Problem 46 - A Complete Resource Book in Chemistry for JEE Main

Question

3.65 grams of \(\mathrm{HCl}\) is dissolved in \(16.2\) grams of water The mole fraction of \(\mathrm{HCl}\) in the resulting solution is (a) \(0.4\) (b) \(0.3\) (c) \(0.2\) (d) \(0.1\)

Step-by-Step Solution

Answer

The mole fraction of HCl is 0.1, which corresponds to option (d).

Step 1: Calculate Moles of HCl

To find the number of moles of HCl, divide the mass of HCl by its molecular weight. The molecular weight of HCl is approximately 36.46 g/mol. Thus, calculate the moles of HCl as follows: \[ \text{moles of HCl} = \frac{3.65 \text{ g}}{36.46 \text{ g/mol}} \approx 0.10 \text{ mol} \]

Step 2: Calculate Moles of Water

Similarly, calculate the moles of water in the solution. The molecular weight of water (H2O) is approximately 18.02 g/mol. Thus, the moles of water are: \[ \text{moles of water} = \frac{16.2 \text{ g}}{18.02 \text{ g/mol}} \approx 0.90 \text{ mol} \]

Step 3: Calculate Total Moles in Solution

Add the moles of both HCl and water to obtain the total moles in the solution: \[ \text{total moles} = 0.10 \text{ mol} + 0.90 \text{ mol} = 1.00 \text{ mol} \]

Step 4: Determine Mole Fraction of HCl

The mole fraction of a component is calculated by dividing the moles of that component by the total moles in the solution. For HCl, the mole fraction is given by: \[ \text{mole fraction of HCl} = \frac{0.10 \text{ mol}}{1.00 \text{ mol}} = 0.10 \]

Step 5: Identify the Correct Answer

Based on the calculated mole fraction of HCl, which is 0.10, the correct option in the given choices is \((d)\).

Key Concepts

Mole Calculation

Mole calculation is a fundamental concept in chemistry often used to convert between mass and the number of particles. To perform a mole calculation, you divide the given mass of a substance by its molecular weight. This gives you the number of moles present. For instance, in the case of hydrochloric acid (HCl), you would calculate its moles as follows:

• First, identify the mass of your sample, such as 3.65 grams of HCl.
• Then, determine its molecular weight, which for HCl, is about 36.46 g/mol.
• Finally, you divide the mass by the molecular weight: \[\text{moles of HCl} = \frac{3.65}{36.46} \approx 0.10 \text{ mol}\]

Each step is crucial to understanding how particles interact in a chemical solution. Whether you're adding substances together or trying to find empirical formulas, mastering mole calculation will help you make precise chemical assessments.

Molecular Weight

Molecular weight, often known as molar mass, is a key concept that helps us translate between the micro world of atoms and the more tangible macro world of grams and liters. Calculating molecular weight involves adding up the atomic weights of all atoms in a molecule. Let's see how it works:

• Identify the chemical formula, for example, HCl for hydrochloric acid.
• Look at each atomic weight: Hydrogen (H) - 1.01 g/mol and Chlorine (Cl) - 35.45 g/mol.
• Add them together to get the total molecular weight: \[\text{Molecular Weight of HCl} = 1.01 + 35.45 \approx 36.46 \text{ g/mol}\]

This value is used in mole calculations to convert grams to moles, allowing for detailed exploration of chemical reactions. It's a foundational skill for any chemistry student aiming to understand how substances combine and react on a molecular level.

Chemical Solution

A chemical solution is a homogeneous mixture where substances (solutes) are dissolved in a solvent. When dealing with solutions, understanding the interplay of components is critical for tasks like calculating concentrations or mole fractions.

• The solvent is usually present in the largest amount; in this context, water acts as the solvent.
• The solute, such as HCl, is the substance dissolved in the solvent.
• To assess the composition of a solution, chemical calculations like mole fractions are used, representing the ratio of moles of a solute to the total moles in the solution.\[\text{Mole fraction} = \frac{\text{moles of solute}}{\text{total moles of solution}} = \frac{0.10}{1.00} = 0.10\]

These calculations are essential for predicting how a solution behaves in different chemical environments, such as dilution or mixing with other solutions. Understanding solutions also lays the groundwork for complex topics like reaction kinetics and equilibrium.