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
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
| \(K_2Cr_2O_7\) | \(CuSO_4\) | |
| Side X | SPM | Side Y |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)