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The equivalent wieght of an acid is equa...

The equivalent wieght of an acid is equal to

A

Molecular weight x acidity

B

Molecular weight x basicity

C

molecular weight/basicity

D

molecular weight /acidity

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To determine the equivalent weight of an acid, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding Equivalent Weight**: The equivalent weight of an acid is defined as the mass of the acid that can donate one mole of protons (H⁺ ions) in a reaction. 2. **Formula for Equivalent Weight**: The formula for calculating the equivalent weight of an acid is: \[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Basicity}} \] where: - Molecular Weight is the sum of the atomic weights of all atoms in the acid. - Basicity is the number of protons (H⁺ ions) that the acid can donate. 3. **Example Calculation**: Let's calculate the equivalent weight of hydrochloric acid (HCl) and sulfuric acid (H₂SO₄) as examples. - **For HCl**: - **Molecular Weight**: - Hydrogen (H) = 1.008 g/mol - Chlorine (Cl) = 35.45 g/mol - Total Molecular Weight of HCl = 1.008 + 35.45 = 36.458 g/mol - **Basicity**: HCl can donate 1 proton (H⁺), so its basicity = 1. - **Equivalent Weight**: \[ \text{Equivalent Weight of HCl} = \frac{36.458}{1} = 36.458 \text{ g/equiv} \] - **For H₂SO₄**: - **Molecular Weight**: - Hydrogen (H) = 1.008 g/mol (2 H atoms) - Sulfur (S) = 32.07 g/mol - Oxygen (O) = 16 g/mol (4 O atoms) - Total Molecular Weight of H₂SO₄ = (2 × 1.008) + 32.07 + (4 × 16) = 98.086 g/mol - **Basicity**: H₂SO₄ can donate 2 protons (H⁺), so its basicity = 2. - **Equivalent Weight**: \[ \text{Equivalent Weight of H₂SO₄} = \frac{98.086}{2} = 49.043 \text{ g/equiv} \] 4. **Conclusion**: - For monobasic acids like HCl, the equivalent weight is equal to the molecular weight. - For dibasic acids like H₂SO₄, the equivalent weight is half of the molecular weight.
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RESONANCE ENGLISH-STRUCTURAL IDENTIFICATION & PRACTICAL ORGANIC CHEMISTRY-Section -B
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  5. The desiccants used for absorbing water during Liebig's method for est...

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  6. 0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for ...

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  7. 0.28 g of nitrogenous compound was subjected to Kjeldahl's process to ...

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  8. The equivalent wieght of an acid is equal to

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  9. 0.30 gm of an organic compound gave 50 ml of nitrogen collected at 300...

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  10. Liebig test is used ot estimate

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  11. Copper wire test is called

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  12. During estimation of nitrogen present in an organic compound by Kjelda...

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  13. In Carius method of estimation of halogen, 0.15 g of an organic compou...

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  14. In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of...

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  15. An organic compound having molecular mass 60 is found to contain C = 2...

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  16. A gaseous hydrocarbon has 85% carbon and vapour density of 28. The pos...

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  17. Two element X ( at . Mass = 75) and Y(at .mass =16) combine to given ...

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