How is 1.0N H₂SO₄ solution is prepared from 10 dm³ of 1.5M H₂SO₄ solution ?

Question

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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.

Why the other options are wrong — (a) 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.  (b) 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.  (d) 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.

Concept: 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.

State PCS 2023 Q62 — Science & Technology PYQ with Answer

State PCS 2023 Q62 — Science & Technology PYQ with Answer