Problem 17
Question
Among the following pairs of compounds, the one that illustrates the law of multiple proportions is (a) \(\mathrm{Cu}\) and \(\mathrm{CuSO}_{4}\) (b) \(\mathrm{CuO}\) and \(\mathrm{Cu}_{2} \mathrm{O}\) (c) \(\mathrm{H}_{2} \mathrm{~S}\) and \(\mathrm{SO}_{2}\) (d) \(\mathrm{NH}_{3}\) and \(\mathrm{NCl}_{3}\)
Step-by-Step Solution
Verified Answer
The pair that illustrates the law of multiple proportions is (b) CuO and Cu₂O.
1Step 1: Understanding Law of Multiple Proportions
The law of multiple proportions states that if two elements form more than one compound, the different masses of one element that combine with a fixed mass of the other are in ratios of small whole numbers.
2Step 2: Analyzing Compound Pairs
We need to look at each pair of compounds to see if they consist of the same two elements combining in different ratios, which is key to the law of multiple proportions.
3Step 3: Examine Pair (a)
Pair (a) consists of Cu and CuSO₄. These are not two compounds with the same elements; thus, they do not illustrate the law of multiple proportions.
4Step 4: Examine Pair (b)
In pair (b), CuO and Cu₂O are both compounds made from copper (Cu) and oxygen (O). CuO has a ratio of 1 copper to 1 oxygen atom, while Cu₂O has a ratio of 2 coppers to 1 oxygen atom. The ratio of copper combining with oxygen in these compounds is 1:2.
5Step 5: Examine Pair (c)
Pair (c) consists of H₂S and SO₂. These compounds have different elements (hydrogen and oxygen) combining with sulfur, thus not illustrating the law of multiple proportions.
6Step 6: Examine Pair (d)
Pair (d) includes NH₃ and NCl₃. These compounds involve nitrogen combining with different other elements (hydrogen and chlorine), not the same element at different ratios.
7Step 7: Conclusion
From the analysis, pair (b) with CuO and Cu₂O demonstrates the law of multiple proportions, as they involve the same two elements (copper and oxygen) combining in different ratios.
Key Concepts
Chemical CompoundsElement RatiosChemical BondingJEE Main Chemistry
Chemical Compounds
Chemical compounds are substances made up of two or more elements that are chemically bonded together. The elements are combined in fixed ratios, creating entirely new substances with unique properties different from the individual elements. For example, water (\(\text{H}_2\text{O}\)) is a compound made from the elements hydrogen and oxygen. The fixed ratio in water is two hydrogen atoms for every one oxygen atom.
Chemical compounds can be classified based on the types of elements they contain or the type of bond holding them together. They can be covalent, where electrons are shared between atoms, ionic where electrons are transferred between atoms, or metallic, where electrons are delocalized across metal ions.
Chemical compounds can be classified based on the types of elements they contain or the type of bond holding them together. They can be covalent, where electrons are shared between atoms, ionic where electrons are transferred between atoms, or metallic, where electrons are delocalized across metal ions.
- Covalent compounds: Typically involve non-metal elements combining.
- Ionic compounds: Generally form between metals and non-metals.
- Metallic compounds: Consist of metal elements bonded together.
Element Ratios
Element ratios refer to the proportion of different elements within a compound. For any given compound, these ratios are consistent and define the chemical formula of the compound. This concept is vital in understanding the law of multiple proportions.
For example, in water (\(\text{H}_2\text{O}\)), the ratio of hydrogen to oxygen is 2:1. In the context of the exercise, understanding element ratios helps in identifying if two compounds follow the law of multiple proportions. Consider the compounds CuO and Cu\(_2\)O:
For example, in water (\(\text{H}_2\text{O}\)), the ratio of hydrogen to oxygen is 2:1. In the context of the exercise, understanding element ratios helps in identifying if two compounds follow the law of multiple proportions. Consider the compounds CuO and Cu\(_2\)O:
- CuO has a copper to oxygen ratio of 1:1.
- Cu\(_2\)O has a copper to oxygen ratio of 2:1.
Chemical Bonding
Chemical bonding is the force that holds atoms together within a compound and is a fundamental concept of chemistry. It determines the structure, properties, and behavior of a compound. There are several types of chemical bonds, each with unique characteristics.
- Covalent Bond: Atoms share pairs of electrons to achieve stability. Common in organic compounds and molecules like water and carbon dioxide.
- Ionic Bond: Involves the transfer of electrons from one atom to another, resulting in positively and negatively charged ions. Found in salts like sodium chloride.
- Metallic Bond: Features a sea of delocalized electrons flowing through a lattice of metal ions, giving metals their conductive properties.
JEE Main Chemistry
JEE Main chemistry is an essential subject for students aspiring to enter engineering programs in India. The exam assesses understanding in various topics of chemistry, including principles like the law of multiple proportions, chemical bonding, and element ratios.
- JEE Main requires thorough comprehension of inorganic, organic, and physical chemistry concepts.
- Understanding and applying core concepts like the law of multiple proportions helps in solving analytical problems efficiently.
- Strengthening problem-solving skills through practice exercises in various chemistry topics is key to success in JEE Main exam.
Other exercises in this chapter
Problem 15
The hydrated salt \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{n} \mathrm{H}_{2} \mathrm{O}\) undergoes \(63 \%\) loss in mass on heating and becomes anhydro
View solution Problem 16
The vapour density of a mixture having \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(27.6 .\) The mole fraction \(\mathrm{NO}_{2}\) in the mixt
View solution Problem 18
How many grams of \(\mathrm{KCl}\) must be added to \(75 \mathrm{~g}\) of water to produce a solution with a molality of \(2.25\) (a) \(1.257 \mathrm{~g}\) (b)
View solution Problem 19
Normality of \(0.04 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is (a) \(0.02 \mathrm{~N}\) (b) \(0.01 \mathrm{~N}\) (c) \(0.04 \mathrm{~N}\) (d) \(0.08 \mathrm
View solution