE-Q01`
Exam / Year / Paper	JEE Advanced · 2026 · Paper 2
Subject	Chemistry (Physical)
Chapter	Electrochemistry
Topic	Conductance of electrolytic solutions
Subtopics	Molar & limiting molar conductivity · Kohlrausch's law of independent migration · Solubility of a sparingly soluble salt from conductance
Question type	Single correct MCQ (+3 / 0 / −1)
Difficulty	Moderate
Answer	(C) 5