Problem 152 - A Complete Resource Book in Chemistry for JEE Main

Question

A compound A of S, Cl and \(O\) has a vapour density of 67.5. It reacts with water to form two acids and reacts with \(\mathrm{KOH}\) to form two salts \(\mathrm{B}\) and \(\mathrm{C}\). while B gives white precipitate with \(\mathrm{AgNO}_{3}\) solution, \(\mathrm{C}\) gives white precipitate with \(\mathrm{BaCl}_{2}\) solution. Identify \(\mathrm{A}\) (a) \(\mathrm{SOCl}_{2}\) (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{SO}_{2} \mathrm{Cl}\) (d) none of these

Step-by-Step Solution

Answer

Compound A is \( \text{SO}_2 \text{Cl}_2 \).

Step 1: Calculate Molar Mass

The vapor density of a compound is given as half of its molar mass. Therefore, the molar mass (M) of compound A is given by \( M = 2 \times \text{vapor density} = 2 \times 67.5 = 135 \text{ g/mol} \).

Step 2: Analyze Chemical Reactions

From the problem statement, compound A reacts with water to form two acids. This suggests the compound is a chlorosulfonic acid derivative. The reaction with \( \text{KOH} \) forming two salts, \( \text{B} \) and \( \text{C} \), gives clues about its structure. \( \text{B} \) forms a white precipitate with \( \text{AgNO}_3 \), indicating a chloride presence, while \( \text{C} \) forms a white precipitate with \( \text{BaCl}_2 \), indicating a sulfate presence.

Step 3: Identify Compound Structure

The given options include \( \text{SOCl}_2 \), \( \text{SO}_2 \text{Cl}_2 \), and \( \text{SO}_2 \text{Cl} \). The molar mass calculation suggests \( \text{SO}_2 \text{Cl}_2 \) (sulfuryl chloride) as a plausible identification since it would match the calculated molar mass: \[ \text{S} (32) + 2\times\text{O} (2\times16) + 2\times\text{Cl} (2\times35.5) = 135 \text{ g/mol} \].

Step 4: Conclude Identification

Given the molar mass and the chemical behavior detailed in the options, \( \text{A} \) can be identified as \( \text{SO}_2 \text{Cl}_2 \) because it fits the mass calculation and the reaction criteria provided in the problem statement.

Key Concepts

Molar Mass Calculation

Molar mass calculation is a foundational step in identifying compounds. By knowing the vapour density of a substance, we can find its molar mass. The relationship between these two is straightforward: the molar mass (M) is simply twice the vapour density of the compound. For instance, with a vapour density of 67.5, the molar mass of the compound is calculated as \( M = 2 \times 67.5 = 135 \text{ g/mol} \). This simple multiplication shows how vapour density helps us ascertain the size of the molecules.

The molar mass is critical because it allows for comparison with known compounds. By calculating the molar mass, you can narrow down potential compounds that fit this characteristic. This step is crucial for determining the chemical identity of a compound in many chemistry problems, especially in competitive exams like JEE.

If you're working on problems like these, keep a periodic table handy for quick reference. It ensures you can quickly compile the atomic masses needed for these calculations.

Vapour Density

Vapour density provides insight into the composition of a gas or vapor. It's a physical property that assists in identifying gaseous compounds. This measurement tells us how much heavier the compound's vapor is compared to hydrogen. Since hydrogen is the lightest element, it is used as a standard.

In practice, vapour density is calculated by dividing the molar mass of a compound by two. For example, if the vapour density is given as 67.5, you can deduce that the molar mass is \(2 \times 67.5 = 135 \text{ g/mol} \). This technique of using vapour density helps in identifying compounds during chemical analysis.

• Understanding vapour density helps predict behaviors of compounds under changes in state.
• It is particularly useful when analyzing gases and volatile substances.
• The measure is vital in stoichiometry for calculating reactant amounts in gaseous reactions.

Chemical Reactions Analysis

Analyzing chemical reactions involves understanding reactants and their transformation products. From the problem, compound A reacts with water to form two acids. Such a reaction suggests the compound might belong to a category known as chlorosulfonic acid derivatives.

Further reacting with \(\text{KOH}\), compound A forms salts B and C. Each salt gives distinctive precipitate reactions:

• Salt B gives a white precipitate with \(\text{AgNO}_3\), indicating it contains chloride ions.
• Salt C gives a white precipitate with \(\text{BaCl}_2\), revealing the presence of sulfate ions.

These observations of both acid and salt formation help deduce the chemical nature and possible structure of compound A. By combining these reactions with molar mass information, we can identify the correct compound as \(\text{SO}_2 \text{Cl}_2\).

Precipitation Reactions

Precipitation reactions occur when two solutions react to form an insoluble solid, known as a precipitate. These reactions are significant in identifying ions present in a solution.

In this example:

• The formation of a white precipitate when salt B if combined with \(\text{AgNO}_3\) implies the presence of chloride ions. Silver chloride \(\text{(AgCl)}\) is a common white precipitate.
• The white precipitate formed when salt C combines with \(\text{BaCl}_2\) indicates the presence of sulfate ions. Barium sulfate \(\text{(BaSO}_4\text{)}\) is another well-known white precipitate.

These reactions are powerful tools in analytical chemistry. They allow chemists to confirm the presence of different ions based on the color and nature of the precipitate formed.