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ILTS-02 · Error notes · Chemistry

One chapter, twenty questions

Every chemistry question saved from ILTS-02, grouped by sub-topic within the chapter. Each carries what was given, what was asked, the concept and formula behind it, the steps in full, the fastest route through, and a figure where one helps.

20
Questions lost
9
Attempted, wrong
11
Left blank
1
Chapter involved

All twenty come from Electrochemistry — specifically the conductance half of the chapter. On NEET marking they were worth 80 marks, and the nine wrong attempts cost 9 more.

Every one of the nine errors is a reading error rather than a chemistry error. Three rest on the same confusion — conductivity measured per volume against molar conductivity measured per mole — and one submitted a correct fact as the answer to a question asking for the wrong statement.

Chapters in this paper  —  red = wrong · amber = blank · green = correct

Electrochemistry

20 questions · 9 wrong · 11 blank

Q141Q142Q143Q147Q149Q158Q159Q161Q165Q166Q167Q168Q171Q172Q173Q175Q177Q178Q179Q180

Chapter

Electrochemistry

Conductance, Kohlrausch's law and Faraday's laws  ·  20 questions · 9 wrong · 11 blank

Conductance quantities and their units

5 questions · 3 wrong · 2 blank
Q142 Units of resistivity, conductance and conductivity Marked wrong

Match Column I with Column II.

Column IColumn II
A. Resistivity1. ohm−1
B. Conductance2. ohm m
C. Conductivity3. S cm−1

build each unit from its own defining equationResistivity ρR = ρL/A → ρ = RA/Lohm mConductance GG = 1/Rohm−1 (siemens)Conductivity κκ = 1/ρS cm−1resistivity and conductivity are reciprocals, so their units are reciprocals too
  •  A – 2; B – 3; C – 1
  •  A – 1; B – 2; C – 3
  • KEYA – 2; B – 1; C – 3
  • MARKEDA – 1; B – 3; C – 2
Given
  • Three quantities and three sets of units.
Asked
The correct matching.
Concept to use
Derive each unit from its own defining equation rather than trying to recall three separate facts. Resistivity comes from R = ρL/A, so ρ = RA/L carries ohm × m. Conductance is simply 1/R, so ohm−1. Conductivity is 1/ρ, so its unit is the reciprocal of ohm m — that is S m−1, or S cm−1 in CGS.
Formula to use
ρ = RA/L → ohm m  |  G = 1/R → ohm−1  |  κ = 1/ρ → S cm−1
Baby steps
  1. A. Resistivity — ρ = RA/L = ohm × m²/m = ohm m → 2.
  2. B. Conductance — the reciprocal of resistance = ohm−1 → 1.
  3. C. Conductivity — the reciprocal of resistivity = (ohm m)−1 = S cm−1 → 3.
  4. Matching: A–2, B–1, C–3.
Answer
A – 2; B – 1; C – 3
Shortcut
Spot the reciprocal pairs: resistivity and conductivity are reciprocals, so their units must be reciprocals too — ohm m and (ohm m)−1. Once those two are paired, conductance takes what is left.
Where it went wrong
The chosen answer gives resistivity ohm−1, which is a conductance unit — resistivity cannot carry an inverse-ohm unit. A quick check: anything with “resist” in its name has ohm on top, anything with “conduct” has ohm underneath. That one rule sorts all three.
Q149 Conductance and temperature Marked wrong

Choose the wrong statement

  •  electrical conductance of an electrolytic conductor increases with increase in temperature
  • KEYelectrical conductance of a metallic conductor increases with increase in temperature
  • MARKEDelectrical conductance of a metallic conductor decreases with increase in temperature
  •  degree of dissociation of an electrolyte increases with dilution
Given
  • Four statements about conductance.
Asked
The wrong statement.
Concept to use
The two kinds of conductor behave oppositely with temperature. In a metal, heating makes the lattice ions vibrate more, which scatters the electrons — so resistance rises and conductance falls. In an electrolyte, heating lowers the viscosity and increases ionic mobility and dissociation — so conductance rises.
Formula to use
Metal: conductance ↓ with T  |  Electrolyte: conductance ↑ with T
Baby steps
  1. Electrolytic conductor, conductance increases with T — correct, ions move faster and dissociate more.
  2. Metallic conductor, conductance increases with T — this is the reverse of the truth. The wrong statement.
  3. Metallic conductor, conductance decreases with T — correct.
  4. Degree of dissociation increases with dilution — correct, this is Ostwald's dilution law.
  5. The wrong statement is the second.
Answer
electrical conductance of a metallic conductor increases with increase in temperature
Shortcut
Two options make opposite claims about the same thing — metallic conductance rising and falling with temperature. Exactly one must be wrong, so the answer is one of those two, and metals are the familiar case: hot wires resist more.
Where it went wrong
The chosen option is the true statement about metals. The question asks for the wrong one, so the correct physics was identified and then submitted as the answer. When two options contradict each other and the question asks for the false one, pick the one that contradicts what you know — not the one that confirms it.
Q159 Which concentration has the highest specific conductance Not attempted

Which of the following solutions of NaCl has the higher specific conductance?

  •  0.001 N
  •  0.01 N
  •  0.1 N
  • KEY1 N
Given
  • NaCl solutions at four different concentrations.
Asked
Which has the highest specific conductance.
Concept to use
Specific conductance (conductivity, κ) is conductance per unit volume, so it counts ions per cubic centimetre. The more concentrated the solution, the more ions in that volume — so κ is highest at the highest concentration. Molar conductance behaves the opposite way, which is what makes this pair confusing.
Formula to use
κ = ions per unit volume → rises with concentration  |  Λm = κ/c → rises on dilution
Baby steps
  1. Specific conductance measures how well a fixed volume of solution conducts.
  2. A 1 N solution has ten times as many ions per cm³ as 0.1 N, and a thousand times as many as 0.001 N.
  3. More charge carriers per unit volume means higher κ.
  4. So the highest specific conductance is at the highest concentration: 1 N.
  5. (By contrast, molar conductance would be highest at 0.001 N.)
Answer
1 N
Shortcut
Ask which quantity is per volume and which is per mole. Specific conductance is per volume → concentrated wins. Molar or equivalent conductance is per mole → dilute wins. That single distinction answers this question and Q167 in the same paper.
The word “specific” is the signal — it means per unit volume, exactly as specific heat means per unit mass. Reading the prefix tells you which way the trend runs before you think about ions at all.
Q171 Relation between specific and equivalent conductance Marked wrong

The specific conductance (k) of an electrolyte of 0.1 N concentration is related to equivalent conductance (Λ) by the following formula in CGS system

  •  Λ = k
  • MARKEDΛ = 10k
  •  Λ = 100k
  • KEYΛ = 10000k
Given
  • Specific conductance k, normality N = 0.1.
  • CGS units.
Asked
The relation between Λ and k.
Concept to use
The standard relation carries a factor of 1000 (converting cm³ to litres) as well as the division by normality. Both act in the same direction here, so they multiply rather than cancel.
Formula to use
Λ = 1000k/N
Baby steps
  1. Start from Λ = 1000k/N.
  2. Substitute N = 0.1: Λ = 1000k/0.1.
  3. Dividing by 0.1 is multiplying by 10.
  4. Λ = 10 000 k.
Answer
Λ = 10000k
Shortcut
Do the two factors separately: the standard 1000, then ÷ 0.1 which is another × 10. Multiplying gives 10 000. Handling them one at a time prevents the common slip of applying only one.
Where it went wrong
10k applies the division by N but drops the factor of 1000 entirely. That 1000 is not decoration — it converts the per-cm³ basis of specific conductance into the per-litre basis of normality, and it appears in every equivalent-conductance formula in the chapter. Q158 and Q161 in this same paper both use it.
Q178 Unit of a proportionality constant Not attempted

Conductance is directly proportional to area and the concentration of solution in it and is inversely proportional to the length of the vessel then the unit of the constant of proportionality is

  •  S m mol−1
  • KEYS m2 mol−1
  •  S−2 m2 mol
  •  S2 m2 mol−2
Given
  • G ∝ A × c / L, with constant of proportionality k.
Asked
The unit of that constant.
Concept to use
Write the relation with the constant in it, rearrange for the constant, and substitute units. The constant here is really the molar conductivity, so recognising the relation is a shortcut in itself — but the dimensional route works even without that insight.
Formula to use
G = k(Ac/L)  →  k = GL/(Ac)
Baby steps
  1. Rearrange: k = GL/(Ac).
  2. Units: G in S, L in m, A in m², c in mol m−3.
  3. k = S × m ÷ (m² × mol m−3).
  4. Denominator: m² × mol m−3 = mol m−1.
  5. k = S m ÷ (mol m−1) = S m² mol−1.
Answer
S m2 mol−1
Shortcut
Recognise the quantity rather than deriving it: conductance per unit concentration, corrected for cell geometry, is molar conductivity — whose unit is S m² mol−1. The two options with squared or negative powers of S can be struck out at once, since a conductance appears to the first power only.
A useful habit for any “unit of the constant” question: solve for the constant explicitly first, then substitute units mechanically. Trying to reason the units out in one step is where sign and power errors creep in.

Conductance numericals

3 questions · 3 blank
Q158 Equivalent conductance from resistance Not attempted

The resistance of N/10 solution is found to be 2.5 × 103 ohm. The equivalent conductance of the solution is (cell constant = 1.25 cm−1)

  •  2.5 ohm−1 cm² equiv−1
  • KEY5 ohm−1 cm² equiv−1
  •  2.5 ohm−1 cm−2 equiv−1
  •  5 ohm−1 cm−2 equiv−1
Given
  • R = 2.5 × 103 ohm.
  • Cell constant = 1.25 cm−1.
  • Normality N = 0.1.
Asked
The equivalent conductance.
Concept to use
A fixed three-step chain: resistance → conductance → conductivity → equivalent conductance. The cell constant converts a measured conductance into a conductivity, and the factor 1000 converts per-cm³ into per-litre so the normality can be divided out.
Formula to use
κ = (1/R) × cell constant  |  Λeq = 1000κ/N
Baby steps
  1. Conductance: G = 1/R = 1/(2.5×103) = 4 × 10−4 S.
  2. Conductivity: κ = G × cell constant = 4×10−4 × 1.25 = 5 × 10−4 S cm−1.
  3. Equivalent conductance: Λeq = 1000κ/N = 1000 × 5×10−4 / 0.1.
  4. = 0.5/0.1 = 5 ohm−1 cm² equiv−1.
Answer
5 ohm−1 cm² equiv−1
Shortcut
Two of the four options carry cm−2, which is dimensionally impossible — equivalent conductance is a conductance times an area per equivalent, so cm² must be positive. Checking the unit halves the list before any arithmetic.
Keep the chain in one line: R → G = 1/R → κ = G × cell constant → Λ = 1000κ/N. Nearly every conductance numerical in this chapter is one of these four steps, or all of them in order.
Q161 Equivalent conductivity from cell dimensions Not attempted

0.5 N solution of a salt placed between two platinum electrodes 2.0 cm apart and of area of cross-section 4.0 sq.cm has a resistance of 25 ohms. The equivalent conductivity of solution (S cm2 eq−1) is

  •  80
  • KEY40
  •  20
  •  10
Given
  • Electrode separation L = 2.0 cm, area A = 4.0 cm².
  • Resistance R = 25 ohm, normality N = 0.5.
Asked
The equivalent conductivity.
Concept to use
Here the cell constant is not given but must be constructed from the geometry: it is the separation divided by the area, L/A. After that the chain is the same as any other conductance numerical.
Formula to use
cell constant = L/A  |  κ = (1/R)(L/A)  |  Λeq = 1000κ/N
Baby steps
  1. Cell constant = L/A = 2.0/4.0 = 0.5 cm−1.
  2. Conductance: G = 1/25 = 0.04 S.
  3. Conductivity: κ = 0.04 × 0.5 = 0.02 S cm−1.
  4. Λeq = 1000 × 0.02 / 0.5 = 20/0.5 = 40 S cm² eq−1.
Answer
40
Shortcut
Cell constant is always length ÷ area — note that it is not area/length, which is the commonest slip and would give 80 here. The unit cm−1 confirms it: length over area gives cm/cm² = cm−1.
80 is exactly what you get from inverting the cell constant, and it sits first on the option list. Writing “cell constant = L/A” before substituting is the guard against it.
Q177 Molar conductivity using a shared cell constant Not attempted

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω. The conductivity of this solution is 1.29 S m−1. Resistance of the same cell when filled with 0.02 M of the same solution is 520 Ω. The molar conductivity of 0.02 M solution in S m2 mol−1

  •  12.4 × 10−4
  • KEY124 × 10−4
  •  1240 × 10−4
  •  1.24 × 10−4
Given
  • 0.1 M: R = 100 Ω, κ = 1.29 S m−1.
  • 0.02 M: R = 520 Ω in the same cell.
  • Answer required in S m² mol−1 (SI).
Asked
Molar conductivity of the 0.02 M solution.
Concept to use
The first solution is given only to calibrate the cell — from its resistance and conductivity you extract the cell constant, which then applies to the second solution since the same cell is used. Note that the answer is wanted in SI, so the concentration must be converted from mol/L to mol m−3.
Formula to use
cell constant = κR  |  κ2 = cell constant/R2  |  Λm = κ/c, with c in mol m−3
Baby steps
  1. Cell constant from the first solution: G* = κR = 1.29 × 100 = 129 m−1.
  2. Conductivity of the second: κ2 = 129/520 = 0.248 S m−1.
  3. Convert the concentration to SI: 0.02 mol/L = 0.02 × 1000 = 20 mol m−3.
  4. Λm = κ2/c = 0.248/20 = 0.0124 S m² mol−1.
  5. Written in the required form: 124 × 10−4.
Answer
124 × 10−4
Shortcut
Recognise the role of each piece of data before calculating: the first solution is there only to give the cell constant. Once G* = 129 m−1 is found, the first solution is finished with and never appears again.
All four options share the digits 124, so the entire question rests on the power of ten — which comes from converting mol/L to mol m−3. Multiplying the molarity by 1000 is the step that decides the mark.

Kohlrausch's law

2 questions · 2 wrong
Q141 Kohlrausch construction for barium hydroxide Marked wrong

Molar conductances at infinite dilution of barium chloride, sodium hydroxide, and sodium chloride are X, Y, and Z, respectively. Molar conductance at infinite dilution of barium hydroxide at the same temperature would be (all measurements expressed in SI units)

cancel the spectator ions and see what is left+BaCl2 → Ba²⁺ + 2Cl⁻X+2 × NaOH → 2Na⁺ + 2OH⁻2Y−2 × NaCl → 2Na⁺ + 2Cl⁻2ZBa(OH)2 = X + 2Y − 2ZNa⁺ : 2 added, 2 removed → 0Cl⁻ : 2 added, 2 removed → 0left with Ba²⁺ + 2OH⁻which is exactly Ba(OH)2the coefficients 2 are needed because barium is divalent — one Ba²⁺ needs two OH⁻
  •  X + Y − Z
  • KEYX + 2Y − 2Z
  • MARKED−X + 2Y + 2Z
  •  X + 2Z − 2Y
Given
  • Λ° BaCl2 = X, Λ° NaOH = Y, Λ° NaCl = Z.
Asked
Λ° of Ba(OH)2.
Concept to use
Kohlrausch's law lets you build any electrolyte from others by adding and subtracting so that the spectator ions cancel. The point that decides this question is the stoichiometry: Ba(OH)2 needs two hydroxide ions per barium, so NaOH and NaCl must both be taken twice.
Formula to use
Λ°[Ba(OH)2] = Λ°[BaCl2] + 2Λ°[NaOH] − 2Λ°[NaCl]
Baby steps
  1. Write what each supplies: BaCl2 → Ba²⁺ + 2Cl⁻.
  2. Add 2 × NaOH → 2Na⁺ + 2OH⁻.
  3. Subtract 2 × NaCl → removes 2Na⁺ + 2Cl⁻.
  4. Check the cancellation: 2Cl⁻ added and 2Cl⁻ removed → 0; 2Na⁺ added and 2Na⁺ removed → 0.
  5. What remains is exactly Ba²⁺ + 2OH⁻, i.e. Ba(OH)2.
  6. So Λ° = X + 2Y − 2Z.
Answer
X + 2Y − 2Z
Shortcut
Write the ions supplied by each compound in a column and cancel by inspection. The coefficient on each term is whatever it takes to make the spectator ions vanish — here two, because barium is divalent and drags two chlorides with it.
Where it went wrong
The chosen answer has −X, which would mean removing the barium — but barium is the ion we need to keep. The compound supplying the ion you want must always enter with a plus sign. Checking that Ba²⁺ survives the cancellation rules out that option immediately.
Q173 Statement of Kohlrausch's law Marked wrong

Kohlrausch's law states that at

  • KEYInfinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever is the nature of the other ion of the electrolyte
  •  Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
  • MARKEDInfinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte
  •  Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte
Given
  • Four candidate statements of Kohlrausch's law.
Asked
The correct statement.
Concept to use
Kohlrausch's law of independent migration of ions makes two essential claims: it holds at infinite dilution, and each ion's contribution is independent of the other ion. That independence is precisely what allows a weak electrolyte to be built from strong ones, as in Q141 of this paper.
Formula to use
Λ° = ν+λ°+ + ν−λ°− — each term independent of its partner
Baby steps
  1. The law applies only at infinite dilution, where ions are far enough apart not to interact — so “finite dilution” is wrong.
  2. Each ion's contribution is independent of what its partner ion is — so “depending on the nature of the other ion” is wrong.
  3. The quantity involved is equivalent (or molar) conductance, not plain conductance — so the fourth option is wrong.
  4. Only the first statement gets all three right.
Answer
Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever is the nature of the other ion of the electrolyte
Shortcut
Check three words in turn: infinite (not finite), whatever (not depending on), and equivalent conductance (not conductance). Each wrong option alters exactly one of the three, so testing them in order eliminates one option at a time.
Where it went wrong
The chosen option says the contribution depends on the other ion — which is the exact opposite of independent migration, and would make the law useless. If each ion's contribution depended on its partner, you could not add and subtract electrolytes as in Q141. The name of the law contains the answer: independent migration.

Weak electrolytes and Ostwald's law

7 questions · 3 wrong · 4 blank
Q143 Dissociation constant from conductance Not attempted

The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mho cm2 eq−1 and at infinite dilution is 400 mho cm2 eq−1. The dissociation constant of this acid is

  • KEY1.25 × 10−5
  •  1.25 × 10−6
  •  1.625 × 10−4
  •  1.25 × 10−4
Given
  • Λ = 8.0 and Λ° = 400 mho cm² eq−1.
  • Concentration c = M/32 = 1/32 M.
Asked
The dissociation constant Ka.
Concept to use
Two steps. The degree of dissociation is the ratio of the measured conductance to the limiting value. Then Ostwald's dilution law converts α and c into Ka. Because α here is small, the (1 − α) in the denominator barely matters.
Formula to use
α = Λ/Λ°  |  Ka = cα²/(1 − α) ≈ cα²
Baby steps
  1. α = 8.0/400 = 0.02.
  2. c = 1/32 = 0.03125 M.
  3. Ka = cα²/(1 − α) = (0.03125)(0.0004)/(0.98).
  4. Numerator: 0.03125 × 4×10−4 = 1.25 × 10−5.
  5. Dividing by 0.98 changes little, so Ka ≈ 1.25 × 10−5.
Answer
1.25 × 10−5
Shortcut
For a weak electrolyte with small α, use Ka = cα² and ignore the (1 − α). Here c is a neat 1/32 and α² is 4 × 10−4, so the product is exact without a calculator: 4/32 = 0.125, giving 1.25 × 10−5.
The distractors differ only in the power of ten, so the arithmetic on the exponent is where the mark sits. Count carefully: 10−2 from c, and 10−4 from α², plus a leading 1.25 — that is 10−5 overall.
Q147 pH of a weak base from conductance Marked wrong

At 298 K, molar conductance of 0.01 M weak monoacidic base, BOH is 39.6 mho cm2 mol−1. Molar conductance of BOH at infinite dilution is 396 mho cm2 mol−1. pH of given BOH solution is

  • KEY11
  •  10.699
  •  12
  • MARKED9.301
Given
  • Λ = 39.6, Λ° = 396 mho cm² mol−1.
  • Concentration c = 0.01 M.
  • BOH is a monoacidic base.
Asked
The pH.
Concept to use
Three steps, and the last one is where marks are lost. The conductance ratio gives α; multiplying by c gives [OH−]; that gives pOH, and pH is 14 minus pOH. Because BOH is a base, the answer must come out above 7.
Formula to use
α = Λ/Λ°  |  [OH−] = cα  |  pOH = −log[OH−]  |  pH = 14 − pOH
Baby steps
  1. α = 39.6/396 = 0.1.
  2. [OH−] = cα = 0.01 × 0.1 = 10−3 M.
  3. pOH = −log(10−3) = 3.
  4. pH = 14 − 3 = 11.
Answer
11
Shortcut
Check the direction before computing: BOH is a base, so pH must exceed 7 — that eliminates 9.301 immediately. Then the numbers are chosen to be clean: α = 0.1 exactly, giving [OH−] = 10−3 and a whole-number pOH.
Where it went wrong
9.301 has the shape of a real answer but sits below 11 and results from mishandling the log or the 14-minus step. The deeper check is the one to build: a base cannot have a pH below 7, so any option under 7 is impossible — and this option is close enough to 7 to look plausible without being checked.
Q166 Identifying strong and weak electrolytes from dilution Marked wrong

When solutions with equal molarities of two electrolytes (X) and (Y) are diluted to the same extent, then ΛM of 'X' increases by 1.2 times, while that of 'Y' increases 20 times. Identify electrolytes 'X' and 'Y' from the following.

  • MARKEDX = CH3COOH, Y = NaCl
  • KEYX = KCl, Y = CH3COOH
  •  X = CH3COOH, Y = NH4OH
  •  X = CH3COONa, Y = NaCl
Given
  • On dilution, ΛM of X rises 1.2 times and of Y rises 20 times.
Asked
Which electrolyte is which.
Concept to use
For a strong electrolyte dissociation is already essentially complete, so dilution changes ΛM only slightly — the small rise comes from reduced inter-ionic attraction. For a weak electrolyte dilution drives dissociation sharply upwards, so ΛM rises steeply. A 1.2-fold rise marks a strong electrolyte; a 20-fold rise marks a weak one.
Formula to use
Strong electrolyte → small increase on dilution  |  Weak electrolyte → large increase
Baby steps
  1. X increases only 1.2 times → almost fully dissociated already → strong electrolyte.
  2. Y increases 20 times → dissociation rises dramatically → weak electrolyte.
  3. Among the options, KCl is a strong electrolyte and CH3COOH is a weak acid.
  4. So X = KCl and Y = CH3COOH.
  5. Check the others: NaCl and CH3COONa are both strong, and NH4OH is weak — no other option pairs a strong X with a weak Y.
Answer
X = KCl, Y = CH3COOH
Shortcut
Read the two numbers as labels rather than data: small factor = strong, large factor = weak. Then scan the options for the pair that puts a strong electrolyte first. Only one does.
Where it went wrong
The chosen answer has the two the wrong way round — acetic acid as X and NaCl as Y — putting the weak electrolyte where the small rise was reported. Both substances were correctly classified; only the assignment was reversed. When a question says “identify X and Y”, write the property beside each letter before looking at the options.
Q167 Statements about conductivity and dilution Marked wrong

The correct statements from the following is
(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolyte.
(B) The number of ions per unit volume that carry current in a solution increases on dilution.
(C) Molar conductivity (Λm) increases with decrease in concentration.
(D) The variation in molar conductivity (Λm) is different for strong and weak electrolyte

the two conductivities move in opposite directions on dilutiondilution →valueκ fallsΛm risesconductivity κions per unit volume fall → κ fallsmolar conductivity Λmper mole, and dissociation risesstatement B is the trap: ions per unit volume decrease on dilution
  •  A, B, C, D
  • MARKEDOnly B, C, D
  • KEYOnly A, C, D
  •  Only B, C
Given
  • Four statements about conductivity on dilution.
Asked
Which statements are correct.
Concept to use
The whole question turns on the difference between per unit volume and per mole. Diluting a solution spreads the same ions through more volume, so the number of ions per unit volume falls — and with it the conductivity. But molar conductivity, measured per mole, rises, because dissociation increases and the ions interfere with each other less.
Formula to use
κ falls on dilution (per volume)  |  Λm rises on dilution (per mole)
Baby steps
  1. (A) Conductivity decreases as concentration decreases, for both types. Correct.
  2. (B) Ions per unit volume decrease on dilution, not increase — the total number of ions may rise, but they occupy far more volume. Incorrect.
  3. (C) Molar conductivity increases as concentration falls. Correct.
  4. (D) Strong electrolytes show a small linear rise; weak ones show a steep rise near infinite dilution. Correct.
  5. Answer: only A, C and D.
Answer
Only A, C, D
Shortcut
Statements A and B contradict each other — if the ions per unit volume rose, the conductivity could not fall. Exactly one of them is true, and A is the standard result. Spotting the contradiction halves the work.
Where it went wrong
Statement B was accepted and A rejected, which is the pair the wrong way round. The phrase to underline is “per unit volume”: dilution always reduces anything measured per volume. This distinction also decides Q159 in the same paper, where the highest specific conductance belongs to the most concentrated solution.
Q172 Slope and intercept of an Ostwald plot Not attempted

Plotting 1/Λm against cΛm for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is

  • KEYKaΛ°m
  •  KaΛ°m/2
  •  2KaΛ°m
  •  1/(KaΛ°m)
Given
  • Plot of 1/Λm against cΛm, giving intercept P and slope S.
Asked
The ratio P/S.
Concept to use
Ostwald's dilution law can be rearranged into a straight-line form. Writing α = Λm/Λ°m in Ka = cα²/(1 − α) and rearranging gives an equation in 1/Λm and cΛm, from which the intercept and slope can simply be read off.
Formula to use
1/Λm = 1/Λ°m + (cΛm)/(Ka(Λ°m)²)
Baby steps
  1. Compare with y = c + mx, where y = 1/Λm and x = cΛm.
  2. Intercept: P = 1/Λ°m.
  3. Slope: S = 1/(Ka(Λ°m)²).
  4. Ratio: P/S = (1/Λ°m) × Ka(Λ°m)².
  5. One power of Λ°m cancels: P/S = KaΛ°m.
Answer
KaΛ°m
Shortcut
Once the linear form is written down, P/S is just a division — and dividing by a reciprocal means multiplying, so the (Λ°m)² in the slope loses one power against the Λ°m in the intercept. Track the powers rather than re-deriving.
This linearised Ostwald plot is how Λ°m for a weak electrolyte is obtained experimentally — it cannot be found by extrapolating a Λm-versus-√c graph the way a strong electrolyte's can, because the curve rises too steeply near zero.
Q175 Degree of dissociation and pH of acetic acid Not attempted

The molar conductivity of acetic acid at infinite dilution is 390.7 S cm2 mol−1 and for 0.01 M acetic acid is 3.907 S cm2 mol−1. The degree of dissociation and pH of the solution is (choose the correct row: 1 → α 0.1, pH 3; 2 → α 0.02, pH 4; 3 → α 0.001, pH 5; 4 → α 0.01, pH 4)

  •  1
  •  2
  •  3
  • KEY4
Given
  • Λ° = 390.7, Λ = 3.907 S cm² mol−1.
  • c = 0.01 M acetic acid.
Asked
The row giving the correct α and pH.
Concept to use
Same three-step chain as Q147 but for an acid, so it ends at pH directly rather than going through pOH. The numbers are arranged so that α comes out as a clean power of ten.
Formula to use
α = Λ/Λ°  |  [H+] = cα  |  pH = −log[H+]
Baby steps
  1. α = 3.907/390.7 = 0.01. (The digits are identical, so the ratio is exactly 1/100.)
  2. [H+] = cα = 0.01 × 0.01 = 10−4 M.
  3. pH = −log(10−4) = 4.
  4. So α = 0.01 and pH = 4, which is row 4.
Answer
4
Shortcut
Notice that 3.907 and 390.7 share the same digits, so the ratio is exactly 10−2 with no division needed. Papers arrange this deliberately — if two given numbers look like the same digits shifted, the ratio is a clean power of ten.
Row 1 pairs α = 0.1 with pH 3, which would be right if Λ were 39.07. Each wrong row is internally consistent, so you cannot eliminate by checking the pH against the α — you must compute α first.
Q179 Molar conductivity at infinite dilution on further dilution Not attempted

A conductivity cell with two electrodes (dark side) are half filled with infinitely dilute aqueous solution of a weak electrolyte. If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will

  •  decrease sharply
  • KEYremain same or cannot be measured accurately
  •  depend upon type of electrolyte
  •  increase sharply
Given
  • A weak electrolyte already at infinite dilution.
  • The volume is then doubled by adding water.
Asked
What happens to the molar conductivity.
Concept to use
The phrase “infinitely dilute” is doing all the work. Molar conductivity rises on dilution only until dissociation is complete; at infinite dilution it has already reached its limiting value Λ°m and cannot rise further. Adding more water changes nothing — though in practice the conductance becomes too small to measure reliably.
Formula to use
Λm rises on dilution up to Λ°m, then stays constant
Baby steps
  1. The solution is stated to be at infinite dilution, so α is already effectively 1 and Λm = Λ°m.
  2. Further dilution cannot increase dissociation, since it is already complete.
  3. So the molar conductivity remains the same.
  4. In practice the measured conductance becomes vanishingly small, so the value cannot be determined accurately — which is why Λ°m for weak electrolytes is obtained indirectly, by Kohlrausch's law.
  5. Answer: remains the same, or cannot be measured accurately.
Answer
remain same or cannot be measured accurately
Shortcut
Treat “infinitely dilute” as meaning “already at the ceiling”. Any question asking what happens on further dilution from that state has the answer “nothing” — the limiting value is by definition the end of the curve.
This is precisely why Kohlrausch's law exists. Λ°m for a weak electrolyte cannot be measured directly by diluting, so it is constructed from strong electrolytes instead — exactly the calculation performed in Q141 of this paper.

Solubility product from conductance

1 question · 1 blank
Q165 Solubility product from conductance Not attempted

Equivalent conductance of saturated BaSO4 is 400 ohm−1 cm2 equiv−1 and specific conductance is 8 × 10−5 ohm−1 cm−1. Hence Ksp of BaSO4 is

  •  4 × 10−8 M²
  • KEY1 × 10−8 M²
  •  2 × 10−4 M²
  •  1 × 10−4 M²
Given
  • Λeq = 400 ohm−1 cm² equiv−1.
  • κ = 8 × 10−5 ohm−1 cm−1.
  • BaSO4 dissolves as Ba²⁺ + SO4²⁻.
Asked
The solubility product.
Concept to use
The conductance data give the normality of the saturated solution, which must then be converted to molarity. For BaSO4 each mole supplies 2 equivalents (the ions are divalent), so molarity is half the normality. Only then is Ksp = [Ba²⁺][SO4²⁻] = S².
Formula to use
N = 1000κ/Λeq  |  M = N/2 for BaSO4  |  Ksp = M²
Baby steps
  1. Normality: N = 1000κ/Λeq = 1000 × 8×10−5 / 400.
  2. = 0.08/400 = 2 × 10−4 N.
  3. Convert to molarity: BaSO4 gives 2 equivalents per mole, so M = N/2 = 1 × 10−4 M.
  4. For BaSO4 ⇌ Ba²⁺ + SO4²⁻, both ions are at the solubility S.
  5. Ksp = S² = (10−4)² = 1 × 10−8 M².
Answer
1 × 10−8 M²
Shortcut
Check the units of the answer to see how far you must go: Ksp is in M², so a squaring is required — the two options given in M−4 are just the solubility itself, unsquared.
The normality-to-molarity step is the one most often skipped. For a salt of two divalent ions the factor is 2; for NaCl it would be 1; for Al2(SO4)3 it would be 6. Ask how many equivalents one mole supplies before dividing.

Faraday's laws of electrolysis

2 questions · 1 wrong · 1 blank
Q168 Faraday equivalence between oxygen and silver Not attempted

The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be

one faraday deposits one gram-equivalent of anythingOXYGEN side5600 mL at STP= 5600/22400 = 0.25 molO2 needs 4 electrons per mole0.25 × 4 = 1 equivalentSILVER sidesame charge = 1 faradayAg⁺ + e⁻ → Ag, so n = 11 equivalent = 108/1 g= 108 gequivalents are always equal at both electrodesn = 4 for O2, 2 for H2 and Cu, 1 for Ag — write the half-reaction first
  •  5.4 g
  •  10.8 g
  •  54.0 g
  • KEY108.0 g
Given
  • 5600 mL of O2 at STP liberated by a given charge.
  • Atomic weight of silver = 108.
  • Molar volume at STP = 22400 mL.
Asked
The mass of silver deposited by the same charge.
Concept to use
The same quantity of electricity deposits the same number of gram-equivalents of every substance. So convert the oxygen to equivalents, then convert that many equivalents back into grams of silver. The key numbers are n = 4 for O2 and n = 1 for Ag.
Formula to use
equivalents of O2 = equivalents of Ag  |  equivalent weight = atomic weight / n
Baby steps
  1. Moles of O2: 5600/22400 = 0.25 mol.
  2. O2 liberation needs 4 electrons per molecule (2H2O → O2 + 4H+ + 4e−), so equivalents = 0.25 × 4 = 1.
  3. The same charge therefore deposits 1 equivalent of silver.
  4. For Ag+ + e− → Ag, n = 1, so the equivalent weight is 108/1 = 108 g.
  5. Mass of silver = 108 g.
Answer
108.0 g
Shortcut
Note that 5600 mL is exactly a quarter of 22400, so the oxygen is 0.25 mol — and with n = 4 that is exactly one faraday. One faraday deposits one equivalent, which for monovalent silver is one whole mole. The numbers are chosen to land on 1.
Getting n right is where these questions are decided: 4 for O2, 2 for H2 and Cu, 1 for Ag and H+. Write the half-reaction before doing any arithmetic; the distractors here all correspond to using a different n.
Q180 Mass deposited by a known current Marked wrong

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulfate solution for 100 seconds is: (Given: Molar mass of Cu = 63 g mol−1, 1F = 96487 C)

  •  31.5 g
  • MARKED0.0315 g
  •  3.15 g
  • KEY0.315 g
Given
  • I = 9.6487 A, t = 100 s.
  • Cu molar mass 63 g mol−1, F = 96487 C.
  • Cu²⁺ + 2e− → Cu, so n = 2.
Asked
The mass of copper deposited.
Concept to use
Faraday's first law in its equivalent-weight form. The charge passed is I × t; dividing by F gives the number of faradays, and multiplying by the equivalent weight (M/n) gives the mass. For copper n = 2, and forgetting that halving is the commonest error after decimal slips.
Formula to use
m = (I t / F) × (M / n), with n = 2 for Cu
Baby steps
  1. Charge: Q = I t = 9.6487 × 100 = 964.87 C.
  2. Faradays: 964.87 / 96487 = 0.01 F. (The numbers are chosen to give exactly one hundredth.)
  3. Equivalent weight of copper: M/n = 63/2 = 31.5 g.
  4. Mass = 0.01 × 31.5 = 0.315 g.
Answer
0.315 g
Shortcut
The current 9.6487 A for 100 s is engineered so that Q/F = 0.01 exactly — the digits of the current match the faraday constant. Spotting that turns the whole problem into 31.5 × 0.01, with no long division.
Where it went wrong
0.0315 is a factor of ten below the answer, which is a decimal slip rather than a conceptual error — and 31.5 (the bare equivalent weight, with the 0.01 forgotten) is also on the list. Note the ladder of options: 31.5, 3.15, 0.315, 0.0315 all differ only by powers of ten, so the paper is testing the exponent arithmetic alone. Write Q/F on its own line and check it is 0.01 before multiplying.

What the twenty have in common

One chapter, nine attempted and missed, eleven blank. Every error is a reading error rather than a chemistry error — and three of them are the same misreading.

1 · “Per volume” confused with “per mole” — Q159, Q167
Conductivity κ is measured per unit volume, so dilution lowers it. Molar conductivity Λm is per mole, so dilution raises it. Q167's statement B claims ions per unit volume increase on dilution — it was accepted while the true statement A was rejected. Q159 rests on the same point. Fix: underline the word specific or molar in the question; it tells you which way the trend runs before any reasoning.

2 · The correct fact submitted as the wrong answer — Q149
The question asked for the wrong statement; the option chosen is the true one about metals. Two options made directly opposite claims, so one had to be false — but the one contradicting known physics was the answer wanted. Fix: when a stem contains “wrong”, “incorrect” or “not true”, circle it before reading the options.

3 · Right content, wrong assignment — Q141, Q142, Q166
Q166 classified both electrolytes correctly and then swapped X and Y. Q142 gave resistivity a conductance unit. Q141 put a minus sign on the compound supplying the ion that had to be kept. In each case the chemistry was known and the placement was not. Fix: write the property beside each label before looking at the options — the same discipline that matching questions need.

4 · The factor of 1000 — Q171, and it recurs in Q158, Q161, Q165, Q177
Q171 gave Λ = 10k, dropping the 1000 that converts a per-cm³ basis into a per-litre one. That constant appears in five questions on this paper. Fix: hold the chain as one item — R → G = 1/R → κ = G × cell constant → Λ = 1000κ/N — and note that the cell constant is length ÷ area, not the reverse.

Six results that cover the chapter

The correct option is marked KEY and the option selected in the test is marked MARKED; questions with no marked option were left unattempted. All figures have been drawn fresh for these notes, and every numerical answer here was checked computationally before being written in.