Match List-I with List-II :
List-I List-II (A) Negatively charged sol (I) Fe2O3.xH2O (B) Macromolecular colloid (II) CdS sol (C) Positively charged sol (III) Starch (D) Cheese (IV) a gel
Choose the correct answer from the options given below:
Match List-I with List-II :
| List-I | List-II |
| (A) Negatively charged sol | (I) Fe2O3.xH2O |
| (B) Macromolecular colloid | (II) CdS sol |
| (C) Positively charged sol | (III) Starch |
| (D) Cheese | (IV) a gel |
Choose the correct answer from the options given below:
- (A) – (II), (B) – (III), (C) – (IV), (D) – (I)
- (A) – (II), (B) – (I), (C) – (III), (D) – (IV)
- (A) – (II), (B) – (III), (C) – (I), (D) – (IV)
- (A) – (I), (B) – (III), (C) – (II), (D) – (IV)
The Correct Option is C
Solution and Explanation
The correct answer is (C) : (A) – (II), (B) – (III), (C) – (I), (D) – (IV)
(A) Negatively charged sol - CdS sol
(B) Macromolecular colloid - Starch
(C) Positively charged sol - Fe2O3.xH2O
(D) Cheese - A gel
Top Questions on Solutions
- If equal volumes of AB and XY (both are salts) aqueous solutions are mixed, which of the following combination will give precipitate of AY, at 300 K?
- Elements A and B form two non-volatile compounds (AB$_2$ and AB$_4$). When 1 g of AB$_2$ is added to 20 g of C$_6$H$_6$ (molar mass =78 gmol$^{-1}$) the freezing point of C$_6$H$_6$ is lowered by 2.3 K. When 1 g of AB$_4$ is added to 20 g of C$_6$H$_6$, the freezing point of C$_6$H$_6$ was lowered by 1.3 K. The atomic masses of A and B are respectively (K$_f$(C$_6$H$_6$)=5.1 K kg mol$^{-1}$)
- Mole fraction of H$_2$SO$_4$ in aqueous solution is 0.9. Find mass % of H$_2$SO$_4$. (H=1, S=32, O=16 u)
- 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? Observe the following data given in the table. (\(K_H\) = Henry's law constant)
Gas CO₂ Ar HCHO CH₄ \(K_H\) (k bar at 298 K) 1.67 40.3 \(1.83 \times 10^{-5}\) 0.413 The correct order of their solubility in water is
Questions Asked in JEE Main exam
Let $f: [0, \infty) \to \mathbb{R}$ be a differentiable function such that $f(x) = 1 - 2x + \int_0^x e^{x-t} f(t) \, dt$ for all $x \in [0, \infty)$. Then the area of the region bounded by $y = f(x)$ and the coordinate axes is
- JEE Main - 2025
- Differential Equations
- If $ \epsilon_0 $ denotes the permittivity of free space and $ \Phi_E $ is the flux of the electric field through the area bounded by the closed surface, then the dimension of $ \epsilon_0 \frac{d\Phi_E}{dt} $ are that of:
- JEE Main - 2025
- Electromagnetic induction
Quantitative analysis of an organic compound (X) shows the following percentage composition.
C: 14.5%
Cl: 64.46%
H: 1.8%
Empirical formula mass of the compound (X) is:- JEE Main - 2025
- Molality and Molarity
If $\lim_{x \to 1} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda + \mu$ is equal to
- The formation enthalpies, \( \Delta H_f^\circ \) for \( \text{H}_2 \) and \( \text{O}_2 \) are 220.0 and 250.0 kJ mol\(^{-1}\), respectively, at 298.15 K, and \( \Delta H_f^\circ \) for \( \text{H}_2\text{O} \) (g) is -242.0 kJ mol\(^{-1}\) at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is:
- JEE Main - 2025
- Thermodynamics
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.