Problem 131 - A Complete Resource Book in Chemistry for JEE Main

Question

The maximum percentage of available chlorine on the basis of \(\mathrm{CaOCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) formula is (a) 40 (b) 35 (c) 49 (d) 45

Step-by-Step Solution

Answer

The maximum percentage of available chlorine is 50%, but given options, it could be closest to (c) 49%.

Step 1: Determine Molar Masses

First, calculate the molar masses of each component in the compound \(\mathrm{CaOCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\). The molar masses are: \(\mathrm{Ca} = 40\), \(\mathrm{O} = 16\), \(\mathrm{Cl} = 35.5\) (times 2 for \(\mathrm{Cl}_{2}\)), and \(\mathrm{H}_{2} \mathrm{O} = 18\).

Step 2: Calculate Molar Mass of the Compound

Sum the molar masses for the entire compound: \[\mathrm{CaOCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O} = 40 + 16 + 2\times35.5 + 18 = 142\]. So the molar mass of \(\mathrm{CaOCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) is \(142\) grams per mole.

Step 3: Calculate Contribution of Chlorine

The total mass contributed by chlorine in \(\mathrm{CaOCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) is \(71\) grams (since \(2 \times 35.5 = 71\)).

Step 4: Determine Percentage of Chlorine

The percentage of chlorine by mass is calculated as \[\frac{71}{142} \times 100\] which equals \(50\) percent.

Key Concepts

Molar Mass Calculation

Molar mass calculation is a crucial skill in chemistry. It involves finding the mass of one mole of a compound by summing the masses of individual elements within it. For example, to determine the molar mass of \(\mathrm{CaOCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\), you need to know the atomic masses of calcium (\(\mathrm{Ca}\)), oxygen (\(\mathrm{O}\)), chlorine (\(\mathrm{Cl}\)), and water (\(\mathrm{H}_{2} \mathrm{O}\)).
- **Calcium**: The atomic mass is 40.
- **Oxygen**: The atomic mass is 16.
- **Chlorine**: The atomic mass is 35.5, and since there are two chlorine atoms in \(\mathrm{Cl}_{2}\), you multiply 35.5 by 2 to get 71.
- **Water**: The atomic mass of water is 18.

Sum these values to find the molar mass of the compound. Thus, \(\mathrm{CaOCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) has a molar mass of \(142\) grams per mole. Proper understanding of molar mass helps in various calculations, such as determining percentages and converting between moles and grams.

Available Chlorine

Available chlorine refers to the amount of chlorine released during a chemical reaction, especially from disinfectants like bleaching powder. It's important because it signifies the active chlorine capable of disinfecting or bleaching.

In \(\mathrm{CaOCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\), the chlorine's role is to act as a sterilizing agent. To calculate the actual percentage of available chlorine, you take the mass of chlorine from the compound (obtained from molar mass calculations) and divide it by the total molar mass of the compound, then multiply by 100 to get a percentage.
This calculation shows the efficiency and strength of a chlorine-containing compound. In our example, the available chlorine percentage is \(\frac{71}{142} \times 100 = 50\)%. This reveals that half of the compound's mass comes from potentially active chlorine.

Chemistry Problem Solving

Chemistry problem solving builds on understanding the principles of chemistry and applying them to derive answers. It requires you to break down complex problems into smaller steps. With the knowledge gained from molar mass calculations and available chlorine, you can solve many chemistry challenges more effectively.

• **Understand the Question**: Know what the problem is asking for. Is it about calculating a percentage, a molar mass, or something else?
• **Gather Data**: Determine the substances involved and their respective molar masses or any given data.
• **Perform Calculations**: Use formulas for molar mass, concentration, and percentage composition to find your answer.
• **Cross-Check**: Verify results by plugging values back into the problem. Ensure all numbers align logically with the theory.

Effective problem-solving improves with practice and attention to detail, making you capable of tackling advanced chemistry problems in the future.