The cycloalkene $(\mathrm{X})$ on bromination consumes one mole of bromine per mole of $(\mathrm{X})$ and gives the product $(\mathrm{Y})$ in which $\mathrm{C}: \mathrm{Br}$ ratio is 3:1. The percentage of bromine in the product $(\mathrm{Y})$ is $\_\_\_\_$ %. (Nearest integer)

(Given : molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{O}: 16, \mathrm{Br}: 80$ )

Dissociation of a gas $\mathrm{A}_2$ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K .

$$ \mathrm{A}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g}) $$

The standard Gibbs energy of formation of the involved substances has been provided below:

$$ \begin{array}{|c|c|} \hline \text { Substance } & \Delta \mathrm{G}_{\mathrm{f}}^{\circ} / \mathrm{kJ} \mathrm{~mol}^{-1} \\ \hline \hline \mathrm{~A}_2 & -100.00 \\ \hline \mathrm{~A} & -50.832 \\ \hline \end{array} $$

The degree of dissociation of $\mathrm{A}_2(\mathrm{~g})$ is given by $\left(x \times 10^{-2}\right)^{1 / 2}$ where $x=$

$\_\_\_\_$ . (Nearest integer).

[ Given : $\mathrm{R}=8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2=0.3010, \log 3=0.48$ ]

Assume degree of dissociation is not negligible.

The temperature at which the rate constants of the given below two gaseous reactions become equal is $\_\_\_\_$ K. (Nearest integer)

$$ \begin{array}{ll} \mathrm{X} \longrightarrow \mathrm{Y}, & \mathrm{k}_1=10^6 e^{\frac{-30000}{\mathrm{~T}}} \\ \mathrm{P} \longrightarrow \mathrm{Q}, & \mathrm{k}_2=10^4 e^{\frac{-24000}{\mathrm{~T}}} \end{array} $$

Given : $\ln 10=2.303$

Consider the following electrochemical cell at 298 K

$\mathrm{Pt}\left|\mathrm{HSnO}_2^{-}(\mathrm{aq})\right| \mathrm{Sn}(\mathrm{OH})_6{ }^{2-}(\mathrm{aq})\left|\mathrm{OH}^{-}(\mathrm{aq})\right| \mathrm{Bi}_2 \mathrm{O}_3(\mathrm{~s}) \mid \mathrm{Bi}(\mathrm{s})$.

If the reaction quotient at a given time is $10^6$, then the cell EMF $\left(\mathrm{E}_{\text {cell }}\right)$ is

$\_\_\_\_$ $\times 10^{-1} \mathrm{~V}$ (Nearest integer).

Given the standard half-cell reduction potential as

$$ \mathrm{E}_{\mathrm{Bi}_2 \mathrm{O}_3 / \mathrm{Bi}, \mathrm{OH}^{-}}^{\circ}=-0.44 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_6^{2-} / \mathrm{HSnO}_2^{-}, \mathrm{OH}^{-}}^{\circ}=-0.90 \mathrm{~V} $$