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If water sample are taken from sea, rivers or lake, they will be found to contain hydrogen and oxygen in the approximate ratio of \(1: 8\). This indicates the law of: (a) Multiple proportion (b) Definite proportion (c) Reciprocal proportions (d) None of these

Short Answer

Expert verified
The correct answer is (b) Definite proportion.

Step by step solution

01

Understanding the Concept of Definite Proportions

The law of definite proportions states that a chemical compound always contains exactly the same proportion of elements by mass. In this case, water (H2O) samples from various sources contain hydrogen and oxygen in a consistent ratio of approximately 1:8 by mass.
02

Analyzing the Options

Based on the law described, we must find the option that best matches the concept of a consistent composition by mass, regardless of the sample's source.
03

Selecting the Correct Option

Since the water samples have consistent hydrogen and oxygen ratios, this reflects the law of definite proportion. Therefore, the correct answer is (b) Definite proportion.

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Key Concepts

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

Understanding Chemical Compound Composition
Chemical compounds are substances composed of two or more different types of atoms bonded together. The specific elements and the number of atoms of each element in a compound determine its composition. For example, water is a compound made up of hydrogen and oxygen atoms in a fixed ratio; specifically, two hydrogen atoms for every one oxygen atom, represented by the chemical formula H2O.

When we refer to the composition of a chemical compound, we're talking about the types and quantities of atoms that constitute the compound. The principle of chemical compound composition is critical, as it underpins the formation of every substance with distinct chemical and physical properties. No matter where we obtain a sample of a compound like water, be it from a sea, river, or lake, its fundamental composition remains unchanged. The consistency of this composition is what fundamentally establishes the identity of any compound.

Consistency Across Samples

In our exercise, different water samples show uniform composition, which illustrates a larger rule in chemistry: no matter the source, pure compounds will always have the same elemental composition, as dictated by their chemical formula. This is a fundamental aspect of chemical science that helps chemists understand and predict the behavior of substances.
The Role of Stoichiometry in Chemical Reactions
Stoichiometry is a branch of chemistry that involves the calculation of the relative quantities of reactants and products in chemical reactions. It's based on the conservation of mass and the law of definite proportions. Stoichiometry uses the balanced chemical equation as a basis for these calculations, where the coefficients represent the molar amounts of each substance involved.

Using stoichiometry, chemists can predict how much product will result from a given amount of reactant and vice versa. This precise calculation is essential in many fields, such as pharmaceuticals, where exact dosages of substances must be used.

Real-world Applications

In the industrial production of compounds, stoichiometry ensures that chemical processes are efficient and cost-effective by minimizing waste. By comprehending stoichiometry, students can better understand the quantitative aspects of chemical reactions, reinforcing their grasp of chemical principles and preparing them for practical applications in scientific and industrial settings.
Mass Ratio in Compounds
The mass ratio in compounds is a concept that describes the relationship between the masses of different elements in a compound. It's intimately linked to the law of definite proportions, which indicates that a chemical compound will always contain the same elements in the same mass ratio, regardless of the compound's quantity or source.

For instance, the mass ratio in water is roughly 1:8 for hydrogen to oxygen, meaning that in any sample of water, the mass of hydrogen found is about one-eighth the mass of oxygen. This fixed ratio assures the uniformity of the compound's identity. Moreover, understanding the mass ratio is vital for performing stoichiometric calculations, as these ratios are used to convert between masses of reactants and products.

Understanding Composition through Mass Ratio

An understanding of the mass ratio enhances a student's ability to deduce the composition of unknown substances through experimental data. In laboratory settings, chemists use the mass ratio to determine empirical formulas, calculate molecular formulas, and analyze elemental composition, making it a cornerstone concept in both educational and professional chemistry practices.

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Most popular questions from this chapter

\(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) solution of 1 molal concentration is present in 1 litre solution of \(2.684 \mathrm{~g} / \mathrm{cc} .\) How many moles of \(\mathrm{BaSO}_{4}\) would be precipitated on adding \(\mathrm{BaCl}_{2}\) in excess ? (a) 2 moles (b) 3 moles (c) 6 moles (d) 12 moles

\(1.25 \mathrm{~g}\) of a solid dibasic acid is completely neutralised by \(25 \mathrm{~mL}\) of \(0.25\) molar \(\mathrm{Ba}(\mathrm{OH})_{2}\) solution. Molecular mass of the acid is : (a) 100 (b) 150 (c) 120

Some older emergency oxygen masks containing potassium superoxide, \(\mathrm{KO}_{2}\) which reacts with \(\mathrm{CO}_{2}\) and water in exhaled air to produce oxygen according to the given equation. If a person exhales \(0.667 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) per minute, how many grams of \(\mathrm{KO}_{2}\) are consumed in \(5.0\) minutes? $$ 4 \mathrm{KO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+4 \mathrm{CO}_{2} \longrightarrow 4 \mathrm{KHCO}_{3}+3 \mathrm{O}_{2} $$ (a) \(10.7\) (b) \(0.0757\) (c) \(1.07\) (d) \(5.38\)

Concentrated \(\mathrm{HNO}_{3}\) is \(63 \% \mathrm{HNO}_{3}\) by mass and has a density of \(1.4 \mathrm{~g} / \mathrm{mL}\). How many millilitres of this solution are required to prepare \(250 \mathrm{~mL}\) of a \(1.20 \mathrm{M} \mathrm{HNO}_{3}\) solution? (a) \(18.0\) (b) \(21.42\) (c) \(20.0\) (d) \(14.21\)

\(\mathrm{KMnO}_{4}\) reacts with oxalic acid according to the equation $$ 2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O} $$ Here, \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{KMnO}_{4}\) is equivalent to : (a) \(120 \mathrm{~mL}\) of \(0.25 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (b) \(150 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (c) \(25 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (d) \(50 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

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