ANAND CLASSES Study Material and Notes to learn Normality formula in chemistry, equivalent mass calculation example, relation between normality and molarity, how to calculate normality from molar mass, normality vs molarity in titration for class 11 Chemistry of chapter Some Basic Concepts of Chemistry.
Q1: What is Normality (N) in chemistry?
A: Normality (N) is a measure of concentration that expresses the number of equivalents of a solute dissolved in one liter (or dm³) of a solution. It is especially useful in acid-base reactions, redox reactions, and titrations.
Q2: What is the formula for Normality?
A: $$\text{Normality (N)} = \frac{\text{Number of equivalents of solute}}{\text{Volume of solution in liters}}$$
This means that normality depends on both the quantity of solute (in equivalents) and the total volume of the solution.
Q3: What is Equivalent Mass, and how is it related to Molar Mass?
A: Equivalent mass is the mass of a substance that reacts with or replaces one mole of H⁺ (for acids), one mole of OH⁻ (for bases), or one mole of electrons (for redox reactions).
It is calculated using the formula: $$\text{Equivalent Mass} (E) = \frac{\text{Molar Mass} (M)}{\text{n-factor}}$$
where:
- Molar Mass (M) is the molecular weight of the substance in grams per mole.
- n-factor (also called valency factor) depends on the type of reaction:
- Acids: Number of H+ ions donated per molecule.
- Bases: Number of OH– ions accepted per molecule.
- Salts: Total positive or negative charge contributed by one formula unit.
- Redox reactions: Number of electrons lost or gained per molecule.
Thus, equivalent mass is always a fraction of molar mass, making it highly specific to reaction type.
Q4: How is Normality related to Equivalent Mass and Molar Mass?
A: We can express Normality in terms of mass and equivalent mass: $$\text{Normality} = \frac{\text{Mass of solute} (W)}{\text{Equivalent mass} (E) \times \text{Volume of solution in liters} (V)}$$
Since $$E = \frac{M}{\text{n-factor}}$$,
We get: $$\text{Normality} = \frac{W}{\left(\frac{M}{\text{n-factor}}\right) \times V}$$
which simplifies to:
$$N = \frac{W \times \text{n-factor}}{M \times V}$$
This formula helps in directly calculating Normality when molar mass and reaction type are known.
Q5: How is Equivalent Mass Different for Acids, Bases, and Redox Compounds?
| Substance Type | Formula for Equivalent Mass (E) | Example Calculation |
|---|---|---|
| Acid | $$E = \frac{M}{\text{Basicity (n-factor)}}$$ | H2SO4 → 2H+ + SO42-$$E = \frac{98}{2} $$ $$E = 49 \:g/equiv$$ |
| Base | $$E = \frac{M}{\text{Acidity (n-factor)}}$$ | NaOH → Na+ + OH–$$E = \frac{40}{1} $$ $$E = 40 g/equiv$$ |
| Redox Compound | $$E = \frac{M}{\text{Electrons Transferred (n-factor)}}$$ | Fe2+ → Fe3+ $$E = \frac{55.8}{1} $$ $$E = 55.8 g/equiv$$ |
Q6: Real-world example of Normality calculation?
A: Let’s calculate the normality of H2SO4 solution where 4.9 g of H2SO4 is dissolved in 250 mL (0.25 L) of solution.
Step 1: Determine the equivalent mass of H2SO4
- Molar mass of H₂SO₄ = 98 g/mol
- Since H2SO4 donates 2 H+ ions per molecule, its n-factor = 2
- Equivalent mass: $$E = \frac{98}{2} = 49 \text{ g/equiv}$$ $$E = 49 g/equiv$$
Step 2: Use the Normality Formula
$$N = \frac{W}{E \times V} = \frac{4.9}{49 \times 0.25}$$
Step 3: Solve
$$N = \frac{4.9}{12.25} = 0.4 \text{ N}$$
Thus, the normality of the solution is 0.4 N.
Q7: How is Normality Different from Molarity?
| Property | Normality (N) | Molarity (M) |
|---|---|---|
| Definition | Equivalents of solute per liter | Moles of solute per liter |
| Dependence | Varies based on reaction type (n-factor) | Independent of reaction type |
| Formula | $$N = \frac{W}{E \times V}$$ | $$M = \frac{W}{M \times V}$$ |
| Use Cases | Acid-base, redox, titrations | General solution concentration |
For substances where n-factor = 1, normality and molarity are the same.
Q8: How Does Temperature Affect Normality?
A: Since volume changes with temperature due to expansion or contraction of the solvent, normality (which depends on volume) also varies with temperature.
Example: A 1 N H2SO4 solution at 25°C may have a different normality at 50°C due to volume expansion.
Q9: How Do You Prepare a Solution of Known Normality?
To prepare a solution of a given normality:
- Calculate the required mass of solute using: $$W = N \times E \times V$$
- Weigh the solute accurately.
- Dissolve in distilled water and mix well.
- Transfer to a volumetric flask and adjust to the desired volume.
- Shake well for uniform distribution.
Example: To prepare 1 N HCl solution, dissolve 36.5 g of HCl in 1 L of water.
Final Takeaway: Why is Normality Important?
- More precise than molarity for reactions where reaction equivalents matter.
- Adapts to different chemical processes, making it useful for titrations and industrial formulations.
- Crucial for stoichiometric calculations in acid-base and redox reactions.
Understanding normality with equivalent mass and molar mass helps ensure accurate solution preparation and chemical analysis! 🎯
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