Suppose \(5 \mathrm{~g}\) of acetic acid are dissolved in one litre of ethanol. Assume no reaction in between them. Calculate molality of resulting solution if density of ethanol is \(0.789 \mathrm{~g} / \mathrm{mL}\).

Understanding molar mass is crucial when delving into chemistry problems, especially when calculating solution concentration. Molar mass, indicated in grams per mole (g/mol), refers to the weight of one mole of a chemical compound or element. One mole, by definition, contains Avogadro's number of particles, which is approximately 6.022 x 1023 particles.

To calculate the molar mass, one simply adds up the atomic masses of all the atoms in the molecule. For instance, acetic acid (CH3COOH) consists of 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms. Using the periodic table, the atomic masses are as follows: Carbon (C) has a molar mass of approximately 12.01 g/mol, Hydrogen (H) 1.01 g/mol, and Oxygen (O) 16.00 g/mol. Therefore, the molar mass of acetic acid is calculated as: \[ (2 \times 12.01\text{ g/mol}) + (4 \times 1.01\text{ g/mol}) + (2 \times 16.00\text{ g/mol}) = 60.05\text{ g/mol} \].

Using the molar mass, one can convert the mass of a substance in grams to its number of moles, crucial for a plethora of chemistry problems. The formula to find the number of moles (\( n \)) is: \[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]. This step is vital for molality calculations and is a typical concept tested in competitive exams like IIT-JEE.

Solution concentration describes how much solute is present in a given amount of solvent. Concentration can be expressed in various ways, including molarity, molality, percent composition, and others. In the context of molality, which is the focus of our exercise, concentration is defined by the number of moles of solute per kilogram of solvent.

The formula for molality (\( m \)) is given as: \[ m = \frac{n_{\text{solute}}}{m_{\text{solvent~in~kg}}} \]. To apply this formula, one must know the number of moles of the solute and the mass of the solvent in kilograms. Molality is particularly useful because, unlike molarity, it is not affected by temperature changes, as it relies on mass rather than volume.

In practice, to find the molality of a solution, first calculate the solvent's mass (in kilograms) using its volume and density, as seen in our example with ethanol. Then determine the number of moles of solute, using the molar mass as previously discussed. Dividing the moles of solute by the solvent's mass gives the molality. Molality problems not only are common in coursework but are also integral in competitive exams such as the IIT-JEE, where precise knowledge and application of solution concentration is essential.

Solving chemistry problems for competitive exams like the IIT-JEE requires a solid understanding of the concepts involved and meticulous application of the principles. The IIT-JEE, known for its challenging questions, often tests students on their ability to apply basic concepts to solve complex problems.

For instance, calculating the molality of a solution may seem straightforward, but it involves several steps that are foundational chemistry knowledge, such as understanding molar mass, converting between units, and effectively using the molality formula. Candidates must not only solve problems accurately but also quickly, which necessitates a high level of proficiency and extensive practice.

Success in the IIT-JEE means being adept at such calculations and having the ability to manipulate and apply various concepts across physical, organic, and inorganic chemistry. Aspiring students should regularly solve sample problems, understand the theory behind the formulas they use, and apply these formulas in various scenarios to build confidence and agility in solving typical IIT-JEE chemistry problems.