Every chemistry question lost on this paper, rebuilt in full. Each one carries what was given, what was asked, the concept behind it, the formula, the steps written out, a table justifying the right option and ruling out each of the others, the fastest route through, and an animated figure wherever seeing the thing settles the answer.
13Questions lost
6Attempted, wrong
7Left blank
58Marks at stake
The shape of this paper
Thirteen questions from Solutions: six attempted and missed, seven left blank. On NEET marking
that is 52 marks not gained plus 6 in negatives.
One important caveat before anything else. On Q156 the answer selected was
chemically correct and the paper's key appears to be wrong. That question is flagged in
full below, and it should not be counted as an error.
Of the remaining five, three — Q161, Q162 and Q173 — come down to reading a
quantity precisely: molarity against particle concentration, i against α, and which
temperature a constant belongs to.
All thirteen come from one chapter, split almost evenly between attempted and blank. Three of the six wrong answers turn on the van't Hoff factor or on counting particles rather than formula units — and one of the six was, in fact, answered correctly.
Q 148Left blankRaoult's law · composition of a boiling mixture
At 80 °C, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture solution of 'A' and 'B' boils at 80°C and 1 atm pressure, the amount of 'A' in the mixture is (1 atm = 760 mm Hg)
52 mol percent
34 mol percent
48 mol percent
Correct50 mol percent
Given
Pₐ° = 520 mm Hg, Pₒ° = 1000 mm Hg at 80 °C.
The mixture boils at 80 °C under 1 atm = 760 mm Hg.
Asked
The mole per cent of A in the mixture.
Concept to use
A liquid boils when its total vapour pressure equals the external pressure. So “boils at 1 atm” is simply the statement P_total = 760 mm Hg. Raoult's law then gives one linear equation in one unknown.
DiagramAnimatedBoiling at 1 atm means the total vapour pressure equals 760 mm.
Formula to useP_total = Pₐ°xₐ + Pₒ°(1 − xₐ)
Baby steps
Write Raoult's law: 520 xₐ + 1000(1 − xₐ) = 760.
Expand: 1000 − 480 xₐ = 760.
480 xₐ = 240.
xₐ = 0.5, i.e. 50 mol per cent.
Answer
50 mol per cent
Why this option and not the others
Option
Verdict
Reason
52 mol per cent
rule out
Would need P_total ≈ 750 mm Hg.
34 mol per cent
rule out
Would give P_total = 837 mm Hg, well above 1 atm.
48 mol per cent
rule out
Gives 769.6 mm Hg — close, but the arithmetic is exact and lands on 0.5.
Notice that 760 sits exactly halfway between 520 and 1000: (520 + 1000)/2 = 760. Since Raoult's law is linear in composition, the halfway pressure means the halfway composition — 50 mol per cent, with no algebra at all. Always check for that midpoint before solving.
Q 149Attempted · wrongTwo colligative effects added together
How many grams of sucrose (molecular weight = 342) should be dissolved in 100 gm water in order to have a solution with sum of lowering of freezing point and elevation of boiling point equal to 5 °C (Kₜ = 0.51, Kᶠ = 1.86)
34.2 gm
Correct72 gm
342 gm
Her answer460 gm
Given
Sucrose, M = 342 g/mol, in 100 g water = 0.1 kg solvent.
ΔTᶠ + ΔTₜ = 5 °C.
Kₜ = 0.51, Kᶠ = 1.86 K kg mol⁻¹.
Asked
The mass of sucrose required.
Concept to use
Both effects are driven by the same molality, so instead of solving twice you can add the two constants first and treat (Kᶠ + Kₜ) as a single number. That turns the whole problem into one division followed by one multiplication.
DiagramAnimatedBoth effects share one molality, so add the constants first.
Formula to useΔTᶠ + ΔTₜ = (Kᶠ + Kₜ) m → m = 5/(1.86 + 0.51)
Baby steps
Add the constants: Kᶠ + Kₜ = 1.86 + 0.51 = 2.37.
Molality: m = 5 / 2.37 = 2.11 mol kg⁻¹.
Moles of sucrose = m × mass of solvent in kg = 2.11 × 0.1 = 0.211 mol.
Mass = 0.211 × 342 = 72 g.
Answer
About 72 g
Why this option and not the others
Option
Verdict
Reason
34.2 gm
rule out
One tenth of the molar mass — a number lifted from the data rather than calculated.
72 gm
keep
0.211 mol of a 342 g/mol solute.
342 gm
rule out
The molar mass itself, quoted back. That would be 1 mol in 0.1 kg, a molality of 10.
460 gm
rule out
Would correspond to a molality of about 13.5, giving a combined effect near 32 °C rather than 5 °C.
Shortcut
Add the constants before doing anything else. Whenever a question asks for the sum of freezing-point depression and boiling-point elevation, (Kᶠ + Kₜ)m is a single step. And remember the solvent mass must be in kilograms — 100 g is 0.1 kg, and forgetting that inflates the answer tenfold.
Where it went wrong
460 g is roughly six times too large, which points at the molality step rather than the chemistry. The most likely route is treating 100 g as 1 kg somewhere, or solving for each effect separately and then adding the two masses instead of adding the constants. The single guard is a sanity check at the end: 460 g of sucrose in 100 g of water is a physically absurd solution — more than four times as much solute as solvent.
Q 153Left blankRelative lowering of vapour pressure
18 g glucose C₆H₁₂O₆ is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is:
76.0
Correct752.4
759.0
7.6
Given
18 g glucose (M = 180), so 0.1 mol.
178.2 g water (M = 18), so 9.9 mol.
Vapour pressure of pure water at 100 °C = 760 torr.
Asked
The vapour pressure of water above the solution.
Concept to use
Raoult's law for a non-volatile solute: the vapour pressure above the solution is the pure value scaled by the mole fraction of the solvent. The numbers here are chosen so the mole fraction comes out at a clean 0.99.
DiagramAnimatedThe solvent's mole fraction scales the pure vapour pressure.
Formula to useP_solution = x_solvent × P°_solvent
Baby steps
Moles of glucose = 18/180 = 0.1.
Moles of water = 178.2/18 = 9.9.
Total = 10.0 mol, so x_water = 9.9/10.0 = 0.99.
P = 0.99 × 760 = 752.4 torr.
Answer
752.4 torr
Why this option and not the others
Option
Verdict
Reason
76.0
rule out
One tenth of 760 — a decimal place lost.
752.4
keep
0.99 × 760. The solute lowers the vapour pressure by just 1%.
759.0
rule out
Would need x_water = 0.9987, which does not follow from 0.1 and 9.9 mol.
7.6
rule out
Two decimal places out. This is 1% of 760, which is the lowering, not the resulting pressure.
Shortcut
Do the sanity check first: the solute is only 1 mole per cent of the mixture, so the vapour pressure must drop by roughly 1% — from 760 to a little over 750. Only one option is in that range, and you can pick it before doing any arithmetic. Note that 7.6 is the lowering — read whether the question wants ΔP or P.
Q 156Attempted · wrongAzeotropes · minimum or maximum boiling
The paper's answer key looks wrong hereThe key marks “maximum boiling azeotrope” correct, but the mixture boils at 312 K, which is below both components (320 K and 329 K). A mixture boiling below both of its components is a minimum boiling azeotrope. Acetone + CS₂ is also NCERT's own example of positive deviation, which always gives a minimum boiling azeotrope. The answer selected was the correct one. This card explains the chemistry so the reasoning is secure either way.
Acetone (BP 329 K) and CS₂ (BP 320 K) are mixed in a definite composition so that the mixture of the two behave like a pure liquid and boils at 312 K, then it is
Not an Azeotrope
CorrectMaximum boiling azeotrope
Her answerMinimum boiling azeotrope
Rectified spirit
Given
Acetone boils at 329 K; CS₂ boils at 320 K.
The mixture boils at a fixed 312 K and behaves like a pure liquid.
312 K is lower than both components.
Asked
What kind of mixture this is.
Concept to use
An azeotrope is a mixture of fixed composition that boils at a constant temperature and distils unchanged. Which kind it is depends on where its boiling point sits relative to the two components. Boiling below both means the molecules escape more easily than Raoult's law predicts — positive deviation, giving a minimum boiling azeotrope.
DiagramAnimatedCompare the mixture's boiling point with both components.
Formula to usepositive deviation → MINIMUM boiling · negative deviation → MAXIMUM boiling
Baby steps
The mixture boils at a constant temperature and behaves like a pure liquid, so it is an azeotrope — option (a) goes.
Compare the boiling points: 312 K is below both 320 K and 329 K.
Boiling below both components means the mixture is more volatile than either — A–B attractions are weaker than A–A and B–B, which is positive deviation.
Positive deviation gives a minimum boiling azeotrope.
Acetone + CS₂ is in fact NCERT's standard example of positive deviation, alongside ethanol + acetone.
Answer
Minimum boiling azeotrope (see the note above — the paper's key says maximum)
Why this option and not the others
Option
Verdict
Reason
Not an azeotrope
rule out
A mixture that boils at a constant temperature and behaves like a pure liquid is exactly what an azeotrope is.
Maximum boiling
rule out
A maximum boiling azeotrope boils above both components — nitric acid + water boils at 393.5 K, higher than either. Here 312 K is below both.
Minimum boiling
keep
Boils below both components, from positive deviation. This is the chemically correct answer, though the paper's key says otherwise.
Rectified spirit
rule out
Rectified spirit is a specific product — 95% ethanol with water — not a category of azeotrope.
Shortcut
Read the three boiling points and nothing else. Mixture boils lower than both → minimum. Higher than both → maximum. You never need to reason about deviations or intermolecular forces if the numbers are given — the comparison alone decides it.
The two examples to hold: ethanol + water and acetone + CS₂ are minimum boiling; nitric acid + water and HCl + water are maximum boiling.
Where it went wrong
Nothing went wrong in the reasoning — the option selected is the chemically correct one, and it is worth being clear about that so the right method is not abandoned. If a similar question appears again, apply the same test: compare the mixture's boiling point with both components and choose on that basis. If a paper's key disagrees, note it and move on rather than losing time.
Q 159Left blankOsmotic pressure of equal masses
The relationship between osmotic pressure at 273 K when 10 g glucose (P₁), 10 g urea (P₂), and 10 g sucrose (P₃) is dissolved in 250 ml of water is
P₁ > P₂ > P₃
P₃ > P₁ > P₂
CorrectP₂ > P₁ > P₃
P₂ > P₃ > P₁
Given
10 g each of glucose (M = 180), urea (M = 60) and sucrose (M = 342).
All in 250 mL of water, all at 273 K.
All three are non-electrolytes, so i = 1 for each.
Asked
The order of their osmotic pressures.
Concept to use
Osmotic pressure is a colligative property, so it counts particles. With equal masses and equal volumes, the number of moles goes as 1/molar mass — so the substance with the smallest molar mass gives the most particles and the highest pressure.
DiagramAnimatedEqual masses, so the lightest molar mass gives the most particles.
Formula to useπ = (n/V)RT with n = mass/M → π ∝ 1/M for equal masses
Baby steps
Glucose: 10/180 = 0.056 mol.
Urea: 10/60 = 0.167 mol — much the largest.
Sucrose: 10/342 = 0.029 mol — the smallest.
Order of moles, and therefore of osmotic pressure: urea > glucose > sucrose.
In the question's labels: P₂ > P₁ > P₃.
Answer
P₂ > P₁ > P₃
Why this option and not the others
Option
Verdict
Reason
P₁ > P₂ > P₃
rule out
Puts glucose above urea, but urea's molar mass is three times smaller so it gives three times the particles.
P₃ > P₁ > P₂
rule out
Exactly reversed — it ranks by molar mass instead of by moles.
Urea first is right, but sucrose and glucose are swapped.
Shortcut
With equal masses in equal volumes, just rank the molar masses and reverse the order. Smallest M gives the biggest π. Urea at 60 is the lightest, sucrose at 342 the heaviest, so urea wins and sucrose loses — no calculation at all. The temperature and volume are common to all three and cancel out of any comparison.
Q 161Attempted · wrongIsotonic solutions · the van't Hoff factor
Consider separate solutions of 0.500 M C₂H₅OH(aq), 0.100 M Mg₃(PO₄)₂(aq), 0.250 M KBr(aq) and 0.125 M Na₃PO₄(aq) at 25°C. Which statement is true about these solutions, assuming all salts are strong electrolytes?
Her answer0.500 M C₂H₅OH(aq) has the highest osmotic pressure
CorrectThey all have the same osmotic pressure
0.100 M Mg₃(PO₄)₂(aq) has the highest osmotic pressure
0.125 M Na₃PO₄(aq) has the highest osmotic pressure
Given
0.500 M ethanol — a non-electrolyte, i = 1.
0.100 M Mg₃(PO₄)₂ — gives 3 Mg²⁺ + 2 PO₄³⁻, so i = 5.
0.250 M KBr — gives K⁺ + Br⁻, so i = 2.
0.125 M Na₃PO₄ — gives 3 Na⁺ + PO₄³⁻, so i = 4.
Asked
Which solution has the highest osmotic pressure.
Concept to use
Osmotic pressure depends on the total concentration of particles, not on the concentration of formula units. So multiply each molarity by its van't Hoff factor i — the number of ions each formula unit produces. The molarities in this question have been chosen so that every product comes out the same.
DiagramAnimatedMolarity times i — and all four land on 0.500.
Formula to useπ = i M R T → compare i × M for each
Baby steps
Ethanol: i = 1, so effective concentration = 0.500 × 1 = 0.500.
Mg₃(PO₄)₂: 3 + 2 = 5 ions, so 0.100 × 5 = 0.500.
KBr: 2 ions, so 0.250 × 2 = 0.500.
Na₃PO₄: 3 + 1 = 4 ions, so 0.125 × 4 = 0.500.
All four give 0.500 M of particles, so all four are isotonic.
Answer
They all have the same osmotic pressure
Why this option and not the others
Option
Verdict
Reason
ethanol highest
rule out
Its molarity is highest, but it produces no ions at all (i = 1). Molarity alone is not the measure.
all the same
keep
Every molarity × i product equals 0.500.
Mg₃(PO₄)₂ highest
rule out
It has the largest i at 5, but the smallest molarity at 0.100. The two cancel exactly.
Na₃PO₄ highest
rule out
Same objection — i = 4 against a molarity of 0.125, giving 0.500 again.
Shortcut
Count the ions in each formula unit and multiply. The moment two or three products come out equal, check them all — the answer is almost certainly “they are the same”. Papers build these deliberately, choosing molarities that are exact reciprocals of the i values. Here 0.5×1, 0.1×5, 0.25×2 and 0.125×4 all give 0.5.
Where it went wrong
The ethanol was chosen because its molarity, 0.500, is the largest number in the list. But molarity counts formula units, and osmotic pressure counts particles — and ethanol is the only solute here that does not break up at all. Writing the i value beside each formula before comparing anything would have shown all four landing on 0.500. Note the link to Q162 on this same paper: both turn on the van't Hoff factor, one for dissociation and one for association.
Q 162Attempted · wrongDegree of association from molar masses
Normal molar mass and observed molar mass of a solute dissolved in a solvent are 60 g and 80 g respectively. True statement about degree of association of the solute in the solution is
Correct50% dimerized
Her answer75% dimerized
60% dimerized
40% dimerized
Given
Normal (formula) molar mass = 60 g/mol.
Observed molar mass = 80 g/mol — larger, so the solute is associating.
Association is into dimers, so n = 2.
Asked
The degree of association α.
Concept to use
Two separate quantities are in play and the question is built on confusing them. i is the ratio of molar masses; α is the fraction of solute that has associated. They are related but not equal. For dimerisation, i = 1 − α/2, so α = 2(1 − i).
DiagramAnimatedi below 1 means association; α is a further step from there.
Formula to usei = M_normal / M_observed ; dimerisation: i = 1 − α/2
Baby steps
i = 60/80 = 0.75. Note i < 1, which confirms association rather than dissociation.
For dimerisation 2A → A₂: i = 1 − α/2.
0.75 = 1 − α/2, so α/2 = 0.25.
α = 0.50, i.e. 50% dimerised.
Answer
50% dimerized
Why this option and not the others
Option
Verdict
Reason
50% dimerized
keep
α = 2(1 − 0.75) = 0.5.
75% dimerized
rule out
This is i, not α. The van't Hoff factor happens to be 0.75, and reading it as a percentage of association is exactly the trap.
60% dimerized
rule out
This is the normal molar mass, 60, read as a percentage.
40% dimerized
rule out
No route to this from i = 0.75.
Shortcut
For dimerisation the relation is beautifully simple: α = 2(1 − i). Here 2(1 − 0.75) = 0.5 in one step.
And a direction check that costs nothing: i < 1 always means association (particles joining, so fewer of them), while i > 1 means dissociation. Observed molar mass larger than normal is the same signal.
Where it went wrong
0.75 was computed correctly — and then reported as the answer. But 0.75 is i, the van't Hoff factor, while the question asks for the degree of association. The two differ by the conversion α = 2(1 − i), which turns 0.75 into 0.50. The habit worth building is to write down what the symbol you have just calculated actually means before matching it to an option — the same read-the-noun failure that appeared in Q177 of the ILTS-04 paper.
Q 164Left blankvan't Hoff factor for a weak acid
For a weak monobasic acid, if pKₐ = 4, then at a concentration of 0.01 M of the acid solution, the van't Hoff factor is
1.01
1.02
Correct1.10
1.20
Given
Weak monobasic acid, pKₐ = 4, so Kₐ = 10⁻⁴.
Concentration C = 0.01 M = 10⁻² M.
Monobasic means it gives 2 particles when fully ionised, so n = 2.
Asked
The van't Hoff factor i.
Concept to use
A weak acid only partly ionises, so i sits just above 1. Two steps: get the degree of dissociation α from Ostwald's dilution law, then feed it into the dissociation form of the van't Hoff factor. The numbers are chosen so α comes out at a clean 0.1.
DiagramAnimatedα from Ostwald's law, then i = 1 + α for a monobasic acid.
Formula to useα = √(Kₐ/C) ; i = 1 + α(n − 1), n = 2
Baby steps
Kₐ = 10−pKₐ = 10⁻⁴.
α = √(Kₐ/C) = √(10⁻⁴ / 10⁻²) = √(10⁻²) = 0.1.
For HA → H⁺ + A⁻, n = 2, so i = 1 + α(2 − 1) = 1 + α.
i = 1 + 0.1 = 1.10.
Answer
i = 1.10
Why this option and not the others
Option
Verdict
Reason
1.01
rule out
Would need α = 0.01, i.e. Kₐ/C = 10⁻⁴.
1.02
rule out
Would need α = 0.02.
1.10
keep
α = 0.1 from Ostwald's law, and i = 1 + α for a monobasic acid.
1.20
rule out
Would need α = 0.2, or i = 1 + 2α which applies to a dibasic acid.
Shortcut
For any monobasic weak acid the relation collapses to i = 1 + α, so the whole question is just finding α. And √(Kₐ/C) is easiest in powers of ten: 10⁻⁴/10⁻² = 10⁻², whose square root is 10⁻¹ = 0.1. Halving the exponent is the entire calculation.
Q 165Left blankMolarity of a pure liquid
Molarity of the liquid HCl if density of the solution is 1.17 g/cc is
36.5
18.25
Correct32.05
42.10
Given
Pure liquid HCl, density 1.17 g/cm³.
Molar mass of HCl = 36.5 g/mol.
Asked
The molarity.
Concept to use
For a pure liquid, one litre of the liquid is the solute — there is no solvent to worry about. So the moles per litre come straight from the mass of a litre divided by the molar mass. The word “liquid” rather than “aqueous” is what makes this a one-step question.
DiagramAnimatedFor a pure liquid there is no solvent — M = 1000d/M.
Formula to useM = 1000 × d / M_molar (for a pure liquid)
Baby steps
Mass of 1 litre = 1000 cm³ × 1.17 g/cm³ = 1170 g.
Moles in that litre = 1170 / 36.5 = 32.05 mol.
Since this is one litre of solution, molarity = 32.05 M.
Answer
32.05 M
Why this option and not the others
Option
Verdict
Reason
36.5
rule out
The molar mass, quoted back as a molarity.
18.25
rule out
Half the molar mass — no step in the calculation produces it.
32.05
keep
1170 g per litre divided by 36.5 g per mole.
42.10
rule out
Would need a density of about 1.54 g/cm³.
Shortcut
Memorise the one-liner for pure liquids: M = 1000d / M_molar. No mass percentage is needed because the liquid is 100% solute. Compare with the aqueous version, M = 1000 × d × w% / (100 × M_molar), where the extra factor is the mass per cent — and here that factor is simply 1.
Q 171Left blankSolubility in a mixed solvent
The solubility of substance 'X' in pure ethanol is 0.1 gm/lit and in water is 0.01 gm/lit. To dissolve 11 gm of dry 'X' we are adding 20 ml of fresh 50% (V/V) ethanol solution in each time on 'X'. How many times are we to add this ethanol solution to dissolve 'X'?
100
10⁶
10³
Correct10⁴
Given
Solubility in ethanol = 0.1 g/L; in water = 0.01 g/L.
Each addition is 20 mL of 50% (V/V) ethanol = 10 mL ethanol + 10 mL water.
Total mass of X to dissolve = 11 g.
Asked
The number of additions required.
Concept to use
Each solvent dissolves its own share, and the two shares add. So work out how much X one 20 mL portion can carry away, then divide the total mass by that amount. The only trap is forgetting that the water contributes too, or forgetting that 20 mL of a 50% mixture contains only 10 mL of each.
DiagramAnimatedBoth solvents dissolve some X, so add the two contributions.
Formula to useper addition: (0.1 × V_ethanol) + (0.01 × V_water), volumes in litres
Baby steps
Each 20 mL portion is 10 mL ethanol + 10 mL water = 0.010 L of each.
Dissolved by the ethanol: 0.1 × 0.010 = 0.0010 g.
Dissolved by the water: 0.01 × 0.010 = 0.0001 g.
Total per addition = 0.0010 + 0.0001 = 0.0011 g.
Number of additions = 11 / 0.0011 = 10⁴.
Answer
10⁴ additions
Why this option and not the others
Option
Verdict
Reason
100
rule out
Off by a hundredfold; would need each portion to dissolve 0.11 g.
10⁶
rule out
A hundred times too many; would need each portion to carry only 1.1 × 10⁻⁵ g.
10³
rule out
Ten times too few — this would leave most of the X undissolved.
10⁴
keep
11 g divided by 0.0011 g per addition.
Shortcut
The 11 in the question is a gift: 11 ÷ 0.0011 = 10⁴ exactly, because 0.0011 is 11 × 10⁻⁴. If your per-addition figure does not divide 11 cleanly, you have probably left out the water's contribution — ethanol alone would give 0.0010 g and an untidy 1.1 × 10⁴.
Q 173Attempted · wrongHenry's constant and temperature
At 293 K Henry's constant for N₂ gas is greater than Henry's constant for O₂. At 303 K Henry's constant for N₂ gas is
Correctgreater than that of Kₐ of O₂ at 303 K
Lesser than that of Kₐ of O₂ at 303 K
Equal to that of Kₐ of O₂ at 303 K
Her answerEqual to that of Kₐ of O₂ at 293 K
Given
At 293 K: Kₐ(N₂) > Kₐ(O₂).
The comparison is now to be made at 303 K, a higher temperature.
Asked
How Kₐ(N₂) at 303 K compares.
Concept to use
Two facts settle this. First, Kₐ rises with temperature for every gas — gases get less soluble when you warm the liquid, which is why warm water holds less dissolved oxygen. Second, the ordering between two gases does not flip: N₂ is less soluble than O₂ at 293 K and remains so at 303 K.
DiagramAnimatedKₐ rises with temperature, and the curves do not cross.
Formula to useKₐ ↑ as T ↑ (solubility falls) ; the N₂/O₂ ordering is preserved
Baby steps
Warming raises Kₐ for both gases, so Kₐ(N₂) at 303 K is greater than Kₐ(N₂) at 293 K.
The relative ordering is unchanged: N₂ still has the larger constant.
So Kₐ(N₂) at 303 K > Kₐ(O₂) at 303 K.
It certainly cannot equal any value at 293 K, since both curves have risen.
Answer
Greater than that of Kₐ of O₂ at 303 K
Why this option and not the others
Option
Verdict
Reason
greater than O₂ at 303 K
keep
Both constants rise with temperature and the N₂ > O₂ ordering is preserved.
lesser than O₂ at 303 K
rule out
Would require the two curves to cross, which warming does not cause.
equal to O₂ at 303 K
rule out
Also requires a crossing.
equal to O₂ at 293 K
rule out
Compares across two different temperatures. Raising T raises Kₐ, so the 303 K value must exceed anything measured at 293 K.
Shortcut
Two things to keep straight, and the question is trivial. Higher Kₐ means LOWER solubility. And Kₐ always increases with temperature. Then check that the option compares like with like — the chosen distractor compares a 303 K value with a 293 K one, which is already suspicious before any chemistry.
Where it went wrong
The option chosen compares Kₐ(N₂) at 303 K with Kₐ(O₂) at 293 K — two different temperatures. Even without knowing the trend, an option that changes two things at once should draw suspicion. And with the trend: warming raises Kₐ, so the 303 K value cannot equal a 293 K value; it must be larger. Reading the temperature attached to each quantity in the option is the guard here.
Q 174Attempted · wrongWhat makes a solution a solution
Statement-I: In solution different parts shown different composition. Statement-II: Solution is homogeneous mixture.
Her answerIf both statements, I and II, are correct.
If both statements, I and statement II, are incorrect.
If statement I correct, but statement II is incorrect.
CorrectIf statement I incorrect, but statement II is correct.
Given
Statement I: different parts of a solution show different composition.
Statement II: a solution is a homogeneous mixture.
Asked
The truth of each statement.
Concept to use
Homogeneous means uniform throughout — take a sample from anywhere in a solution and it has the same composition as any other sample. That is the defining property. Statement II states it correctly; Statement I asserts the exact opposite, and so describes a suspension or a heterogeneous mixture instead.
DiagramAnimatedA solution is uniform; a suspension is not.
Formula to usesolution = HOMOGENEOUS = same composition everywhere
Baby steps
Statement II. This is the textbook definition of a solution. Correct.
Statement I. If different parts had different composition, the mixture would be heterogeneous — a suspension, not a solution. Incorrect.
The two statements directly contradict each other, so they cannot both be true.
So Statement I is incorrect and Statement II is correct.
Answer
Statement I is incorrect, but statement II is correct
Why this option and not the others
Option
Verdict
Reason
Both correct
rule out
The two statements contradict each other. If a solution is homogeneous, its parts cannot differ in composition.
Both incorrect
rule out
Statement II is the standard definition and is correct.
I correct, II incorrect
rule out
The reverse of the truth.
I incorrect, II correct
keep
Homogeneity is the defining property; Statement I denies it.
Shortcut
Read the two statements against each other before judging either against chemistry. Here they are direct opposites, so “both correct” and “both incorrect” are impossible from the start — that halves the option list in about three seconds. The same structural check decides Q105 and Q111 on the physics half of this paper.
Where it went wrong
Both statements were accepted, but they cannot both be true — one says the composition varies from place to place and the other says the mixture is homogeneous. Statement I is probably being read as describing solute and solvent being different substances, which is true but is not what it says. It says different parts of the solution differ, and that is false by definition.
Q 175Left blankMole fraction from a mass percentage
The mole fraction of glucose in 10% (w/w) aqueous solution is approximately
0.18
Correct0.011
0.1
0.017
Given
10% (w/w) glucose means 10 g glucose in 100 g of solution.
So the water is 100 − 10 = 90 g.
M(glucose) = 180, M(water) = 18.
Asked
The mole fraction of glucose.
Concept to use
The one thing to get right is what w/w means: 10 g of solute in 100 g of solution, not in 100 g of solvent. After that it is moles over total moles — and note how heavily the water dominates, because its molar mass is ten times smaller.
DiagramAnimatedw/w means grams per 100 g of SOLUTION, not of solvent.
Formula to usex_glucose = n_glucose / (n_glucose + n_water)
Baby steps
Take 100 g of solution: 10 g glucose and 90 g water.
n_glucose = 10/180 = 0.0556 mol.
n_water = 90/18 = 5.0 mol.
x = 0.0556 / (0.0556 + 5.0) = 0.0556 / 5.0556.
x = 0.011.
Answer
x ≈ 0.011
Why this option and not the others
Option
Verdict
Reason
0.18
rule out
Would need the two mole counts to be far closer. This looks like a misplaced 180.
0.011
keep
0.0556 mol of glucose against 5.0 mol of water.
0.1
rule out
This is the mass fraction, 10%, mistaken for the mole fraction. They are very different because the molar masses differ by a factor of ten.
0.017
rule out
Would follow from taking 90 g of solution and 10 g of water, or from an arithmetic slip in the division.
Shortcut
Estimate before calculating. Water's molar mass is ten times smaller than glucose's, so even at only nine times the mass it supplies about ninety times the moles. The mole fraction of glucose must therefore be around 1%, not 10% — which picks 0.011 immediately and exposes 0.1 as the mass fraction in disguise.
What the thirteen have in common
Reading the paper as a whole
Q156 — the answer given was right
Acetone boils at 329 K and CS₂ at 320 K; the mixture boils at 312 K, which is
below both. A mixture boiling below both of its components is a minimum boiling azeotrope,
arising from positive deviation — and acetone + CS₂ is NCERT's own example of
positive deviation.
The paper marked “maximum boiling” correct. That does not survive checking. The
method used was sound and should be kept: compare the mixture's boiling point with both
components, and choose on that comparison alone.
Three questions, one idea: count the particles
Q
What was compared
What should be compared
161
Molarity — 0.500 M ethanol looked biggest
Molarity × i. All four give 0.500, so all are isotonic.
162
i = 0.75, reported as the answer
α = 2(1 − i) = 0.50. i and α are different quantities.
159
(left blank)
Equal masses → moles go as 1/M → urea wins.
Every colligative property counts particles in solution, never grams and never formula
units. Writing the i value beside each formula before comparing anything is the fix, and it takes
about ten seconds.
The blanks were mostly short
Q148 — 760 sits exactly halfway between 520 and 1000, so the composition is
halfway too: 50 mol per cent, with no algebra.
Q153 — the solute is 1 mole per cent, so the vapour pressure drops by about 1%:
just over 750. One option is in range.
Q165 — for a pure liquid, M = 1000d/M_molar. One division.
Q175 — water's molar mass is ten times smaller, so its mole fraction dominates;
glucose must be around 1%, not 10%.
Q164 — √(10⁻⁴/10⁻²) = 0.1, then i = 1 + α.
Five questions, 20 marks, and none needing more than a minute.
Two checks worth carrying
i < 1 means association; i > 1 means dissociation. And i is not α —
for dimerisation, α = 2(1 − i). Q162 turns entirely on that conversion.
Read the temperature attached to each quantity in an option. The distractor chosen in
Q173 compared a value at 303 K with one at 293 K. An option that changes two things at once
should draw suspicion before any chemistry is applied.