MPSC 2023 Prelims Paper-I · Question 62 of 100

Correct — option (3). The question is a dilution problem with one extra step hidden at the front, and that hidden step is the whole of the difficulty: the starting solution is described in molarity and the target in normality, so the two must be brought onto the same scale before any dilution formula can be applied. The bridge between them is the n-factor, the number of replaceable hydrogen ions the acid supplies per molecule, through the relation Normality = Molarity × n-factor. Sulphuric acid, H₂SO₄, is dibasic — each molecule can furnish two hydrogen ions — so its n-factor as an acid is 2, and the given 1.5 M solution is therefore 1.5 × 2 = 3.0 N. That single conversion is where most of the marks on this item are won or lost. With both solutions now expressed in normality, apply the dilution law N₁V₁ = N₂V₂, which simply states that adding water changes the volume but not the number of gram-equivalents of acid present. Substituting the known quantities gives 3.0 N × 10 dm³ = 1.0 N × V₂, so V₂ = 30 dm³. The final solution must therefore occupy thirty cubic decimetres. The last step is the one candidates most often fumble even after computing V₂ correctly: the question does not ask for the final volume, it asks what must be added, and since the solution already occupies ten cubic decimetres the volume of water to be added is 30 – 10 = 20 dm³. That is what option (3) states. It is worth confirming the answer from the other direction as a check on the arithmetic: 10 dm³ of 3.0 N acid contains 30 gram-equivalents of H₂SO₄; those same 30 gram-equivalents spread through 30 dm³ give exactly 1.0 gram-equivalent per cubic decimetre, which is 1.0 N. Note also that dm³ is simply another name for the litre, so nothing in this problem requires a unit conversion, and that the calculation assumes volumes are additive on dilution, the standard approximation at this level. Option (3) is the answer.

• (1)By adding 5 dm³ of water to it — This is the answer produced by the single commonest error on this question — skipping the conversion from molarity to normality and treating the 1.5 M solution as though it were 1.5 N. On that mistaken footing the dilution law gives 1.5 × 10 = 1.0 × V₂, so V₂ = 15 dm³ and the water to be added is 15 – 10 = 5 dm³, exactly this option. The option is placed first for that reason: it is what a candidate arrives at by doing the dilution correctly while forgetting that sulphuric acid is dibasic. Checking the result exposes the error at once, because 10 dm³ of 3.0 N acid diluted to 15 dm³ gives 30 ÷ 15 = 2.0 N, twice the strength the question asks for. Whenever a problem mixes the two ways of stating concentration, the n-factor has to be applied before anything else is done.
• (2)By adding 294 g of H₂SO₄ to it — This option moves in the wrong direction entirely. The starting solution is 3.0 N and the target is 1.0 N, so the solution has to be made weaker, and the only way to weaken it is to add solvent; adding more sulphuric acid to it can only raise the concentration further. The figure is not arbitrary — 294 g of H₂SO₄ is exactly three moles, since the molar mass is 2 + 32 + 64 = 98 g per mole — and that tidiness is what makes the option tempting to a candidate who has calculated something involving three and is looking for a place to put it. But no addition of solute, of any mass whatever, can bring a 3.0 N solution down to 1.0 N. Reading the two concentrations and asking in which direction the change must go eliminates this option and option (4) together before any arithmetic is attempted.
• (4)By adding 174 g of H₂SO₄ to it — Like option (2), this proposes to dilute a solution by adding more of the very substance dissolved in it, which is impossible: putting sulphuric acid into sulphuric acid raises the normality, and the question requires it to fall from 3.0 N to 1.0 N. The mass offered here does not even correspond to a whole number of moles — 174 g against a molar mass of 98 g per mole is about 1.78 moles — so it does not arise from any natural step of the calculation and functions purely as a second wrong-direction option beside option (2). The lesson worth taking from the pair is procedural: in any concentration problem, settle the direction of change from the two stated strengths first, which here disposes of half the option set immediately and leaves only the two volumes of water to choose between.

Molarity and normality are two ways of measuring the same thing — how much solute sits in a given volume of solution — but they count the solute differently, and the whole of this question turns on the difference. Molarity counts moles of solute per litre of solution. Normality counts gram-equivalents of solute per litre, where a gram-equivalent is the mass that supplies one mole of the reactive unit: one mole of H⁺ for an acid, one mole of OH⁻ for a base, one mole of electrons for a redox reagent. The two are linked by the n-factor, the number of such reactive units each formula unit provides, through Normality = Molarity × n-factor. For a monobasic acid such as HCl the n-factor is one and the two numbers coincide, which is why candidates who learn on hydrochloric acid are caught out by sulphuric acid, where the n-factor as an acid is two and the normality is double the molarity. The n-factor is a property of the reaction and not merely of the formula: sulphuric acid neutralised completely counts as two, and the same distinction gives three for phosphoric acid in full neutralisation and two for calcium hydroxide as a base. Dilution is the other half of the topic. Adding water to a solution changes its volume while leaving the quantity of solute untouched, so the product of concentration and volume is conserved: M₁V₁ = M₂V₂ if both sides are in molarity, N₁V₁ = N₂V₂ if both are in normality. The formula is valid only when both sides use the same measure, which is exactly why the conversion has to come first. Normality has fallen out of use in international practice, where IUPAC prefers molar concentration, but it survives in Indian syllabi and in volumetric analysis, and it remains a standard examination topic for that reason.

This is a two-step numerical dressed as a one-step one, and MPSC uses the shape deliberately. Every quantity needed is printed in the stem, the arithmetic involves nothing harder than dividing thirty by one, and the item is still failed at scale — because a candidate who reaches for N₁V₁ = N₂V₂ without noticing that the given concentration is a molarity produces a clean, confident and wrong answer that is waiting for them among the options. The Commission's science questions frequently carry exactly one such conceptual gate, and the efficient way to read a numerical here is to look first at the units of what is given against the units of what is asked, and only then to choose a formula. The second habit this question rewards is finishing the question that was asked. Having computed a final volume of 30 dm³, a candidate under time pressure may look for thirty among the options; the item asks what must be added, and the subtraction of the original ten cubic decimetres is a real step that carries a real mark. The printed English of this stem is defective — it reads 'How is 1.0N H₂SO₄ solution is prepared', with the verb duplicated — and that is reproduced here as printed, since the paper's wording is not the card's to correct; the meaning is not in doubt and a candidate should not lose time over it.

• Normality equals molarity multiplied by the n-factor, the number of reactive units supplied per formula unit; for sulphuric acid acting as an acid the n-factor is 2, so a 1.5 M H₂SO₄ solution is 3.0 N.
• The dilution law N₁V₁ = N₂V₂ holds because adding solvent changes the volume of a solution while leaving the number of gram-equivalents of solute unchanged; both sides must be expressed in the same measure of concentration.
• Diluting 10 dm³ of 3.0 N H₂SO₄ to 1.0 N requires a final volume of 30 dm³, so the volume of water to be added is 20 dm³, the difference between the final and the original volume.
• One cubic decimetre is exactly one litre, so concentrations quoted per litre may be used directly with volumes quoted in dm³ without conversion.
• The molar mass of H₂SO₄ is 98 g per mole, being 2 for the two hydrogen atoms, 32 for sulphur and 64 for the four oxygen atoms; its equivalent mass as an acid is therefore 49 g.

• Treating a molarity as though it were a normality, which for a dibasic acid such as H₂SO₄ misstates the concentration by a factor of two and leads straight to a wrong option that has been planted for it
• Reporting the final volume when the question asks for the volume of water to be added, and so omitting the subtraction of the original volume
• Attempting to dilute a solution by adding more solute, when the direction of the required change alone rules such options out
• Using n-factor 2 for sulphuric acid in every context, when the n-factor is fixed by the reaction the acid is undergoing and not by its formula
• Forgetting that molarity and normality are defined per litre of solution and not per litre of solvent, so the added water changes the total volume of the solution

Solution-concentration numericals are a standing fixture of MPSC's science section, and they come in a narrow set of shapes: convert between molarity and normality, dilute a solution to a stated strength, compute the mass of solute needed for a given volume and concentration, or find the strength of a solution after two are mixed. The Commission's preferred difficulty is not heavier arithmetic but an extra conversion buried in the statement of the problem, most often a change of measure between what is given and what is asked, and dibasic sulphuric acid is the reagent it uses most for that purpose. A candidate who prepares the n-factors of the common acids, bases and oxidising agents on a single page, and who memorises the two conservation formulae M₁V₁ = M₂V₂ and N₁V₁ = N₂V₂ together with the condition that both sides carry the same measure, can answer the whole family within a minute each.

No directly related past PYQ was found.

• practice — not a real PYQ

  What is the normality of a 0.5 M solution of H₂SO₄ when it is completely neutralised by a base ?

  • (a)0.25 N
  • (b)0.5 N
  • (c)1.0 N
  • (d)2.0 N

  Answer(c) 1.0 N — sulphuric acid is dibasic and supplies two hydrogen ions per molecule, so its n-factor in complete neutralisation is 2, and Normality = Molarity × n-factor gives 0.5 × 2 = 1.0 N. The answer of 0.5 N comes from assuming molarity and normality are the same, which holds only for monobasic acids such as HCl, and 0.25 N comes from dividing by the n-factor instead of multiplying by it.

• practice — not a real PYQ

  To what final volume must 250 mL of a 2.0 N solution be diluted in order to obtain a 0.5 N solution ?

  • (a)500 mL
  • (b)750 mL
  • (c)1000 mL
  • (d)1250 mL

  Answer(c) 1000 mL — by the dilution law N₁V₁ = N₂V₂, the number of gram-equivalents is unchanged by adding water, so 2.0 × 250 = 0.5 × V₂ and V₂ = 1000 mL. Note that the volume of water actually added is 750 mL, the difference between the final and the original volume, which is the figure a question phrased as 'how much water must be added' would be asking for.