Which of the following properties will change when system containing solution 1 will become solution 2? 10 mol of solute x + 10 L of water (Solution 1) -> 1 L of solution 1 + 1 mol of solute x + 1 L of water (Solution 2)

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q53 – JEE-MAIN 2025

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Home Chemistry SolutionsQuestion #583

Which of the following properties will change when system containing solution 1 will become solution 2?

\begin{matrix} 10 mol of solute x \\ + \\ 10 L of water \\ (Solution 1) \end{matrix} ⟶ \begin{matrix} 1 L of solution 1 \\ + \\ 1 mol of solute x \\ + \\ 1 L of water \\ (Solution 2) \end{matrix}

Clue 1 of 3: Identify Property Types

Distinguish between intensive properties (independent of amount) and extensive properties (dependent on amount).

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Ninja Strategy•Intensive vs. Extensive Properties

Quickly identify which properties are intensive (independent of amount) and which are extensive (dependent on amount). Only extensive properties will change if the total amount of matter in the system changes, even if concentration remains constant.

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Step 1: Determine the composition and concentration of Solution 1.✦ Active

Solution 1 is formed by mixing 10 mol of solute x with 10 L of water. The total moles of solute is 10 mol, and the total volume of water is 10 L. Assuming negligible volume change upon mixing, the total volume of Solution 1 is approximately 10 L. The concentration of Solution 1 ( $C_{1}$ ) is calculated as:

C_{1} = \frac{10 mol}{10 L} = 1 M

Step 2: Determine the composition and concentration of Solution 2.○ Expand

Solution 2 is formed by taking 1 L of Solution 1, then adding 1 mol of solute x and 1 L of water. From 1 L of Solution 1, we get 1 mol of solute x ( $1 L \times 1 M$ ) and 1 L of water. The total components for Solution 2 are:

Total moles of solute = $1 mol (from 1 L of Solution 1) + 1 mol (added) = 2 mol$

Total volume of water = $1 L (from 1 L of Solution 1) + 1 L (added) = 2 L$

Assuming negligible volume change, the total volume of Solution 2 is approximately 2 L. The concentration of Solution 2 ( $C_{2}$ ) is calculated as:

C_{2} = \frac{2 mol}{2 L} = 1 M

Step 3: Compare properties based on changes in amount and concentration.○ Expand

1. **Concentration:** Both $C_{1}$ and $C_{2}$ are $1 M$ . Thus, concentration does not change.

2. **Density:** Density is an intensive property. Since the concentration, temperature, and pressure are assumed constant, the density of the solution will not change.

3. **Molar heat capacity:** Molar heat capacity is an intensive property. Since the concentration is constant, the molar heat capacity of the solution will not change.

4. **Gibbs free energy:** Gibbs free energy ( $G$ ) is an extensive property, meaning it depends on the total amount of substance in the system. System 1 contains 10 mol of solute and 10 L of water, while System 2 contains 2 mol of solute and 2 L of water. Since the total amount of solute and solvent in the system changes, the total Gibbs free energy of the system will change.

Therefore, Gibbs free energy is the property that will change.

💡 Teacher's Secret Hint

Remember that extensive properties depend on the amount of matter, while intensive properties do not.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 53 – JEE-MAIN 2025

Clue 1 of 3: Identify Property Types

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