Problem 17 - A Complete Resource Book in Chemistry for JEE Main

Question

The number of moles of sodium hydroxide present in \(2.5 \mathrm{~L}\) and \(0.5 \mathrm{M}\) aqueous solution will be (a) \(1.25\) (b) \(0.5\) (c) \(12.5\) (d) 5

Step-by-Step Solution

Answer

The number of moles is 1.25.

Step 1: Understand Molarity Definition

Molarity (M) is defined as the number of moles of solute per liter of solution. It is given by the formula \( M = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume in liters.

Step 2: Identify Given Values

From the problem statement, we are given: \( V = 2.5 \) L and \( M = 0.5 \) M. We need to find the number of moles, \( n \), of sodium hydroxide.

Step 3: Apply Molarity Formula

Using the formula \( M = \frac{n}{V} \), we rearrange it to find \( n \): \( n = M \times V \).

Step 4: Substitute the Values

Substitute the given values into the equation: \( n = 0.5 \times 2.5 \).

Step 5: Calculate the Result

Calculate \( n = 0.5 \times 2.5 = 1.25 \). Thus, there are 1.25 moles of sodium hydroxide in the solution.

Key Concepts

Moles of Solute

The concept of moles is central to understanding chemistry. "Moles of solute" refer to the amount of a substance, measured in moles, present in a solution. A mole is a unit used to quantify the number of atoms, ions, or molecules in a given sample. It is based on Avogadro's number, which is approximately \(6.022 \times 10^{23}\). This means one mole of a substance contains roughly \(6.022 \times 10^{23}\) particles of that substance.

In the context of our exercise, the solute is sodium hydroxide, and its quantity is measured in moles. To find the number of moles of solute present in a solution, one often follows a systematic approach. It involves using the formula for molarity, which interconnects moles, liters, and molarity (\(M\)).

• First, determine the given values from a problem (volume and molarity).
• Then use these to calculate the total moles of solute.

This ensures a comprehensive understanding of how many moles of sodium hydroxide are in the solution.

Concentration Calculation

Calculating the concentration of a solution is crucial in chemistry. Concentration refers to how much solute is present in a given volume of solution. It is commonly expressed in terms of "molarity," denoted by \(M\). Molarity is defined by the formula \(M = \frac{n}{V}\), where \(n\) represents the number of moles of solute and \(V\) is the volume of the solution in liters.

To find how concentrated a solution is, follow these simple steps:

• Identify the volume of the solution and its molarity from the problem statement.
• Reorganize the formula to solve for the number of moles: \(n = M \times V\).
• Insert the given values and solve to find \(n\).

For instance, in our example with sodium hydroxide, the concentration is straightforwardly calculated by multiplying \(0.5 \text{ M}\) by \(2.5 \text{ L}\), thus yielding \(1.25\) moles. This method highlights the straightforwardness and practicality of concentration calculations in real-world applications.

Aqueous Solution

An aqueous solution is one where water serves as the solvent. This means that the solute, such as sodium hydroxide in our problem, is dissolved in water. Water is known as the "universal solvent" because it can dissolve a wide variety of substances. The ability of water to dissolve so many compounds makes it invaluable in chemical reactions, biological functions, and industrial processes.

When talking about aqueous solutions, it is important to:

• Understand that the solvent is always water.
• Recognize how the solute's properties, like its molarity, affect the solution.

In the given exercise, understanding this concept helps comprehend how sodium hydroxide, the solute, interacts with water to form the solution. It also allows one to properly calculate its concentration and understand the role and behavior of substances in aqueous environments. By recognizing these factors, students can better predict the outcomes of chemical reactions and experiments in aqueous solutions.