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JEE Main 2024 (Online) 8th April Morning Shift
Numerical
+4
-1

A solution containing $$10 \mathrm{~g}$$ of an electrolyte $$\mathrm{AB}_2$$ in $$100 \mathrm{~g}$$ of water boils at $$100.52^{\circ} \mathrm{C}$$. The degree of ionization of the electrolyte $$(\alpha)$$ is _________ $$\times 10^{-1}$$. (nearest integer)

[Given : Molar mass of $$\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{b}}$$ (molal boiling point elevation const. of water) $$=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$$, boiling point of water $$=100^{\circ} \mathrm{C} ; \mathrm{AB}_2$$ ionises as $$\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}]$$

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2
JEE Main 2024 (Online) 1st February Evening Shift
Numerical
+4
-1
Change Language
The amount of electricity in Coulomb required for the oxidation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$ is __________ $\times 10^5 \mathrm{C}$.
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3
JEE Main 2024 (Online) 1st February Evening Shift
Numerical
+4
-1
Change Language
Consider the following redox reaction :

$$ \mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 $$

The standard reduction potentials are given as below $\left(\mathrm{E}_{\text {red }}^0\right)$ :

$$ \begin{aligned} & \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V} \\\\ & \mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V} \end{aligned} $$

If the equilibrium constant of the above reaction is given as $\mathrm{K}_{\mathrm{eq}}=10^x$, then the value of $x=$ __________ (nearest integer)
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4
JEE Main 2024 (Online) 1st February Morning Shift
Numerical
+4
-1
Change Language
The potential for the given half cell at $298 \mathrm{~K}$ is (-) __________ $\times 10^{-2} \mathrm{~V}$

$$ \begin{aligned} & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g}) \\\\ & {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}} \end{aligned} $$

(Given : $2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3$ )
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