Problem 89
Question
In a depression in freezing point experiment, it is found that (1) the vapour pressure of the solution is less than that of pure solvent (2) the vapour pressure of the solution is more than that of pure solvent (3) only solute molecules solidify at the freezing point (4) only solvent molecules solidify at the freezing point (a) 1,2 (b) 2,3 (c) 1,4 (d) 1,3
Step-by-Step Solution
Verified Answer
Option (c) 1,4 is correct.
1Step 1: Understanding the Freezing Point Depression
Freezing point depression is a colligative property that occurs when a solute is dissolved in a solvent, resulting in a lower freezing point compared to the pure solvent. This means that the presence of solute particles disrupts the formation of the solid phase (ice or crystal) of the solvent, thus lowering the freezing point.
2Step 2: Vapour Pressure Relation with Solute and Solvent
When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases compared to that of the pure solvent because the solute particles occupy surface positions and reduce the number of solvent molecules that can escape into the vapour phase. Therefore, the statement where the vapour pressure of the solution is less than that of the pure solvent is correct (Option 1).
3Step 3: Solidification Process at Freezing Point
At the freezing point of the solution, the solvent molecules primarily solidify because solute particles generally remain in the liquid phase. Solute molecules can only solidify if they reach saturation or specific conditions, which is not usually the case in a freezing point depression scenario. Thus, only solvent molecules solidify is the accurate statement (Option 4).
4Step 4: Selecting the Correct Option
From the analysis, the accurate statements are: the vapour pressure of the solution is less than that of the pure solvent (1) and only solvent molecules solidify at the freezing point (4). Therefore, the correct option is (c) 1,4.
Key Concepts
Colligative PropertiesVapour PressureSolidification ProcessSolute and Solvent Interaction
Colligative Properties
Colligative properties are an interesting aspect of chemistry that depend solely on the number of solute particles in a solution, not their identity. Whether you're dealing with salt or sugar, if you add these to water, you'll notice changes that are purely due to the amount present, not what they are.
These properties include:
Understanding these changes is vital in various applications, from antifreeze in cars to preserving food.
These properties include:
- Freezing point depression
- Boiling point elevation
- Vapour pressure lowering
- Osmotic pressure
Understanding these changes is vital in various applications, from antifreeze in cars to preserving food.
Vapour Pressure
Vapour pressure is defined as the pressure exerted by a vapour that is in equilibrium with its liquid or solid form. In simple terms, it's how many molecules are escaping from the liquid to the air.
When you put in a non-volatile solute, like sugar, it affects the vapour pressure by reducing it. This happens because the solute particles take up some of the space at the surface of the liquid. With fewer surface molecules that can escape into the air, the vapour pressure drops.
When you put in a non-volatile solute, like sugar, it affects the vapour pressure by reducing it. This happens because the solute particles take up some of the space at the surface of the liquid. With fewer surface molecules that can escape into the air, the vapour pressure drops.
- This is why solutions often have a lower vapour pressure compared to their pure solvents.
- The effect is based on particle number, not type, demonstrating the essence of colligative properties.
Solidification Process
The solidification process in a solution during freezing involves only the solvent molecules, typically remaining clear of any solute. In the context of freezing point depression:
Understanding this process is crucial because it explains why solutions freeze at a lower temperature than pure solvents. Recognizing this principle has practical uses, such as preventing pipes from freezing in winter.
- The solvent molecules gather to form a solid, like ice, when cooled down.
- Solute particles stay dissolved in the liquid part because they disrupt the orderly formation of the solid phase.
Understanding this process is crucial because it explains why solutions freeze at a lower temperature than pure solvents. Recognizing this principle has practical uses, such as preventing pipes from freezing in winter.
Solute and Solvent Interaction
Understanding the interaction between solute and solvent is essential in explaining many phenomena, including freezing point depression.
When a solute is introduced into a solvent, it disrupts the systematic arrangement of solvent molecules, altering their movement and interaction. This interaction changes several physical properties. Key points include:
When a solute is introduced into a solvent, it disrupts the systematic arrangement of solvent molecules, altering their movement and interaction. This interaction changes several physical properties. Key points include:
- Solute particles interfere with the ability of solvent molecules to crystallize and form a solid.
- This disturbance in formation results in a lower freezing point for the solution.
- The vapour pressure is also affected because fewer solvent molecules are in the position to escape into the vapour phase.
Other exercises in this chapter
Problem 84
Two liquids having vapour pressures \(\mathrm{P}_{1}^{0}\) and \(\mathrm{P}_{2}^{0}\) in pure state in the ratio of \(2: 1\) are mixed in a molar ratio of \(1:
View solution Problem 87
Equal volumes of \(0.1 \mathrm{M} \mathrm{AgNO}_{3}\) and \(0.2 \mathrm{M} \mathrm{NaCl}\) are mixed. The concentration of nitrate ions in the mixture will be (
View solution Problem 90
On mixing \(3 \mathrm{~g}\) of non-volatile solute in \(200 \mathrm{~mL}\) of water its boiling point \(\left(100^{\circ} \mathrm{C}\right)\) becomes \(100.52^{
View solution Problem 92
The freezing point of equimolal aqueous solution will be highest for (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-}\)(anilin
View solution