Problem 62 - A Complete Resource Book in Chemistry for JEE Main

Question

The rise in the boiling point of a solution containing \(1.8 \mathrm{~g}\) of glucose in \(100 \mathrm{~g}\) of solvent is \(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is (a) \(1 \mathrm{~K} / \mathrm{m}\) (b) \(0.1 \mathrm{~K} / \mathrm{m}\) (c) \(0.01 \mathrm{~K} / \mathrm{m}\) (d) \(10 \mathrm{~K} / \mathrm{m}\)

Step-by-Step Solution

Answer

(a) 1 K/m

Step 1: Determine the molality of the solution

The molality \(m\) of a solution is given by the formula \(m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\). First, calculate the moles of glucose. The molecular weight (molar mass) of glucose \(\text{C}_6\text{H}_{12}\text{O}_6\) is approximately \(180\, \text{g/mol}\). So, the moles of glucose are \(\frac{1.8\, \text{g}}{180\, \text{g/mol}} = 0.01\, \text{mol}\). The mass of the solvent (which is the same as the weight given) is \(0.1\, \text{kg}\). Thus, molality \(m\) is \(\frac{0.01}{0.1} = 0.1\, \text{m}\).

Step 2: Apply the boiling point elevation formula

The boiling point elevation \(\Delta T_b\) can be calculated using the formula \(\Delta T_b = K_b \cdot m\), where \(K_b\) is the ebullioscopic constant (molal elevation constant) and \(m\) is the molality. We know \(\Delta T_b = 0.1^{\circ}\, \text{C}\), and from Step 1, \(m = 0.1\, \text{m}\). Thus, \(0.1 = K_b \cdot 0.1\).

Step 3: Solve for the molal elevation constant \(K_b\)

Rearrange the formula from Step 2 to solve for \(K_b\): \(K_b = \frac{0.1}{0.1} = 1\). This means the molal elevation constant \(K_b\) is \(1\, \text{K/m}\).

Key Concepts

Molal Elevation Constant

In the context of boiling point elevation, the molal elevation constant, often represented as \(K_b\), plays a vital role. This constant varies depending on the solvent being used. It represents the change in the boiling point of the solvent for a 1 molal solution of a non-volatile, non-ionizing solute. A higher \(K_b\) value indicates that the solvent's boiling point will increase more significantly when a solute is added. This constant is typically given in units of K/m (kelvin per molality).
Calculating the boiling point elevation involves multiplying the molal elevation constant by the molality of the solution, as seen in the formula \(\Delta T_b = K_b \cdot m\). In the example provided, where the solution's boiling point rises by \(0.1^{\circ}C\), we rearranged the formula to find that \(K_b = 1\, \text{K/m}\). This means that in this solvent, a 1 molal solution would elevate the boiling point by \(1^{\circ}C\).
This universal principle underlies heating processes and other applications across chemistry, giving students insight into why salt on roads prevents freezing or how anti-freeze works.

Molality Calculation

Molality, symbolized by \(m\), is a measure of a solute's concentration in a solution. It differs from molarity, as it is defined without regard to volume changes, which can occur with temperature shifts. The formula to calculate molality is \(m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\).
Let's apply this concept to a practical example to illustrate how it's calculated. Suppose you dissolve \(1.8 \text{ g}\) of glucose in \(100 \text{ g}\) of a solvent. First, convert the mass of glucose into moles using its molar mass. Knowing glucose has a molar mass of approximately \(180 \text{ g/mol}\), calculate moles by using the formula \(\frac{1.8 \text{ g}}{180 \text{ g/mol}}\), resulting in \(0.01 \text{ mol}\).
Transforming the solvent's weight (\(100 \text{ g}\)) into kilograms gives \(0.1 \text{ kg}\). Substituting these values into the molality formula yields \(0.1 \text{ m}\), demonstrating the concentration of the glucose solution as used in calculating boiling point elevation.

Glucose Molar Mass

Glucose is a fundamental sugar molecule in biochemistry, with the chemical formula \(\text{C}_6\text{H}_{12}\text{O}_6\). Its molar mass is an essential value, especially when determining the amount of glucose needed in various chemical reactions or solutions.
The calculation of glucose’s molar mass involves summing up the atomic masses of all the atoms present in a molecule of glucose. Carbon (C) has an atomic mass of about \(12, \text{g/mol}\), hydrogen (H) is approximately \(1, \text{g/mol}\), and oxygen (O) \(16, \text{g/mol}\). Accordingly, the molar mass of glucose is calculated as:

• 6 Carbons: \(6 \times 12 = 72 \text{ g/mol}\)
• 12 Hydrogens: \(12 \times 1 = 12 \text{ g/mol}\)
• 6 Oxygens: \(6 \times 16 = 96 \text{ g/mol}\)

The total molar mass of glucose, therefore, is \(72 + 12 + 96 = 180 \text{ g/mol}\).
Understanding glucose's molar mass is critical for preparing solutions with precise molalities, which are vital in experiments, ensuring proper concentration levels and maintaining accurate results in scientific research.