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

Question 14 Marks

The number of paramagnetic complex among
$\left[ FeF _6\right]^{3-}, \left[ Fe ( CN )_6\right]^{3-}, \left[ Mn ( CN )_6\right]^{3-}, \left[ Co \left( C _2 O _4\right)_3\right]^{3-}$, $\left[ MnCl _6\right]^{3-}$ and $\left[ CoF _6\right]^{3-}$, which involved $d ^2 sp ^3$ hybridization is __________ .

Answer

$\left[\mathrm{FeF}_{6}\right]^{3-}$	Paramagnetic	$\mathrm{sp}^{3} \mathrm{~d}^{2}$
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$	Paramagnetic	$\mathrm{d}^{2} \mathrm{sp}^{3}$
$\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$	Paramagnetic	$\mathrm{d}^{2} \mathrm{sp}^{3}$
$\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$	Diamagnetic	$\mathrm{d}^{2} \mathrm{sp}^{3}$
$\left[\mathrm{MnCl}_{6}\right]^{3-}$	Paramagnetic	$\mathrm{sp}^{3} \mathrm{~d}^{2}$
$\left[\mathrm{CoF}_{6}\right]^{3-}$	Paramagnetic	$\mathrm{sp}^{3} \mathrm{~d}^{2}$

Only $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} $ and $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$ are paramagnetic and $\mathrm{d}^{2} \mathrm{sp}^{3}$ hybridisation of metal.

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Question 24 Marks

The percentage dissociation of a salt $\left(\mathrm{MX}_{3}\right)$ solution at given temperature (van't Hoff factor $\mathrm{i}=$ $2)$ is. __________ % (Nearest integer)

Answer

33
$\mathrm{MX}_{3} \rightarrow \mathrm{M}^{+3}+3 \mathrm{X}^{\ominus}$
$\mathrm{i}=1+(\mathrm{n}-1) \alpha$
$\mathrm{i}=1+(4-1) \alpha=2$
$\alpha=\frac{1}{3}=33.33 \% \approx 33 \%$

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Question 34 Marks

1 Faraday electricity was passed through $\mathrm{Cu}^{2+}(1.5$ $\mathrm{M}, 1 \mathrm{~L}) / \mathrm{Cu}$ and 0.1 Faraday was passed through $\mathrm{Ag}^{+}(0.2 \mathrm{M}, 1 \mathrm{~L}) / \mathrm{Ag}$ electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is-

$\begin{aligned} \text { Given : } & E _{ Cu ^2 / Cu }^0=0.34 V \\ & E _{ Ag ^{+} / Ag }^0=0.8 V \\ \frac{2.303 RT }{ F } & =0.06 V\end{aligned}$

Answer

400
$\mathrm{Cu}^{+2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$
(1 faraday $=$ charge on 1 mole electron)
$\mathrm{t}=0 \quad 1.5 \quad 1$ mole
$\mathrm{t}=\mathrm{t} \quad 1 \quad\quad\quad-\quad\quad 0.5$ mole
$\left[\mathrm{Cu}^{+2}\right]=1 \mathrm{M}$ after electrolysis
$\mathrm{Ag} \oplus+$ $\mathrm{e}^{-} \longrightarrow \mathrm{Ag}$
$\mathrm{t}=0 \quad 0.2 \quad 0.1$ mole
$\mathrm{t}=\mathrm{t} \quad 0.1\quad\quad-\quad\quad\quad-$
$\left[\mathrm{Ag}^{+}\right]=0.1 \mathrm{M}$ after electrolysis
Cell $\quad Cu _{( s )}+2 Ag _{( aq )}^{+} \rightarrow Cu _{(aq)}^{+2}+2 Ag _{( s )}$
reaction
$\begin{array}{l} E = E ^{\circ}-\frac{0.06}{ n } \log \frac{\left[ Cu ^{+2}\right]}{\left[ Ag ^{+}\right]^2} \\ E =(0.8-0.34)-\frac{0.06}{2} \log \frac{1}{(0.1)^2}=0.4 V\end{array}$
Correct answer $=400 \mathrm{mV}$

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Question 44 Marks

Thyroxine, the hormone has given below structure

The percentage of iodine in thyroxine is __________ %. (nearest integer)
(Given molar mass in $g$ $mol^{-1}$ C : 12, H : 1, O : 16, N : 14, I : 127)

Answer

$\rightarrow$ Molecular formula of Thyroxine $\Rightarrow$ $\mathrm{C}_{15} \mathrm{H}_{11} \mathrm{O}_{4} \mathrm{NI}_{4}$
$\rightarrow$ Molecular mass of Thyroxine -
$\mathrm{C} \rightarrow 15 \times 12=180$
$\mathrm{H} \rightarrow 11 \times 1=11$
$\mathrm{O} \rightarrow 16 \times 4=64$
$\mathrm{N} \rightarrow 14 \times 1=14$
I $\rightarrow 127 \times 4=508$
$\rightarrow$ Molecular mass of Thyroxine $\Rightarrow 777$
$\rightarrow \%$ of Iodine $=\frac{508}{777} \times 100$
$=65.38 \%$
Nearest integer $=65$

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Question 54 Marks

An organic compound weighing 500 mg, produced 220 mg of $\mathrm{CO}_{2}$. on complete combustion. The percentage composition of carbon in the compound is __________ $\%$. (nearest integer)
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{C}: 12, \mathrm{O}: 16$)

Answer

12
Organic compound $\xrightarrow[\Delta]{\mathrm{CuO}} \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$
$\mathrm{n}_{\mathrm{CO}_{2}}=\frac{220 \times 10^{-3}}{44}=5 \times 10^{-3}$ moles
$\mathrm{m}_{\mathrm{C}}=5 \times 10^{-3} \times 12$
$\% \mathrm{~m}$ carbon $=\frac{5 \times 10^{-3} \times 12}{500 \times 10^{-3}} \times 100=12 \%$
Correct answer is 12

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JEE SECTION - B [CHEMISTY - NUMERIC]

SECTION - B [CHEMISTY - NUMERIC]

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