SECTION - B [CHEMISTY - NUMERIC] - JEE STD 11 Science Questions

Question 14 Marks

Consider the following half cell reaction
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+6 \mathrm{e}^{-}+14 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ The reaction was conducted with the ratio of $\frac{\left[\mathrm{Cr}^{3+}\right]^{2}}{\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]}=10^{-6}$. The pH value at which the EMF of the half cell will become zero is __________ . (nearest integer value)
[Given : standard half cell reduction potential
$\left.\mathrm{E}_{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}^{+} / \mathrm{Cr}^{3+}}^{\mathrm{o}}=1.33 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right]$

Answer

(10)
$\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{-2}+14 \mathrm{H}_{\text {(aq) }}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}_{\text {(aq) }}^{+3}+7 \mathrm{H}_{2} \mathrm{O}_{(\ell)}$
$\mathrm{E}_{\mathrm{R}}=\mathrm{E}_{\mathrm{R}}^{0}-\frac{0.059}{6} \log \frac{\left[\mathrm{Cr}^{+3}\right]^{2}}{\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}\right]\left[\mathrm{H}^{+}\right]^{14}}$
$0=1.33-\frac{0.059}{6} \log \frac{10^{-6}}{\left[\mathrm{H}^{+}\right]^{14}}$
$\frac{1.33 \times 6}{0.059}=\log \frac{10^{-6}}{[\mathrm{H}]^{14}}$
$135.254=-6-14 \log \left[\mathrm{H}^{+}\right]$
$141.254=14 \mathrm{pH}$
$\mathrm{pH}=\frac{141.254}{14}=10.08$

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Question 24 Marks

The equilibrium constant for decomposition of $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
$\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\left(\Delta \mathrm{G}^{\circ}=92.34 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ is $8.0 \times 10^{-3}$ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation $(\alpha)$ of water is __________ $\times 10^{-2}$ (nearest integer value).
[Assume $\alpha$ is negligible with respect to 1]

Answer

(5)
$H _2 O ( g ) \rightleftharpoons H _{2(g)}+\frac{1}{2} O _{2(g)}$
$\begin{array}{llll} t =0 & 1 \text { mole } \\ t = t _{ eq } & 1-\alpha & \alpha & \frac{\alpha}{2}\end{array}$
$n _{ T }=1+\frac{\alpha}{2} \simeq 1(\alpha < <1)$
$k _{ P }=\frac{ P _{ H _2} \cdot P _{ O _2}^{1 / 2}}{ P _{ H _2 O }}=\frac{(\alpha \cdot P )\left(\frac{\alpha}{2} P \right)^{\frac{1}{2}}}{(1-\alpha) P }$
$P =1$
$8 \times 10^{-3}=\frac{\alpha^{3 / 2}}{\sqrt{2}}$
$\begin{array}{l}\alpha^{3 / 2}=8 \sqrt{2} \times 10^{-3} \\ \alpha^3=128 \times 10^{-6} \\ \alpha=\sqrt[3]{128} \times 10^{-2} \\ =5.03 \times 10^{-2}\end{array}$

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Question 34 Marks

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is __________ M. (Nearest Integer value)
(Given : $\mathrm{Na}=23, \mathrm{I}=127, \mathrm{Ag}=108, \mathrm{~N}=14$, $\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Answer

1
$\begin{array}{ll} NaI _{( aq )}+ AgNO _{3( aq )} \rightarrow & AgI _{( s )}+ NaNO _3( aq ) \\ M , 20 ml \text { excess } & 4.74 g\end{array}$
Moles of $\mathrm{I}^{-}$in $\mathrm{NaI}=$ Moles of $\left(\mathrm{I}^{-}\right)$in $\mathrm{AgI}=\frac{4.74}{235}$
Moles of $\mathrm{NaI}=\frac{4.74}{235}$
Molarity $[\mathrm{NaI}]=\frac{4.74}{235 \times 0.02}=1.008$

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Question 44 Marks

The energy of an electron in first Bohr orbit of H-atom is -13.6 eV . The magnitude of energy value of electron in the first excited state of $\mathrm{Be}^{3+}$ is __________ eV . (nearest integer value)

Answer

(54)
$E_{T}=-13.6 \frac{z^{2}}{n^{2}} e v$
For energy of H-atom, energy of $1^{\text {st }}$ Bohr orbit $\mathrm{E}_{1}=-13.6 \mathrm{eV}[\mathrm{z}=1, \mathrm{n}=1]$
For $\mathrm{Be}^{+3}$ ion, energy of $\mathrm{I}^{\text {st }}$ E.S. $[\mathrm{z}=4, \mathrm{n}=2]$
$\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{Be}^{+3}}}=\frac{\mathrm{z}_{1}^{2}}{\mathrm{n}_{1}^{2}} \times \frac{\mathrm{n}_{2}^{2}}{\mathrm{z}_{2}^{2}}$
$\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{Be}^{+3}}}=\frac{1}{1} \times \frac{4}{16}$
$\mathrm{E}_{\mathrm{Be}^{+3}}=-13.6 \times 4=-54.4 \mathrm{eV}$
$\left|\mathrm{E}_{\mathrm{Be}^{+3}}\right|=54.4 \mathrm{eV}$

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Question 54 Marks

Resonance in $X_{2} Y$ can be represented as

The enthalpy of formation of $\mathrm{X}_{2} \mathrm{Y}\left(\mathrm{X} \equiv \mathrm{X}(\mathrm{g})+\frac{1}{2} \mathrm{Y}=\mathrm{Y}(\mathrm{g}) \rightarrow \mathrm{X}_{2} \mathrm{Y}(\mathrm{g})\right)$ is $80 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The magnitude of resonance energy of $\mathrm{X}_{2} \mathrm{Y}$ is __________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)
Given : Bond energies of $\mathrm{X} \equiv \mathrm{X}, \mathrm{X}=\mathrm{X}, \mathrm{Y}=\mathrm{Y}$ and $\mathrm{X}=\mathrm{Y}$ are 940, 410, 500 and 602 $\mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
valence $X : 3, Y : 2$

Answer

(98)
$\Delta \mathrm{H}_{\text {R.E }}=\Delta \mathrm{H}_{\mathrm{f}(\exp )}-\Delta \mathrm{H}_{\mathrm{f} \text { (Theo) }}$
$\Delta \mathrm{H}_{\mathrm{f}(\mathrm{exp})}$ for $\mathrm{X}_{2} \mathrm{Y}_{(\mathrm{g})}=80 \mathrm{~kJ} / \mathrm{mole}$
for $\Delta \mathrm{H}_{\mathrm{f}}$ (Theo)
$\mathrm{X}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Y}_{2(\mathrm{~g})} \rightarrow \mathrm{X}_{2} \mathrm{Y}_{(\mathrm{g})} \Delta \mathrm{H}_{\mathrm{f}}=$ ?
$\Delta \mathrm{H}_{\mathrm{f}(\text { Theo })}=\left(\mathrm{BE}_{\mathrm{X}=\mathrm{X}}+\frac{1}{2} \mathrm{BE}_{\mathrm{Y}=\mathrm{Y}}\right)-\left(\mathrm{BE}_{\mathrm{X}=\mathrm{X}}+\mathrm{BE}_{\mathrm{X}=\mathrm{Y}}\right)$
$=\left(940+\frac{1}{2} \times 500\right)-(410+602)$
$=178 \mathrm{~kJ} / \mathrm{mole}$
$\Delta H_{\text {R.E }}=80-178$
$=-98 \mathrm{~kJ} / \mathrm{mol}$
$\left|\Delta \mathrm{H}_{\mathrm{R} . \mathrm{E}}\right|=98$

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JEE SECTION - B [CHEMISTY - NUMERIC]

SECTION - B [CHEMISTY - NUMERIC]

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