Problem 85
Question
The volume strength of \(1.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{O}_{2}\) solution is (a) \(4.8\) (b) \(8.4\) (c) \(4.0\) (d) \(8.0\)
Step-by-Step Solution
Verified Answer
The volume strength of the solution is 8.4.
1Step 1: Understanding Volume Strength
Volume strength refers to the volume of oxygen gas (in liters) released at STP (standard temperature and pressure) by one liter of hydrogen peroxide solution. Volume strength can be calculated using the formula: \[ \text{Volume Strength} = \text{Normality} \times 5.6 \] where the factor 5.6 is derived from the reaction of hydrogen peroxide decomposing to produce oxygen.
2Step 2: Calculating Volume Strength
Given the normality (N) of the hydrogen peroxide solution is \(1.5 \text{ N}\), apply the formula for volume strength. \[ \text{Volume Strength} = 1.5 \times 5.6 = 8.4 \] Thus, the volume strength of the solution is calculated to be 8.4.
Key Concepts
NormalitySTP (Standard Temperature and Pressure)Chemical Decomposition Reaction
Normality
Normality is a vital concept in chemistry that measures the concentration of a solution. It is defined as the number of equivalents of a solute per liter of solution. The formula to determine normality is:
This measure of concentration is particularly useful when dealing with reactions where the stoichiometry isn't straightforward, like oxidation-reduction reactions.
When calculating volume strength from normality, as in the given problem, normality directly influences how much oxygen can be released from the solution under specific conditions.
- Normality (N) = \( \frac{\text{Equivalents of Solute}}{\text{Liters of Solution}} \)
This measure of concentration is particularly useful when dealing with reactions where the stoichiometry isn't straightforward, like oxidation-reduction reactions.
When calculating volume strength from normality, as in the given problem, normality directly influences how much oxygen can be released from the solution under specific conditions.
STP (Standard Temperature and Pressure)
STP stands for Standard Temperature and Pressure, a set of conditions often used in gas calculations to provide a standard reference point. At STP:
When volume strength is discussed, it typically refers to the volume of gas released under these standard conditions. By knowing this standard, chemists can predict and compare how much gas a reaction produces without changes in temperature or pressure affecting the results.
Thus, STP provides a consistent basis for calculating parameters such as volume strength, allowing for accurate predictions in experimental and industrial settings.
- Temperature: 0°C or 273.15 K
- Pressure: 1 atm or 101.3 kPa
When volume strength is discussed, it typically refers to the volume of gas released under these standard conditions. By knowing this standard, chemists can predict and compare how much gas a reaction produces without changes in temperature or pressure affecting the results.
Thus, STP provides a consistent basis for calculating parameters such as volume strength, allowing for accurate predictions in experimental and industrial settings.
Chemical Decomposition Reaction
A chemical decomposition reaction occurs when a single compound breaks down into two or more simpler substances. In the exercise we're considering here, hydrogen peroxide (H_2O_2) undergoes decomposition to produce water and oxygen gas:
Throughout the process, the rate at which H_2O_2 decomposes can be influenced by several factors such as concentration, temperature, presence of catalysts, and so on.
Understanding these reactions helps in predicting how solutions will behave under different conditions and is essential in both laboratory and industrial applications for producing or controlling oxygen flows safely and efficiently.
- \[2 H_2O_2 \rightarrow 2 H_2O + O_2\]
Throughout the process, the rate at which H_2O_2 decomposes can be influenced by several factors such as concentration, temperature, presence of catalysts, and so on.
Understanding these reactions helps in predicting how solutions will behave under different conditions and is essential in both laboratory and industrial applications for producing or controlling oxygen flows safely and efficiently.
Other exercises in this chapter
Problem 81
The molarity of pure water at \(277 \mathrm{~K}\) is (a) \(1 \mathrm{M}\) (b) \(5 \mathrm{M}\) (c) \(55.5 \mathrm{M}\) (d) \(2.5 \mathrm{M}\)
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The percentage by weight of hydrogen in \(\mathrm{H}_{2} \mathrm{O}_{2}\) is (a) \(6.5\) (b) \(5.88\) (c) 25 (d) 50
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