Problem 83 - A Complete Resource Book in Chemistry for JEE Main

Question

\(0.25 \mathrm{~mol}\) of \(\mathrm{P}_{4}\) molecules contains atoms. (a) \(1.764 \times 10^{23}\) (b) \(6.02 \times 10^{19}\) (c) \(6.023 \times 10^{23}\) (d) \(8.086 \times 10^{25}\)

Step-by-Step Solution

Answer

The answer is (c) \(6.023 \times 10^{23}\).

Step 1: Understanding the Problem

We need to determine the total number of atoms in \(0.25\, \text{mol}\) of \(\text{P}_4\) molecules. Each \(\text{P}_4\) molecule consists of 4 phosphorus atoms.

Step 2: Calculate Number of Molecules

Using Avogadro's number \(6.022 \times 10^{23}\), which gives the number of units (molecules, atoms, etc.) per mole, we find the number of \(\text{P}_4\) molecules in \(0.25\, \text{mol}\): \[ 0.25 \, \text{mol} \times 6.022 \times 10^{23} \frac{\text{molecules}}{\text{mol}} = 1.5055 \times 10^{23} \text{ molecules} \]

Step 3: Calculate Number of Atoms

Since each \(\text{P}_4\) molecule has 4 atoms, we multiply the number of molecules by 4 to get the total number of atoms:\[ 1.5055 \times 10^{23} \text{ molecules} \times 4 \text{ atoms/molecule} = 6.022 \times 10^{23} \text{ atoms} \]

Step 4: Match the Answer with Options

Comparing our result, \(6.022 \times 10^{23}\) atoms, with the given options, we see that it closely matches option (c) \(6.023 \times 10^{23}\).

Key Concepts

Avogadro's number

Avogadro's number is a fundamental constant in chemistry that is widely used in a variety of molecular calculations. It is defined as the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. This number is approximately \(6.022 \times 10^{23}\). Avogadro's number provides a bridge between the macroscopic scale we experience and the atomic scale, which is beyond direct perception.

By using Avogadro's number, we can easily transition from dealing with grams of substances to the number of molecules or atoms. This is essential for ensuring accurate stoichiometry in chemical reactions, allowing chemists to predict and measure how reactions occur on a molecular level.

Phosphorus atoms

With the chemical symbol \(\text{P}\), phosphorus is an essential element in many biological molecules and industrial applications. In the context of molecules like \(\text{P}_4\), phosphorus atoms form a tetrahedral structure containing four atoms.

Identifying the number of phosphorus atoms in a compound involves understanding the molecular formula. For example, \(\text{P}_4\) shows that each molecule is composed of four phosphorus atoms. This understanding is crucial for calculating the total number of atoms when initially given the amount in moles. Counting the atoms in a molecule helps chemists in determining the proportions needed for reactions and their successful outcomes.

Molecular calculation

Molecular calculation involves using moles to determine quantities related to molecules. Calculations often necessitate the conversion of moles to molecules, and further to atoms when needed. In these calculations, Avogadro's number plays a central role.

For instance, determining the number of \(\text{P}_4\) molecules in \(0.25\, \text{mol}\) requires multiplying the moles by Avogadro's number. Calculating the number of molecules allows us to further dissect into atomic levels, enhancing our understanding of the compound at a fundamental scale. These calculations are fundamental in fields like biochemistry and pharmacology, where precise molecular impacts define outcomes.

Stoichiometry

Stoichiometry is the method used to calculate the relative quantities of reactants and products in chemical reactions. It hinges on the principle of conservation of mass and the mole concept. It ensures balanced chemical equations are used to figure out how much of each substance participates in the reaction.

Understanding stoichiometry involves knowing the ratio of molecules or atoms involved in reactions. For example, in the calculation of phosphorus atoms in \(\text{P}_4\), knowing that \(0.25\, \text{mol}\) contains \(6.022 \times 10^{23}\) atoms illustrates the practical application of stoichiometry in determining quantities. It enables chemists to calculate the precise amounts required or produced, indispensable for industrial applications and laboratory work.