Problem 31 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

How much amount of \(\mathrm{NaCl}\) should be added to \(600 \mathrm{~g}\) of water \((\mathrm{r}=\) \(1.00 \mathrm{~g} / \mathrm{mL}\) ) to decrease the freezing point of water to \(-0.2^{\circ} \mathrm{C}\) ? (The freezing point depression constant for water \(=2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) )

Step-by-Step Solution

Answer

Add 1.75 grams of \(\mathrm{NaCl}\).

Step 1: Identify the formula for freezing point depression

The formula to calculate freezing point depression is given by \[ \Delta T_f = i \cdot K_f \cdot m \]where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant, and \( m \) is the molality of the solution.

Step 2: Determine the van't Hoff factor for NaCl

Sodium chloride \((\mathrm{NaCl})\) dissociates into two ions: Na+ and Cl-. Thus, the van't Hoff factor \( i \) for NaCl is 2.

Step 3: Calculate the change in freezing point

Given that the freezing point is lowered to \(-0.2^{\circ} \mathrm{C}\), the change in freezing point \( \Delta T_f \) is \[ \Delta T_f = 0^{\circ} \mathrm{C} - (-0.2^{\circ} \mathrm{C}) = 0.2 \mathrm{~C} \]

Step 4: Rearrange the formula to solve for molality

Using the formula from Step 1, rearrange it to solve for the molality \( m \):\[ m = \frac{\Delta T_f}{i \cdot K_f} = \frac{0.2}{2 \cdot 2} \]

Step 5: Calculate molality

Substitute the known values into the rearranged formula:\[ m = \frac{0.2}{4} = 0.05 \mathrm{~mol/kg} \]

Step 6: Convert molality to moles of NaCl

Since molality is moles of solute per kilogram of solvent, and we have 0.6 kg of water:\[ \text{moles of } \mathrm{NaCl} = 0.05 \times 0.6 = 0.03 \mathrm{~mol} \]

Step 7: Calculate the mass of NaCl needed

Find the molar mass of NaCl, which is 58.44 g/mol, and calculate the mass required:\[ \text{mass of } \mathrm{NaCl} = 0.03 \times 58.44 = 1.7532 \mathrm{~g} \]

Step 8: Final Answer

Approximately 1.75 grams of \( \mathrm{NaCl} \) should be added to the water to decrease its freezing point to \(-0.2^{\circ} \mathrm{C}\).

Key Concepts

Understanding the Van't Hoff Factor

The van't Hoff factor is an essential concept in solution chemistry when determining how the presence of a solute influences properties like boiling and freezing points. It represents the number of particles a compound dissociates into when dissolved in solution. For NaCl, a simple ionic compound, dissolution in water allows it to separate into two ions: sodium (Na+) and chloride (Cl-). This means the van't Hoff factor for NaCl is 2, since each formula unit of NaCl becomes two particles in solution.

It’s important to understand that the van't Hoff factor affects colligative properties, which depend on the number of dissolved particles rather than their nature. So for solutes that dissociate in solution, the van't Hoff factor provides a corrected measure of concentration when calculating these properties, like the freezing point depression.

In exercises like this one, identifying the correct van't Hoff factor is critical because it directly affects calculations related to the change in temperature of the solution.

The Process of Molality Calculation

Molality is a concentration unit used extensively in solution chemistry. It is defined as moles of solute per kilogram of solvent. Unlike molarity, which involves solution volume, molality depends solely on the mass, making it valuable in temperature-sensitive calculations, such as freezing point depression.

To find molality, we use the rearranged form of the freezing point depression formula: \[ m = \frac{\Delta T_f}{i \cdot K_f} \]where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, and \( K_f \) is the freezing point depression constant. This provides the means to calculate how much solute is dissolved per unit mass of solvent.

In this exercise, with water as the solvent and NaCl as the solute, we calculate the molality as 0.05 mol/kg. Given we have 0.6 kg of water, knowing the molality helps us determine the moles of NaCl required, ensuring accurate results in lowering the freezing point by the desired amount.

Exploring Solution Chemistry

Solution chemistry revolves around understanding how substances dissolve and interact in a liquid to form solutions. The exercise on freezing point depression is a prime example of how different factors, such as solute concentration and dissociation, affect the physical properties of a solution.

In solution chemistry, we often encounter terms like solute, solvent, and solution concentration, each critical to predicting how solutions will behave. When NaCl dissolves in water, it becomes the solute, while water acts as the solvent. The concentration is indicated by terms like molality, which reports the ratio of solute in relation to the solvent's mass.

Understanding such concepts helps predict and calculate changes in solution behavior when a solute is added, impacting freezing or boiling points, among other properties. This foundational knowledge is critical in fields ranging from chemistry to environmental science and engineering.