$4.9\ \text{g}$ of sulphuric acid ($\text{H}_2\text{SO}_4$, molar mass $98.09\ \text{g mol}^{-1}$, $n\text{-factor} = 2$) is dissolved in water to make $500\ \text{mL}$ of solution. Calculate the normality of the solution.

4.9g of sulphuric acid (H₂SO₄, molar mass 98.09g mol⁻¹, n-factor = 2) is dissolved in water to make 500mL of s

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4.9\ \text{g}

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Concept understanding — Concentration Terms

The concentration of a solution says how much solute is dissolved in how much solvent or solution. Several measures exist because different situations demand different reference quantities; this concept defines each one and moves between them.

1 — Molarity (M). Molarity = moles of solute / litres of solution (mol/L). It is the workhorse of volumetric chemistry. From a mass: M = (mass / molar mass) / volume in L. Because it uses the volume of solution, molarity changes with temperature (the solution expands or contracts).

2 — Molality (m). Molality = moles of solute / kilograms of solvent (mol/kg). Note the denominator is the mass of solvent, not solution. Because it uses mass, molality is independent of temperature — the reason colligative-property work prefers it.

3 — Mole fraction (x). For a two-component solution, x_solute = n_solute / (n_solute + n_solvent) and x_solvent = n_solvent / (n_solute + n_solvent); the two sum to 1. It is a pure ratio of moles, dimensionless, and temperature-independent.

4 — Mass percent, ppm. Mass percent (w/w) = (mass of solute / mass of solution) × 100. For very dilute solutions use parts per million: ppm = (mass of solute / mass of solution) × 10⁶. The denominator is the mass of the whole solution (solute + solvent).

5 — Normality and equivalents. Normality = gram-equivalents of solute / litres of solution, and Normality = n-factor × Molarity, where the n-factor is the acidity/basicity or electrons exchanged (H₂SO₄ has n-factor 2, so N = 2M; NaOH has 1, so N = M; for KMnO₄ in acid it is 5). Equivalents let you use the neutralisation/redox rule N₁V₁ = N₂V₂.

6 — Interconversion (the density link). The bridge between a volume-based measure (molarity) and a mass-based one (molality, mass %, mole fraction) is the density of the solution:

mass % → molarity: M = (10 × mass% × density) / molar mass (density in g/mL);
molarity → molality: take 1 L of solution, find its mass from density, subtract the solute mass to get the solvent mass, then m = moles of solute / kg of solvent;
molarity ↔ mole fraction: convert to moles of solute and moles of solvent (via density and the solvent's molar mass) and take the ratio. Whenever a problem mixes a volume-based and a mass-based term, density is the missing link you must use. …

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Moles of $\text{H}_2\text{SO}_4 = \dfrac{4.9}{98.09} = 0.04997 \approx 0.0500\ \text{mol}$

Since $\text{H}_2\text{SO}_4$ has 2 replaceable $\text{H}^+$ ions, its $n$ -factor is 2, so:

Gram equivalents $= n \times n\text{-factor} = 0.0500 \times 2 = 0.100\ \text{eq}$

Volume of solution $= 500\ \text{mL} = 0.500\ \text{L}$

$N = \frac{\text{gram equivalents}}{\text{volume in L}} = \frac{0.100}{0.500} = 0.200\ \text{N}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The expression of concentration of solution which is temperature dependent is$
(a) molarity
(b) molality
(c) mole fraction
(d) None of the above
›Reveal solutionSolution

Only molarity is defined using the volume of the solution, and volume is the one quantity in these definitions that genuinely changes with temperature.

Molarity, $M=\dfrac{n_{solute}}{V_{solution}(\text{L})}$ : as temperature rises, the solution expands (volume $V$ increases), so for the same fixed amount of solute the molarity numerically decreases — molarity is temperature-dependent.

Molality, $m=\dfrac{n_{solute}}{\text{mass of solvent (kg)}}$ : mass does not change with temperature, so molality of a given solution stays fixed regardless of temperature.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The molarity (mol.L-1) of pure water at 25 degC is$
(a) 5.555
(b) 0.55
(c) 55.55
(d) 55
›Reveal solutionSolution

Molarity of pure water = (mass of 1 L of water) / (molar mass of water), using its density of ~1 g/mL.

Calculation: Take 1 litre (1000 mL) of pure water. Since the density of water is ~1 g/mL, this has a mass of 1000 g.

Moles of water $=\dfrac{1000\text{ g}}{18\text{ g/mol}}=55.55\text{ mol}$

Since this is the number of moles present in 1 litre, the molarity is: …

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CBSE 2020Set NC1 mark
$Q. Name two concentration terms which have no effect on temperature change.$
›Reveal solutionSolution
Concentration terms defined using masses (or moles) rather than the solution's volume are temperature-independent, because mass does not change when a liquid expands or contracts on heating/cooling.

Concentration terms like molarity and normality are defined per unit volume of solution. Since the volume of a liquid changes with temperature (thermal expansion/contraction), these values change slightly with temperature even though the actual amount of solute present is unchanged.

By contrast:

Molality ( $m$ ) = moles of solute per kilogram of solvent (a mass-based ratio).

Mole fraction ( $x$ ) = moles of one component divided by total moles of all components (a ratio of amounts of substance). …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. Increasing the temperature of an aqueous solution will cause$
(a) decrease in molality
(b) decrease in molarity
(c) decrease in mole fraction
(d) decrease in mass percent
›Reveal solutionSolution
Molarity is volume-dependent (moles per litre of solution), so it falls as the solution expands on heating; mass-based measures (molality, mole fraction, mass %) are temperature-independent.

On heating an aqueous solution, the volume of the solution increases (thermal expansion) while the number of moles of solute and the total mass remain unchanged.

Molarity $= \dfrac{\text{moles of solute}}{\text{volume of solution (L)}}$ — since volume increases, molarity decreases. …
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CBSE 2017Set ANNUAL1 markMCQ
$Q. Which is the correct value of concentration of copper in drinking water?$
(a) 0.3 PPm
(b) 0.2 PPm
(c) 3.0 PPm
(d) 2 PPm
›Reveal solutionSolution

The WHO guideline (safe) limit for copper in drinking water is 2 ppm (2 mg/L).

Background

Concentration in parts per million (ppm) expresses the mass of solute per $10^6$ parts of solution — e.g. 1 ppm = 1 mg of solute per litre of water (since 1 L of dilute aqueous solution ≈ 1000 g). This unit is commonly used for trace pollutant/metal-ion levels in drinking water because the concentrations involved are extremely small.

The specific limit …

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