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
Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals:
Designate whether each of the following compounds is aromatic or not aromatic.

The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)