Karnataka PUC 12th Chemistry Question Paper 2025

The depression in freezing point ($\Delta T_f$) is a colligative property given by $\Delta T_f = i \cdot K_f \cdot m$. For glucose, which is a non-electrolyte, the van't Hoff factor $i = 1$. For $MgCl_2$, which is an electrolyte, it dissociates as $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$, giving $i = 3$. Since the molality ($m$) is the same for both, $\Delta T_f$ for $MgCl_2$ will be approximately 3 times that of glucose.

During the electrolysis of aqueous $NaCl$ (brine), water is reduced at the cathode to form $H_2$ gas and $OH^-$ ions. Chloride ions are oxidized at the anode to form $Cl_2$ gas. The remaining $Na^+$ and $OH^-$ ions in the solution form $NaOH$. Thus, the products are $H_2$, $Cl_2$, and $NaOH$.

Radioactive decay is a spontaneous process where the rate of disintegration is directly proportional to the number of radioactive nuclei present at that time. This follows first-order kinetics, described by the rate law $Rate = -dN/dt = \lambda N$.

Activation energy ($E_a$) is defined as the minimum extra amount of energy absorbed by a reactant system to reach the threshold energy level, which is required for the formation of the activated complex during a chemical reaction.

Copper ($Cu$) has a positive reduction potential ($E^\circ_{Cu^{2+}/Cu} = +0.34V$), meaning it is a weaker reducing agent than hydrogen. Therefore, it cannot displace hydrogen from dilute acids. The other options are correct: $KMnO_4$ is a strong oxidizer, $MnO_4^-$ is diamagnetic ($d^0$ configuration), and $Ti^{2+}$ is a strong reducing agent.