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91Ó°ÊÓ

Match the following Column-I (a) 10 volumes (b) 20 volumes (c) 30 volumes (d) \(11.2\) volumes Column-II (p) \(5.358 \mathrm{~N}\) (q) \(3.036 \%(\mathrm{w} / \mathrm{v})\) (r) \(3.4 \mathrm{gm} \mathrm{H}_{2} \mathrm{O}_{2} / 100\) \(\mathrm{ml}\) solution (s) \(1.785 \mathrm{M}\)

Short Answer

Expert verified
(a)-(q), (b)-(s), (c)-(r), (d)-(p).

Step by step solution

01

Understanding the Problem

We have two columns, Column-I with volume values and Column-II with different units of measurements related to chemical concentration. Our task is to match each item in Column-I to an appropriate unit in Column-II based on hydrogen peroxide (Hâ‚‚Oâ‚‚) concentration.
02

Analyzing 10 Volumes

10 volumes mean that 1 volume of a solution can release 10 volumes of oxygen at standard conditions. This corresponds to a certain percentage concentration of Hâ‚‚Oâ‚‚, needing to find the match in concentration terms from Column-II.
03

Analyzing 20 Volumes

Similarly, 20 volumes of Hâ‚‚Oâ‚‚ mean that 1 volume of the solution can release 20 volumes of oxygen. More oxygen corresponds to a higher concentration, but we need to analyze which of the units accurately represents these concentrations.
04

Analyzing 30 Volumes

30 volumes indicate a stronger solution than both 10 and 20 volumes since it can release 30 times its volume of oxygen, demanding an appropriate match in Column-II.
05

Analyzing 11.2 Volumes

11.2 volumes suggest a specific yet lower concentration compared to 20 and 30 volumes, so it needs to be matched accordingly with the correct chemical concentration unit from Column-II.
06

Matching Column-I and Column-II

- For 10 volumes, the direct match is typically around 3% w/v. Hence, (a) corresponds to (q). - For 20 volumes, it represents a more concentrated solution corresponding nearby 1.785 M, so (b) maps to (s). - For 30 volumes, the higher concentration matches chemical compositions like 3.4 g Hâ‚‚Oâ‚‚/100 ml, thus (c) is (r). - The closeness of 11.2 volumes to 10 volumes in concentration suggests it corresponds to pressure notation, so (d) relates to (p).
07

Verification of Matches

Reconfirm each pairing by considering the typical usage and chemical interpretations in Hâ‚‚Oâ‚‚ solutions. This process solidifies that (a)-(q), (b)-(s), (c)-(r), and (d)-(p) are logical based on the conventions and the calculations of concentrations provided.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volumes of Oxygen
When discussing hydrogen peroxide (Hâ‚‚Oâ‚‚), the term "volumes of oxygen" refers to how much oxygen gas a certain amount of Hâ‚‚Oâ‚‚ solution can release. Each number indicates how many volumes of oxygen gas are produced from one volume of the solution under standard conditions.
This measure is crucial for understanding how concentrated the solution is:
  • 10 volumes means one volume of the solution releases 10 volumes of oxygen.
  • 20 volumes release 20 volumes of oxygen from one volume of solution.
  • The higher the number, the more concentrated the solution is.

Understanding this concept helps match volume descriptions to chemical concentrations, linking practical effects with scientific measurements.
Percentage Concentration
Percentage concentration tells us how much solute is present in a specific amount of solution and is often expressed as weight/volume percentage ( % w/v) or weight/weight percentage ( % w/w).
For hydrogen peroxide, the % w/v is commonly used to describe how much Hâ‚‚Oâ‚‚ is in 100 ml of the solution.

Here's a simple breakdown:
  • A 3% w/v solution contains 3 grams of Hâ‚‚Oâ‚‚ per 100 ml of solution.
  • Percentage concentration directly influences how much oxygen is released in the volumes of oxygen, as it tells you the direct proportion of Hâ‚‚Oâ‚‚ in the solution.

This metric is not only essential for do-it-yourself projects but also plays a pivotal role in laboratory settings and industrial applications.
Molarity
Molarity ( M) is a way to express concentration, defined as the number of moles of solute per liter of solution. In the context of hydrogen peroxide, this unit of concentration signifies its potential reactivity and effects based on chemical formulas.

For instance:
  • A solution that is 1.0 M means there is one mole of Hâ‚‚Oâ‚‚ per liter of solution.
  • It aids scientists and engineers in calculating how substances will react when mixed or used, crucial for laboratory tasks.
Molarity provides precise information on the concentration of a chemical, essential for detailed chemical equations and analyses.
Weight/Volume Percentage
The weight/volume percentage ( % w/v) is another concentration measure that reveals how much solute is in a set volume of solution, typically expressed as grams per 100 ml.
This is especially useful for solutions like hydrogen peroxide used in various settings for disinfection and bleaching.

To simplify:
  • A % w/v concentrates on the relationship between the solvent's weight and the solution's volume.
  • It's a quick way to understand how concentrated a solution is by directly correlating weight to volume.
This method allows easy calculations for preparing solutions and understanding their strength in practical applications.

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Most popular questions from this chapter

The bonds present in \(\mathrm{N}_{2} \mathrm{O}_{5}\) are (a) only ionic (b) covalent and coordinate (c) only covalent (d) covalent and ionic

About \(20 \mathrm{~km}\) above the earth, is the ozone layer. Which one of the following statements about ozone and ozone layer is true? (a) conversion of \(\mathrm{O}_{3}\) to \(\mathrm{O}_{2}\) is an endothermic reaction (b) it is beneficial to us as it stops ultraviolet radiation (c) ozone has a triatomic linear molecule (d) it is harmful as it stops useful radiations

A variety of water which contains soluble salts of \(\mathrm{Ca}\) and \(\mathrm{Mg}\) is known as (a) soft water (b) heavy water (c) conductivity water (d) hard water

Which of the following equations are correctly formulated here? (1) \(4 \mathrm{Sn}+10 \mathrm{HNO}_{3}\) (dilute) \(\longrightarrow 4 \mathrm{Sn}\left(\mathrm{NO}_{3}\right)_{2}\) \(+\mathrm{NH}_{4} \mathrm{NO}_{3}+\) \(3 \mathrm{H}_{2} \mathrm{O}\) (2) \(3 \mathrm{Zn}+8 \mathrm{HNO}_{3}\) (very dilute) \(\longrightarrow 3 \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) \(+2 \mathrm{NO}+4 \mathrm{H}_{2} \mathrm{O}\) (3) \(3 \mathrm{Cu}+8 \mathrm{HNO}_{3}\) (dilute) \longrightarrow \(3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) \(+2 \mathrm{NO}+4 \mathrm{H}_{2} \mathrm{O}\) (4) \(\mathrm{As}+3 \mathrm{HNO}_{3}\) (dilute) \(\longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{3}+3 \mathrm{NO}_{2}\) (a) 1,2 and 3 (b) 1,3 and 4 (c) 2,3 and 4 (d) 1,2 and 4

It is possible to obtain oxygen from air by fractional distillation because (a) oxygen has a lower density than nitrogen (b) oxygen has high boiling point than nitrogen (c) oxygen is more reactive than nitrogen (d) oxygen is in a different group of the periodic table from nitrogen.

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