Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Pure water freezes at \(273 \mathrm{~K}\) and 1 bar. The addition of \(34.5 \mathrm{~g}\) of ethanol to \(500 \mathrm{~g}\) of water changes the freezing point of the solution. Use the freezing point depression constant of water as \(2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\). The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is \(\left.46 \mathrm{~g} \mathrm{~mol}^{-1}\right]\) Among the following, the option representing change in the freezing point is (a) (b) (c) (d)

Step-by-Step Solution

Answer

The freezing point is lowered to 270 K, indicating a 3 K decrease.

Step 1: Calculate the number of moles of ethanol

First, we need to determine how many moles of ethanol are present.The formula for moles is given by:\[ \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \]Substituting the given values:\[ \text{moles of ethanol} = \frac{34.5\, \text{g}}{46\, \text{g/mol}} \approx 0.75\, \text{mol} \]

Step 2: Calculate Molality

Molality (\(m\)) is defined as the number of moles of solute (ethanol in this case) per kilogram of solvent (water).To calculate molality:\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]The mass of water is given as 500 g, which is equivalent to 0.5 kg. Therefore, the molality is:\[ m = \frac{0.75\, \text{mol}}{0.5\, \text{kg}} = 1.5\, \text{mol/kg} \]

Step 3: Use Freezing Point Depression Formula

The freezing point depression formula is:\[ \Delta T_f = K_f \cdot m \]where \(\Delta T_f\) is the change in the freezing point, \(K_f\) is the freezing point depression constant, and \(m\) is the molality calculated earlier.Substituting the known values:\[ \Delta T_f = 2\, \text{K} \cdot 1.5\, \text{mol/kg} = 3\, \text{K} \]

Step 4: Determine the new freezing point

The change in freezing point means the freezing point has decreased by the calculated \( \Delta T_f \).The original freezing point of water is 273 K.The new freezing point is:\[ 273\, \text{K} - 3\, \text{K} = 270\, \text{K} \]

Step 5: Select the correct option

Now, based on the calculated new freezing point of 270 K, select the option from the given choices that best represents a change corresponding to a decrease of 3 K. Since the option selections (a, b, c, d) are not shown here, you would choose the option indicating a freezing point at 270 K.

Key Concepts

Understanding Molality

Molality is a crucial concept when discussing solutions, especially when dealing with properties like freezing point depression. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality is not affected by changes in temperature or pressure because it is based on mass, not volume.

• Formula: Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \)
• Units: \( \text{mol/kg} \)

In the context of our exercise, the solvent is water, and the solute is ethanol. We calculated the molality as 1.5 mol/kg, using the fact that there are 0.75 moles of ethanol dissolved in 0.5 kg of water. This calculation is important because it is directly used to determine the change in freezing point using the freezing point depression constant.

Exploring Vapor Pressure

Vapor pressure is another important property to consider when studying solutions. It refers to the pressure exerted by a vapor in equilibrium with its liquid or solid phase. This is crucial when studying colligative properties, which include boiling point elevation and freezing point depression, as these depend on the presence of solute particles rather than the identity of those particles.

• When you add a solute to a solvent, the vapor pressure of the solvent generally decreases.
• This happens because the solute particles occupy surface space, reducing the number of solvent molecules at the surface that can evaporate.

In our exercise, even though vapor pressure is not directly measured or adjusted, understanding its role helps us grasp how solutes affect the physical properties of a solvent, such as changing the freezing point through vapor pressure lowering.

Mole Calculations Made Easy

Mole calculations are foundational for understanding many topics in chemistry, including freezing point depression. These calculations are essential for determining the molality of a solution and involve using the molar mass of a substance.

• Formula to calculate moles: \( \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \)
• This allows conversion from grams, a more accessible measure, to moles, which is a chemical standard.

In the exercise, we calculated moles of ethanol by dividing its mass (34.5 g) by its molecular weight (46 g/mol), resulting in approximately 0.75 mol. Understanding this step is vital because it directly leads to calculating the molality, which is further used in applying the freezing point depression formula.