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 : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,