1
JEE Main 2023 (Online) 31st January Evening Shift
Numerical
+4
-1
The resistivity of a $0.8 \mathrm{M}$ solution of an electrolyte is $5 \times 10^{-3} \Omega~ \mathrm{cm}$.

Its molar conductivity is _________ $\times 10^{4}~ \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. (Nearest integer)
2
JEE Main 2023 (Online) 31st January Morning Shift
Numerical
+4
-1

The logarithm of equilibrium constant for the reaction $$\mathrm{Pd}^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{PdCl}_{4}^{2-}$$ is ___________ (Nearest integer)

Given : $$\frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.06 \mathrm{~V}$$

$$\mathrm{Pd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}) \quad \mathrm{E}^{\ominus}=0.83 \mathrm{~V}$$

\begin{aligned} & \mathrm{PdCl}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \mathrm{E}^{\ominus}=0.65 \mathrm{~V} \end{aligned}

3
JEE Main 2023 (Online) 30th January Evening Shift
Numerical
+4
-1
The electrode potential of the following half cell at $298 \mathrm{~K}$

$\mathrm{X}\left|\mathrm{X}^{2+}(0.001 \mathrm{M}) \| \mathrm{Y}^{2+}(0.01 \mathrm{M})\right| \mathrm{Y}$ is _______ $\times 10^{-2} \mathrm{~V}$ (Nearest integer)

Given: $\mathrm{E}^{0} _ {\mathrm{X}^{2+} \mid \mathrm{X}}=-2.36 \mathrm{~V}$

$\mathrm{E}_{\mathrm{Y}^{2+} \mid \mathrm{Y}}^{0}=+0.36 \mathrm{~V}$

$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$
4
JEE Main 2023 (Online) 30th January Morning Shift
Numerical
+4
-1

Consider the cell

$$\mathrm{Pt}_{(\mathrm{s})}\left|\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})|| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \operatorname{Pt}(\mathrm{s})$$

When the potential of the cell is $$0.712 \mathrm{~V}$$ at $$298 \mathrm{~K}$$, the ratio $$\left[\mathrm{Fe}^{2+}\right] /\left[\mathrm{Fe}^{3+}\right]$$ is _____________. (Nearest integer)

Given : $$\mathrm{Fe}^{3+}+\mathrm{e}^{-}=\mathrm{Fe}^{2+}, \mathrm{E}^{\theta} \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+} \mid \mathrm{Pt}=0.771$$

$$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$$

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