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1800-102-2727Glittering white teeth gives charming personality. Hence teeth whitening has grown in popularity in recent years, owing to the increased availability of products on the market.
However, a lot of these solutions can be fairly pricey, which prompts consumers to search for less-priced treatments.
Therefore, the primary component of most teeth-whitening products, hydrogen peroxide, is the cheapest method at home for dazzling teeth whitening.
What you need to know is this; The majority of bottles of hydrogen peroxide you may purchase at a drugstore or grocery shop are diluted to about 3 per cent. Could you determine how much hydrogen peroxide is present in this three per cent volume of solution? If not, let's explore.
Table of Contents
Chemically, hydrogen peroxide has the formula H_{2}O_{2}. It is a very light blue liquid when it is pure, somewhat more viscous than water.
Characteristics of Hydrogen Peroxides
In all ratios, H_{2}O_{2} and water can be mixed to form a hydrate of H_{2}O_{2}.H_{2}O
Volume is a typical method for expressing the strength of hydrogen peroxide. This is a reference to the amount of oxygen that an H_{2}O_{2} solution will produce.
In the case of hydrogen peroxide, a "20 volume" means that 1 L of this solution will produce 20 L of oxygen at N.T.P or 1m L of this solution will produce 20mL of oxygen at N.T.P.
Now, If we want to calculate the amount of Hydrogen peroxide in 20 Volume,
Follow these steps
(i) Decomposition of Hydrogen peroxide can be written as
${2H}_{2}{O}_{2}\left(l\right)\to 2{H}_{2}O\left(l\right)+{O}_{2}\left(g\right)$
68 g 22.7 L
1 mole of O_{2}(g) = 22.7 L or 32 g
1 mole of H_{2}O_{2}(l) = 22.7 L or 34 g
2 moles of H_{2}O_{2}(l) = 45.4L or 68 g
(ii) Mole Concept
1 mole of O_{2}(g)is formed on the decomposition of 2 moles of H_{2}O_{2}(l)
22.7 L of O_{2}(g) is formed on the decomposition of 68 g of H_{2}O_{2}(l)
1 L of O_{2}(g) is formed on the decomposition of = $\frac{68g}{22.7L}$ of H_{2}O_{2}(l)
20 L of O_{2}(g) is formed on the decomposition of = $\frac{68g}{22.7L}\times 20Lof{H}_{2}{O}_{2}\left(l\right)=59.9gof{H}_{2}{O}_{2}\left(l\right)$
Hence, 20 volume contains 59.9 g or 60 g of H_{2}O_{2}(l).
(iii) Strength of H_{2}O_{2}(l) in 20 volume of 1L solution
Strength = $\frac{60g}{1L}$ = $60\frac{g}{L}$
(iv) % Strength of H_{2}O_{2}(l) in 20 volume of 1L solution
% Strength = $\frac{60g}{1000mL}\times 100$ = 6 %
Q1. Calculate the volume of O_{2}(g) in 3% solution of Hydrogen Peroxide.
A. 3 L
B. 30 L
C. 10 L
D. 15 L
Answer: (C)
Solution: 3% solution of Hydrogen Peroxide means the strength of the solution is $\frac{30g}{L}$
Hence, it is given that 30 g of Hydrogen peroxide is present in 1L of solution.
Decomposition of Hydrogen peroxide can be written as
${2H}_{2}{O}_{2}\left(l\right)\to 2{H}_{2}O\left(l\right)+{O}_{2}\left(g\right)$
68 g 22.7 L
2 moles of H_{2}O_{2}(l) decompose to give 1 mole of O_{2}(g)
68 g of H_{2}O_{2}(l) decompose to give 22.7 L of O_{2}(g)
1 g of H_{2}O_{2}(l) decompose to give = $\frac{22.7L}{68g}$of O_{2}(g)
30 g of H2O2(l) decompose to give = $\frac{22.7L}{68g}\times 30gof{O}_{2}\left(g\right)=10Lof{O}_{2}\left(g\right)$
Hence, the correct answer is option (C).
Q2. Calculate the volume strength of 3N H_{2}O_{2}(l) solution.
A. 17 volumes
B. 25 volumes
C. 51 volumes
D. Invalid information
Answer: (A)
Solution: 3N means 3 gram equivalents of H_{2}O_{2}(l) is present in 1L of solution.
Equivalent Weight of H_{2}O_{2}(l) = $\frac{MolarMass}{2}$ = $\frac{34}{2}$ = 17 g
$GramEquivalents=\frac{Weight}{EquivalentWeight}$
$Weight=GramEquivalents\times EquivalentWeight$
$Weightof{H}_{2}{O}_{2}\left(l\right)=3\times 17=51g$
Decomposition of Hydrogen peroxide can be written as
${2H}_{2}{O}_{2}\left(l\right)\to 2{H}_{2}O\left(l\right)+{O}_{2}\left(g\right)$
68 g 22.7 L
2 moles of H_{2}O_{2}(l) decompose to give 1 mole of O_{2}(g)
68 g of H_{2}O_{2}(l) decompose to give 22.7 L of O_{2}(g)
1 g of H_{2}O_{2}(l) decompose to give = $\frac{22.7L}{68g}$ gof O_{2}(g)
51 g of H_{2}O_{2}(l) decompose to give = $\frac{22.7L}{68g}\times 51g$ of O_{2}(g) = 17 L of O_{2}(g)
1 L of this solution will produce 17 L of oxygen at N.T.P, hence the volume strength of H_{2}O_{2}(l) is 17 volumes
Hence, the correct answer is option (A).
Q3. Calculate the weight of H_{2}O_{2}(l) present in 25 volumes of H_{2}O_{2}(l).
A. 25 g
B. 50 g
C. 64 g
D. 75 g
Answer: (D)
Solution: 25 volume means that 1 L of this solution will produce 25 L of oxygen at N.T.P.
Decomposition of Hydrogen peroxide can be written as
${2H}_{2}{O}_{2}\left(l\right)\to 2{H}_{2}O\left(l\right)+{O}_{2}\left(g\right)$
68 g 22.7 L
1 mole of O_{2}(g)is formed on the decomposition of 2 moles of H_{2}O_{2}(l)
22.7 L of O_{2}(g) is formed on the decomposition of 648g of H_{2}O_{2}(l)
1 L of O_{2}(g) is formed on the decomposition of = 68 g22.7 L of H2O2(l)
25 L of O2(g) is formed on the decomposition of = $\frac{68g}{22.7L}\times 25L$ of H_{2}O_{2}(l) = 74.9 g or 75 g of H_{2}O_{2}(l)
Hence, 20 volume contains 75 g of H_{2}O_{2}(l).
The correct option is (D).
Q4. A KI solution that has been acidified can release 1 g of iodine when 1.34 mL of H_{2}O_{2}(l) is added. Calculate the volume strength of H_{2}O_{2}(l) solution at STP.
A. 33 Volumes
B. 3.3 Volumes
C. 4.48 Volumes
D. 0.33 Volumes
Answer: (A)
Solution: H_{2}O_{2}(l) reacts with acidified KI solution as
$2KI+{H}_{2}{SO}_{4}+{H}_{2}{O}_{2}\to {K}_{2}{SO}_{4}+{I}_{2}+2{H}_{2}O$
3.4 mL 1g
254 g of I_{2} liberates = 34 g of H2O2
1 g of I_{2} liberates= $\frac{34g}{254g}$= 0.134 g of H_{2}O_{2}
1.34 mL of H_{2}O_{2}(l) contains = 0.134 g
1 mL of H_{2}O_{2}(l) contains = $\frac{0.134}{1.34}$=0.1 g
Decomposition of Hydrogen peroxide can be written as
${2H}_{2}{O}_{2}\left(l\right)\to 2{H}_{2}O\left(l\right)+{O}_{2}\left(g\right)$
68 g 22.7 L
2 moles of H_{2}O_{2}(l) decompose to give 1 mole of O_{2}(g)
68g of H_{2}O_{2}(l) decompose to give 22.7 L of O_{2}(g)
1 g of H_{2}O_{2}(l) decompose to give = $\frac{22.7L}{68g}$of O_{2}(g)
0.1 g of H_{2}O_{2}(l) decompose to give = $\frac{22.7L}{68g}\times 0.1g$ of O_{2}(g) = 0.033 L of O_{2}(g)
Hence, 1 mL of H_{2}O_{2}(l) contains 0.033 L of O_{2}(g)
Also, 1 mL of H_{2}O_{2}(l) contains 33 mL of O_{2}(g)
Volume Strength = 33 Volumes
Q1. Where in your daily life can you find hydrogen peroxide?
Answer: Hydrogen peroxide functions as an oxidising ingredient in personal care and home goods like toothpaste, mouthwash, bathroom cleaners, and laundry stain removers, providing a lightening and whitening effect.
Q2. How could hydrogen peroxide affect a person's body?
Answer: Hydrogen peroxide is unlikely to result in chronic toxicity because it breaks down quickly in the body. However, repeated inhalations of hydrogen peroxide fumes in chemistry labs may result in chronic respiratory tract irritation and partial or total lung collapse.
Q3. What is the purpose of hydrogen peroxide in medicine?
Answer: For minor burns, scrapes, and wounds, hydrogen peroxide is a gentle antibacterial that can be applied directly to the skin. Additionally, it can be used as a mouthwash to assist clear mucus or soothe mild oral irritation. Additionally, it is utilised as a nasal spray and to remove earwax.
Q4. Is blood a stain that hydrogen peroxide can remove?
Answer: Unfortunately, they are, and for this job, something more powerful than soap and water is frequently required. The best technique is to employ hydrogen peroxide, an oxidising substance that breaks down and eliminates old blood stains through a chemical reaction.