The solution from the following with highest depression in freezing point/lowest freezing point is
- 180 g of benzoic acid dissolved in benzene
- 180 g of acetic acid dissolved in benzene
- 180 g of acetic acid dissolved in water
- 180 g of glucose dissolved in water
The Correct Option is C
Solution and Explanation
To determine which solution has the highest depression in freezing point, we need to consider the colligative properties of solutions, specifically the freezing point depression.
Colligative Properties: Freezing point depression is given by the formula:
ΔTf = i · Kf · m
where ΔTf is the depression in freezing point, i is the van 't Hoff factor (number of particles the solute breaks into), Kf is the cryoscopic constant of the solvent, and m is the molality of the solution.
Analyzing Each Option:
- Option (3): 180 g of acetic acid in water. Acetic acid partially ionizes in water (i ≈ 1). The molality will be relatively high as the solvent is water.
- Option (2): 180 g of acetic acid in benzene. Acetic acid does not ionize in benzene (i = 1). Benzene has a lower Kf than water, leading to a lesser depression.
- Option (1): 180 g of benzoic acid in benzene. Similar to option (2), it does not ionize, and its effect will be limited by benzene’s lower Kf.
- Option (4): 180 g of glucose in water. Glucose does not ionize (i = 1), but its molar mass is higher, leading to a lower molality compared to acetic acid.
Conclusion: Given that acetic acid has a higher molar mass than glucose and ionizes in water, it will result in a greater freezing point depression.
Thus, the solution with the highest depression in freezing point is: 180 g of acetic acid dissolved in water.
Top Questions on Solutions
According to the generally accepted definition of the ideal solution there are equal interaction forces acting between molecules belonging to the same or different species. (This is equivalent to the statement that the activity of the components equals the concentration.) Strictly speaking, this concept is valid in ecological systems (isotopic mixtures of an element, hydrocarbons mixtures, etc.). It is still usual to talk about ideal solutions as limiting cases in reality since very dilute solutions behave ideally with respect to the solvent. This law is further supported by the fact that Raoult’s law empirically found for describing the behaviour of the solvent in dilute solutions can be deduced thermodynamically via the assumption of ideal behaviour of the solvent.
Answer the following questions:
(a) Give one example of miscible liquid pair which shows negative deviation from Raoult’s law. What is the reason for such deviation?
(b) (i) State Raoult’s law for a solution containing volatile components.
OR
(ii) Raoult’s law is a special case of Henry’s law. Comment.
(c) Write two characteristics of an ideal solution.- A solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water \(K_f\) and \(K_b\) are 1.86 K kg mol\(^{-1}\) and 0.512 K kg mol\(^{-1}\) respectively.
- Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law? Give an example.
- Give reasons:
(a) Cooking is faster in a pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? - An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because __________
Questions Asked in JEE Main exam
- Consider a long thin conducting wire carrying a uniform current \( I \). A particle having mass \( M \) and charge \( q \) is released at a distance \( a \) from the wire with a speed \( v_0 \) along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance \( x \) from the wire. The value of \( x \) is:
- JEE Main - 2025
- Magnetic Field
- Two equal sides of an isosceles triangle are along \( -x + 2y = 4 \) and \( x + y = 4 \). If \( m \) is the slope of its third side, then the sum of all possible distinct values of \( m \) is:
- JEE Main - 2025
- 3D Geometry
- Let the position vectors of the vertices A, B, and C of a tetrahedron ABCD be \( \hat{i} + 2\hat{j} + \hat{k} \), \( \hat{i} + 3\hat{j} - 2\hat{k} \), and \( 2\hat{i} + \hat{j} - \hat{k} \) respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is \( \frac{\sqrt{10}}{3} \) and the volume of the tetrahedron is \( \frac{\sqrt{805}}{6\sqrt{2}} \), then the position vector of E is:
- JEE Main - 2025
- Geometry and Vectors
- If \( x_1, x_2, x_3, x_4 \) are in GP (Geometric Progression), then we subtract 2, 4, 7, and 8 from \( x_1, x_2, x_3, x_4 \) respectively, then the resultant numbers are in AP (Arithmetic Progression). Then the value of \( \frac{1}{24} (x_1 \cdot x_2 \cdot x_3 \cdot x_4) \) is:
- JEE Main - 2025
- Arithmetic Progression
- Let the area of a triangle \( \triangle PQR \) with vertices \( P(5, 4) \), \( Q(-2, 4) \), and \( R(a, b) \) be 35 square units. If its orthocenter and centroid are \( O(2, \frac{14}{5}) \) and \( C(c, d) \) respectively, then \( c + 2d \) is equal to:
- JEE Main - 2025
- Geometry and Vectors
Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.