Problem 19 - A Complete Resource Book in Chemistry for JEE Main

Question

A one litre flask is full of brown bromine vapours. The intensity of brown colour of vapour will not decrease appreciably on adding to the flask some (a) carbon disulphide (b) piece of marble (c) carbon tetrachloride (d) animal charcoal powder

Step-by-Step Solution

Answer

Adding a piece of marble will not decrease the brown color intensity appreciably.

Step 1: Understanding the Problem

We have a one litre flask full of brown bromine vapours. We need to determine which substance, when added, will not appreciably decrease the brown colour intensity of the vapour.

Step 2: Analyzing the Options

Each option represents a substance. We need to consider how each interacts with bromine vapours. Carbon disulphide and carbon tetrachloride are solvents that could dissolve bromine, potentially decreasing the color intensity. Animal charcoal powder acts as an adsorbent which could also reduce color intensity by removing bromine vapours. A piece of marble is chemically inert and will not interact with bromine vapours significantly.

Step 3: Selecting the Inert Substance

Given the chemical properties, the piece of marble (contains calcium carbonate) is least likely to interact with bromine vapours. It won't dissolve or adsorb the bromine, hence the color intensity shouldn't decrease.

Key Concepts

Bromine Vapours

Bromine is a dark reddish-brown liquid at room temperature, and it easily evaporates to form bromine vapours, which are also brown. This is why a flask filled with bromine vapours appears dark or brown. These vapours are quite dense and heavier than air, with a tendency to stay low unless disturbed. The visibility and intensity of the brown color are directly related to the concentration of bromine vapours in the air. One important characteristic of bromine vapours is that they are reactive and can change their concentration if interacted with certain substances. When considering how to maintain or change the intensity of bromine vapours, it's crucial to understand the reactivity and solubility of bromine with other materials.

Chemical Interactions

Chemical interactions refer to how different substances interact at the molecular or atomic level. When considering bromine vapours, it is essential to understand that they can interact with certain materials, altering their concentration and, therefore, their visible intensity. For example: - **Solvents:** Substances like carbon disulphide and carbon tetrachloride can dissolve bromine. This means they can capture bromine molecules into themselves, reducing the concentration and intensity of bromine vapours in the air. - **Adsorbents:** Materials such as animal charcoal can adsorb bromine onto their surfaces. This is a process where bromine molecules are trapped on the external surface of the solid, again decreasing the amount of bromine vapours present in the air. The key to preventing these interactions is utilizing substances that are chemically inert to bromine, like marble (calcium carbonate), which do not dissolve or adsorb bromine effectively.

Solvents and Adsorbents

Both solvents and adsorbents play a crucial role in the manipulation of chemical vapours such as bromine. Solvents, like carbon disulphide or carbon tetrachloride, work by dissolving bromine, which means they mix with bromine at the molecular level and can effectively reduce its vapour concentration. This leads to a decrease in the visible intensity of bromine vapours. On the other hand, adsorbents such as animal charcoal work by attracting and holding bromine molecules on their surfaces. This does not involve dissolving; instead, it binds bromine vapours to the surface of the adsorbent material, reducing the free-floating bromine in the environment. Understanding which materials act as solvents or adsorbents helps in controlling the presence and intensity of bromine vapours, and in the exercise scenario, aids in identifying substances that minimize interaction with bromine to keep its concentration and intensity constant.