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1
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
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An aqueous solution is prepared by dissolving $0.1 \mathrm{~mol}$ of an ionic salt in $1.8 \mathrm{~kg}$ of water at $35^{\circ} \mathrm{C}$. The salt remains $90 \%$ dissociated in the solution. The vapour pressure of the solution is $59.724 \mathrm{~mm}$ of Hg. Vapor pressure of water at $35{ }^{\circ} \mathrm{C}$ is $60.000 \mathrm{~mm}$ of $\mathrm{Hg}$. The number of ions present per formula unit of the ionic salt is _________.
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2
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
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Consider the strong electrolytes $Z_{m} X_{n}, U_{m} Y_{p}$ and $V_{m} X_{n}$. Limiting molar conductivity ( $\Lambda^{0}$ ) of $\mathrm{U}_{\mathrm{m}} \mathrm{Y}_{\mathrm{p}}$ and $\mathrm{V}_{\mathrm{m}} \mathrm{X}_{\mathrm{n}}$ are 250 and $440 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, respectively. The value of $(\mathrm{m}+\mathrm{n}+\mathrm{p})$ is

Given:

Ion $\mathrm{Z}^{\mathrm{n}+}$ $\mathrm{U}^{\mathrm{p}+}$ $\mathrm{V}^{\mathrm{n}+}$ $\mathrm{X}^{\mathrm{m}-}$ $\mathrm{Y}^{\mathrm{m}-}$
$\lambda^{0}\left(\mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\right)$ $50.0$ $25.0$ $100.0$ $80.0$ $100.0$

$\lambda^{0}$ is the limiting molar conductivity of ions

The plot of molar conductivity ( $\Lambda$ ) of $\mathrm{Z}_{\mathrm{m}} \mathrm{X}_{\mathrm{n}} v s\, \mathrm{c}^{1 / 2}$ is given below.

JEE Advanced 2022 Paper 2 Online Chemistry - Electrochemistry Question 19 English
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3
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
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The reaction of $\mathrm{Xe}$ and $\mathrm{O}_{2}{F}_{2}$ gives a $\mathrm{Xe}$ compound $\mathbf{P}$. The number of moles of $\mathrm{HF}$ produced by the complete hydrolysis of $1 \mathrm{~mol}$ of $\mathbf{P}$ is __________.
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4
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
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Thermal decomposition of $\mathrm{AgNO}_{3}$ produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired electrons is ____________.
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