The question focuses on the effect of adding a small quantity of naphthalene to benzene on its freezing point. This concept is related to colligative properties in chemistry, specifically freezing point depression. Here’s a detailed explanation:
The freezing point of a pure substance is the temperature at which it changes from a liquid to a solid. When a solute is dissolved in a solvent, the freezing point of the solvent decreases. This phenomenon is known as freezing point depression, a colligative property which depends on the number of solute particles in a solvent, not the nature of the solute itself.
The depression in freezing point can be calculated using the formula:
\(\Delta T_f = i \cdot K_f \cdot m\)
When naphthalene is added to benzene:
Based on the concept of freezing point depression, when a small quantity of naphthalene is added to benzene, the freezing point of benzene decreases.
Correct Answer: Decreases
Two p-n junction diodes \(D_1\) and \(D_2\) are connected as shown in the figure. \(A\) and \(B\) are input signals and \(C\) is the output. The given circuit will function as a _______. 
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.