Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

The elevation in boiling point for \(13.44 \mathrm{~g}\) of \(\mathrm{CuCl}_{2}\) dissolved in \(1 \mathrm{~kg}\) of water as solvent will be \(\left(\mathrm{K}_{\mathrm{b}}=0.52\right.\) \(\mathrm{kg} / \mathrm{J}\), molar mass of \(\left.\mathrm{CuCl}_{2}=134.4 \mathrm{~g} / \mathrm{mol}\right)\) (a) \(0.05\) (b) \(0.10\) (c) \(0.16\) (d) \(0.20\).

Step-by-Step Solution

Answer

The elevation in boiling point is 0.16, corresponding to option (c).

Step 1: Identify Known Values

First, identify all the given values in the problem. We are given that the mass of CuCl\(_2\) is 13.44 g, the mass of water is 1 kg, K\(_b\) is 0.52 kg/J, and the molar mass of CuCl\(_2\) is 134.4 g/mol.

Step 2: Calculate Moles of Solute

Next, calculate the moles of CuCl\(_2\). Use the formula for moles, \(n = \frac{\text{mass}}{\text{molar mass}}\). Therefore, \(n = \frac{13.44 \text{ g}}{134.4 \text{ g/mol}} = 0.1 \text{ mol}\).

Step 3: Determine Van't Hoff Factor

CuCl\(_2\) dissociates into 3 ions (one Cu\(^{2+}\) and two Cl\(^-\)) in solution. Thus, the Van't Hoff factor, \(i\), is 3.

Step 4: Use Boiling Point Elevation Formula

Use the boiling point elevation formula: \(\Delta T_b = i \cdot K_b \cdot m\), where \(m\) is the molality. Here, \(m = \frac{0.1 \text{ mol}}{1 \text{ kg}} = 0.1 \text{ mol/kg}\). So, \(\Delta T_b = 3 \times 0.52 \times 0.1 = 0.156\).

Step 5: Round to Nearest Given Option

The calculated elevation in boiling point is 0.156, which rounds to 0.16 when compared with the options provided. Therefore, the solution matches option (c) 0.16.

Key Concepts

Van't Hoff Factor

The Van't Hoff factor, often represented by the symbol \( i \), is a crucial component in understanding how solutes affect the properties of solvents. This factor is linked to the number of particles a compound forms when it dissolves. It essentially counts how many separate particles or ions are in the solution after dissociation.

For example, if you dissolve a non-ionic compound like glucose, it doesn't separate into ions, so it retains its identity as one particle, making its Van't Hoff factor \( i = 1 \). In contrast, when calcium chloride (\( \text{CaCl}_2 \)) is dissolved, it splits into one \( \text{Ca}^{2+} \) ion and two \( \text{Cl}^- \) ions, giving a Van't Hoff factor of \( i = 3 \). This factor influences the colligative properties, such as boiling point elevation and freezing point depression, by determining the effective concentration of particles in a solution.

Molality

Molality is a measure of the concentration of a solute in a solution. Unlike molarity, which is based on volume, molality relates to mass. It is defined as the number of moles of solute per kilogram of solvent, represented by \( m \).

To find the molality, the formula used is:

• \( m = \frac{n}{mass \text{ of solvent (kg)}} \)

Where \( n \) is the number of moles of solute.

Molality is especially useful in calculations involving temperature because it does not change with temperature variations as volume can. For example, when given that 0.1 moles of a substance are dissolved in 1 kg of solvent, the molality is straightforwardly \( 0.1 \text{ mol/kg} \). Using molality in calculations related to boiling point elevation or freezing point depression ensures accuracy.

Molar Mass Calculation

Determining the molar mass is a fundamental step in many chemistry calculations, such as when converting between grams and moles. The molar mass, usually expressed in grams per mole (g/mol), is the mass of one mole of a given substance.

The process to calculate molar mass follows straightforward principles:

• Identify the chemical formula of the compound.
• Sum the atomic masses of each element in the molecule. These atomic masses are usually found on the periodic table.
• Multiply by the number of atoms of each element present.

For example, in \( \text{CuCl}_2 \), copper (Cu) has an atomic mass of approximately 63.5 g/mol, and chlorine (Cl) has an atomic mass of about 35.5 g/mol. Since there are two chlorine atoms, the calculation would be:

• \( 63.5 + 2 \times 35.5 = 134.5 \text{ g/mol} \)

This method is vital for accurately performing calculations that involve moles, such as determining the amount of solute needed for a desired concentration.