Problem 139 - A Complete Resource Book in Chemistry for JEE Main

Question

A molecule Bn dissolves in water and is non- volatile. A solution of certain molality showed a depression of \(0.93 \mathrm{~K}\) in freezing point. The same solution boiled at \(100.26^{\circ} \mathrm{C}\). When \(7.9 \mathrm{~g}\) of \(\mathrm{Bn}\) was dissolved in \(100 \mathrm{~g}\) water, the solution boiled at \(100.44^{\circ} \mathrm{C}\). Given \(K_{\mathrm{f}}\) for water \(=1.86 \mathrm{~K} \mathrm{~mol}^{-1} \mathrm{~kg}\) and Atomic mass of \(\mathrm{B}=31\) The value of ' \(n\) ' is

Step-by-Step Solution

Answer

The value of 'n' is 3.

Step 1: Calculate Molality from Depression in Freezing Point

The depression in freezing point (ΔTf) is given as 0.93 K. Using the formula for depression in freezing point, \[ \Delta T_f = K_f \cdot m \]where \( K_f \) is the cryoscopic constant and \( m \) is the molality, we can plug in the values:\[ 0.93 = 1.86 \cdot m \]From this, solve for \( m \) (molality):\[ m = \frac{0.93}{1.86} = 0.5 \text{ mol/kg} \]

Step 2: Calculate Molality from Elevation in Boiling Point

We have two boiling points: one for the initial solution (100.26°C), and one when 7.9 g of Bn is dissolved in 100 g of water, causing the boiling point to rise to 100.44°C.Calculate the molality for the initial solution from the elevation in boiling point:\[ 0.26 = K_b \cdot 0.5 \text{ molality} \]Assuming \( K_b = 0.52 \text{ K mol}^{-1} \text{ kg} \) (constant derived for boiling point elevation), check if the molality is consistent. However, here we need to define it: Let the molality during the experiment when additional Bn is added be 'm1.'

Step 3: Determine Molality from Additional Solvent

For the mixture with additional Bn:\[ \Delta T_b = (100.44 - 100) = 0.44 \text{ K} \]Now use the formula:\[ \Delta T_b = K_b \cdot m \]Plugging the values:\[ 0.44 = 0.52 \cdot m \]Solving for \( m \) gives:\[ m = \frac{0.44}{0.52} \approx 0.846 \text{ mol/kg} \]

Step 4: Determine Molar Mass using Molality

With molality calculated, calculate the molar mass of the compound Bn:Given 7.9 g of Bn in 100 g of water (0.1 kg):\[ m = \frac{\text{moles of } Bn}{0.1} = 0.846 \]Solving for moles of Bn:\[ \text{moles of } Bn = 0.846 \times 0.1 = 0.0846 \]Molar mass of Bn (M):\[ M = \frac{7.9}{0.0846} = 93.3 \text{ g/mol} \]

Step 5: Calculate 'n' from Molar Mass

Knowing the atomic mass of B is 31, the formula Bn indicates B's molar mass is 31n.Hence:\[ 31n = 93.3 \]Solving for 'n':\[ n = \frac{93.3}{31} \approx 3 \]Thus, 'n' can be rounded to the nearest whole number being 3.

Key Concepts

Depression in Freezing Point

Depression in freezing point is a fascinating colligative property that describes how the presence of a solute lowers the freezing point of a solvent. But why does this happen? When a non-volatile solute, like the molecule Bn, is dissolved in a solvent such as water, it disrupts the extent to which water molecules can organize into a solid structure (ice). This disruption requires a lower temperature to form ice, hence, the freezing point is depressed.

Understanding the calculation of depression in freezing point involves the formula:

• \( \Delta T_f = K_f \cdot m \)

where:

• \( \Delta T_f \) is the depression in freezing point (0.93 K in this case)
• \( K_f \) is the cryoscopic constant of the solvent (1.86 K mol^-1 kg for water)
• \( m \) is the molality of the solution

By substituting the known values, you can find that the molality, or concentration, of the solution is 0.5 mol/kg. This calculation helps us understand how the addition of solute particles affects the thermodynamic properties of solvents.

Elevation in Boiling Point

Just like depression in freezing point, elevation in boiling point is another key colligative property. It refers to the increase in boiling temperature of a solvent when a solute is dissolved. This occurs because solute particles dissolve among the solvent molecules, requiring additional energy to overcome intermolecular forces for the solvent to boil.

The formula used to determine the elevation in boiling point is:

• \( \Delta T_b = K_b \cdot m \)

Here:

• \( \Delta T_b \) represents the elevation in boiling point
• \( K_b \) is the ebullioscopic constant (assumed here as 0.52 K mol^-1 kg)
• \( m \) is the molality

In the exercise, the boiling point of the solution initially rose to 100.26°C, and further to 100.44°C when additional Bn was added. By solving the equations based on the changes in temperature, you can derive the molalities before and after the additional Bn was dissolved, helping to understand how solutes impact the boiling point behavior of solvents.

Molality

Molality is a measure of the concentration of a solution, representing the moles of solute per kilogram of solvent. It is particularly useful for colligative properties because it does not change with temperature or pressure. Unlike molarity which depends on the volume, molality is purely mass-based where the formula is:

• \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \)

In the exercise, when 7.9 grams of Bn were dissolved in 100 grams of water, the concept of molality was key in calculating the concentration of the solute. You take the mass of the solvent (0.1 kg in this case), together with the number of moles derived from the mass of Bn and its molar mass, to find the molality.
Understanding molality is essential, as it's the basis for accurately determining how solutes will alter the physical properties of a solution under varying conditions of temperature and pressure.