Problem 25 - A Complete Resource Book in Chemistry for JEE Main

Question

The number of grams of a dibasic acid (molecula weight 200 ) present is \(100 \mathrm{~mL}\) of its aqueous solutior to give decinormal strength is (a) \(1 \mathrm{~g}\) (b) \(2 \mathrm{~g}\) (c) \(3 \mathrm{~g}\) (d) \(4 \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct answer is (a) 1 gram.

Step 1: Understanding the Given Information

We have a dibasic acid with a molecular weight of 200. The solution is decinormal, which implies it has a normality of 0.1 N. Decinormal means 1/10 of normal. We need to find out how many grams of this acid are present in 100 mL of solution.

Step 2: Understanding Equivalent Weight in Context

For a dibasic acid, which can donate two protons per molecule, the equivalent weight is half of the molecular weight. Thus, the equivalent weight of our acid is \( \frac{200}{2} = 100 \).

Step 3: Calculating Gram Equivalent

Using the concept of normality, 1 equivalent of the substance corresponds to its equivalent weight in grams present in 1 liter (1000 mL) of solution. For a 0.1 N solution, 0.1 equivalent is present in 1 liter (1000 mL), or equivalently, \(0.1 \times 100 = 10\) grams is present in 1 liter of a decinormal solution.

Step 4: Adjusting Calculation to 100 mL

Since the problem asks for the amount in 100 mL, calculate how many grams this corresponds to: \( \text{grams in 100 mL} = \frac{10 \text{ grams}}{1000 \text{ mL}} \times 100 \text{ mL} = 1 \text{ gram} \).

Step 5: Conclusion

Thus, the amount of dibasic acid required is 1 gram.

Key Concepts

Dibasic Acid

Dibasic acids are a specific type of acid that can donate two protons (hydrogen ions) per molecule in an aqueous solution. This characteristic is crucial because it directly influences the acid's behavior in chemical reactions and calculations involving normality and equivalent weight.
Dibasic acids, such as sulfuric acid (H₂SO₄), are important in various industrial processes and laboratory applications. When considering dibasic acids in solution, their ability to donate two protons must be reflected in calculations involving quantities like normality and equivalent weight.
In the context of the given exercise, understanding that the acid is dibasic helps us determine its equivalent weight, which is exactly half of its molecular weight. This is because each acid molecule provides two equivalents of protons when fully ionized.

Equivalent Weight

The concept of equivalent weight is essential for understanding reactions involving acids and bases. For acids, the equivalent weight is derived by dividing the molecular weight by the number of protons the acid can donate.
In our case, the dibasic acid has a molecular weight of 200 and can donate two protons. Thus, the equivalent weight is:

• Equivalent weight = \[\frac{200}{2} = 100\]

This tells us that 100 grams of the dibasic acid will provide one mole of hydrogen ion equivalents. Equivalent weight helps simplify calculations involving normality, as it directly links to how many grams of a substance will supply one equivalent of reacting capacity.
When dealing with solutions and their concentrations, the equivalent weight provides a straightforward manner to determine the concentration in terms of normality.

Molecular Weight Calculation

Molecular weight, also known as molecular mass, is the sum of the atomic masses of all atoms in a molecule. It's an important concept in chemistry, often reported in atomic mass units (amu) or grams per mole (g/mol).
In this exercise, the molecular weight of the dibasic acid is given as 200 g/mol. This value is crucial as it forms the basis for calculating equivalent weight and normality when the acid is in solution.
Calculating molecular weight involves:

• Identifying all the elements present in the compound and their respective quantities.
• Using the periodic table to find the atomic mass of each element.
• Summing the products of the quantities and their atomic masses.

Understanding molecular weight aids in determining how many grams of a compound are needed to achieve a certain molarity or normality in solution, making it foundational for solution chemistry.