Question

The relation between molarity \((M)\) and molality \((m)\) is given by : \(\left(\rho=\right.\) density of solution \((\mathrm{mg} / \mathrm{mL}), M_{1}=\) molecular weight of solute) (a) \(m=\frac{1000 M}{1000 \rho-M_{1}}\) (b) \(m=\frac{1000 \rho M}{1000 \rho-M M_{1}}\) (c) \(m=\frac{1000 \mathrm{MM}}{1000 \rho-M M_{1}}\) (d) \(m=\frac{1000 M}{1000 \rho-M M_{1}}\)

Short answer

The correct relation between molarity (M) and molality (m) that accounts for density and molecular weight is: (d) \(m=\frac{1000 M}{1000 \rho - M M_{1}}\).

Step-by-step solution

- Understand Molarity and Molality

Molarity (M) is defined as the number of moles of solute per liter of solution, whereas molality (m) is the number of moles of solute per kilogram of solvent. Recognize that molarity involves the total volume of the solution, and molality focuses on the weight of the solvent alone.

- Analyze the Equations

To find the correct relationship between molarity and molality, there must be terms that account for the volume and the density of the solution, as well as compensate for the molecular weight of the solute.

- Conversion between Mass and Volume

We know that mass (in grams) can be converted to volume (in milliliters) using density (\(\rho\)) as follows: Mass = \(\rho\) x Volume. To consider the mass of the solvent specifically, we must subtract the mass contributed by the solute, which is the number of moles of solute (Molarity x Volume) times its molecular weight (\(M_1\)).

- Derive Molality from Molarity

Using the relationship Mass = \(\rho\) x Volume, and knowing that molality is moles of solute per kilogram of solvent, we need an expression that gives the moles of solute per 1000 grams (1 kg) of solvent - not the solution. The correct formula must have a numerator of 1000 times the molarity (to give moles) and a denominator that subtracts the mass contribution of the solute from the total mass of the solution.

- Select the Correct Formula

Out of the given options, the correct formula should correctly transform molarity into molality using the density, the molecular weight, and the mass-volume relationship. Thus, the correct equation is (d): \(m=\frac{1000 M}{1000 \rho - M M_{1}}\), which represents the shift from the total solution's volume to the solvent's mass, adjusting for the solute's contribution.

Key concepts

Understanding Molarity in Physical Chemistry

Molarity is a central concept in physical chemistry that measures the concentration of a solution by indicating how many moles of solute are present per liter of solution. As a tool crucial for predicting the outcomes of chemical reactions, molarity is often the starting point for stoichiometric calculations and helps chemists understand reaction rates and equilibrium states.

In the context of JEE Chemistry, a clear understanding of molarity is vital not only for theoretical questions but also for practical laboratory applications. Molarity calculations are imperative when preparing solutions for titrations, an essential technique to determine unknown concentrations. Therefore, molarity isn't just a concept to be memorized; it's a practical tool that bridges classroom learning with real-world laboratory applications.

Molality and its Significance in Solution Concentration

Molality, on the other hand, measures solution concentration in terms of the number of moles of solute per kilogram of solvent. Unlike molarity, molality is not affected by temperature as it relies on the mass of the solvent, not its volume. This characteristic makes molality particularly useful in situations where temperature fluctuates, such as in boiling point elevation or freezing point depression studies.

Molality is another key concept tested in the JEE Chemistry syllabus. Understanding its application is fundamental when exploring colligative properties, which are properties depending on the number of particles in a solution. As colligative properties are part of the JEE Chemistry and other competitive exams, mastering molality is a crucial stepping stone for any chemistry student.

Relating Molarity to Molality for Solution Concentration

In solving problems related to molarity and molality, the transition from one measure to another must take into account the density of the solution and the molecular weight of the solute. This process involves a precise mathematical relationship that allows chemistry students to contextually apply these concepts.

The relation given in the exercise underlines the importance of formula manipulation in physical chemistry. A clear understanding of the correct formula – which, through meticulous deduction, is determined as option (d) – enables students to seamlessly interconvert these concentration units. This skill in mathematical manipulation is not only tested in the JEE exams but also reflects a deeper chemical understanding essential for any aspiring chemist.

Chemical Education and Its Approach to Teaching Solution Concepts

Chemical education is tasked with the challenge of making complex concepts like molarity and molality accessible and understandable to students. A key educational strategy is to provide step-by-step solutions that lay out logical reasoning paths from question to conclusion. By dissecting the relationships between different units of concentration, educators can better equip students with the problem-solving skills needed to excel in competitive exams like the JEE and in their future academic and professional careers in chemistry.

Incorporating these step-by-step explanations not only helps students with their homework but also builds their foundational knowledge. Explaining the practical applications of molarity and molality in various chemical procedures enhances students' understanding and appreciation of chemistry's role in the real world.