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

Question 14 Marks

Consider the following reaction occurring in the blast furnace.
$\mathrm{Fe}_{3} \mathrm{O}_{4(\mathrm{~s})}+4 \mathrm{CO}_{(\mathrm{g})} \rightarrow 3 \mathrm{Fe}_{(\mathrm{l})}+4 \mathrm{CO}_{2(\mathrm{~g})}$
' $x$ ' kg of iron is produced when $2.32 \times 10^{3} \mathrm{~kg}$ $\mathrm{Fe}_{3} \mathrm{O}_{4}$ and $2.8 \times 10^{2} \mathrm{~kg} \mathrm{CO}$ are brought together in the furnace. The value of ' $x$ ' is
________________ .
(nearest integer)
{Given :
Molar mass of $\mathrm{Fe}_{3} \mathrm{O}_{4}=232 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\left.\mathrm{Fe}=56 \mathrm{~g} \mathrm{~mol}^{-1}\right\}$

Answer

(420)
Sol.
moles of $\mathrm{Fe}_{3} \mathrm{O}_{4}=\frac{2.32 \times 10^{3} \times 10^{3}}{232}=10000 \mathrm{~mol}$
moles of $\mathrm{CO}=\frac{2.8 \times 10^{2} \times 10^{3}}{28}=10000 \mathrm{~mol}$
$\mathrm{Fe}_{3} \mathrm{O}_{4}+4 \mathrm{CO} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{CO}_{2}$
$10^{4} \mathrm{~mol} \quad 10^{4} \mathrm{~mol}$
CO is L.R.
mole of $\mathrm{Fe}=\frac{3}{4} \times 10^{4}$
mass of $\mathrm{Fe}=\frac{3}{4} \times \frac{10^{4} \times 56}{1000} \mathrm{~kg}=420 \mathrm{~kg}$

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Question 24 Marks

Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is : $\mathrm{Cu}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Ba}^{2+}, \mathrm{Ca}^{2+}, \mathrm{NH}_{4}^{+}, \mathrm{Mg}^{2+}, \mathrm{Zn}^{2+}$

Answer

(3)
Sol.
Only $\mathrm{Cu}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Ca}^{2+} \& \mathrm{Zn}^{2+}$ form precipitate with $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

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Question 34 Marks

Xg of benzoic acid on reaction with aq. $\mathrm{NaHCO}_{3}$ release $\mathrm{CO}_{2}$ that occupied 11.2 L volume at STP. X is _______________ g.

Answer

(61)
Sol.
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{NaHCO}_{3} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}$ $+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$
x gm
mole of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}=$ mole of $\mathrm{CO}_{2}=\frac{11.2}{22.4}=0.5$
mass of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}=\mathrm{x}=0.5 \times 122=61 \mathrm{gm}$
Ans. 61

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Question 44 Marks

Standard entropies of $X_{2}, Y_{2}$ and $X Y_{5}$ are 70,50 and $110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature in Kelvin at which the reaction $\frac{1}{2} \mathrm{X}_{2}+\frac{5}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{5} \Delta \mathrm{H}^{-}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Will be at equilibrium is ____________ (Nearest integer)

Answer

(700)
Sol.
$\frac{1}{2} \mathrm{X}_{2}+\frac{5}{2} \mathrm{Y}_{2} \rightleftharpoons \mathrm{XY}_{5}$
$\Delta \mathrm{S}_{\mathrm{R} \times n}^{0}=110-\left[\left(\frac{1}{2} \times 70\right)+\left(\frac{5}{2} \times 50\right)\right]$
$=110-160=-50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
$\Delta G^{0}=0$ at eqb
$\Delta \mathrm{G}^{0}=\Delta \mathrm{H}^{0}-\mathrm{T} \Delta \mathrm{S}^{0}$
$0=-35000-\mathrm{T}(-50)$
$\mathrm{T}=700$ Kelvin
Ans. 700

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Question 54 Marks

$37.8 \mathrm{~g} \mathrm{~N}_{2} \mathrm{O}_{5}$ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K
$2 \mathrm{~N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
The total pressure at equilibrium was found to be
18.65 bar.
Then, $\mathrm{Kp}=$ ____________ $\times 10^{-2}$ [nearest integer]
Assume $\mathrm{N}_{2} \mathrm{O}_{5}$ to behave ideally under these conditions
Given : $\mathrm{R}=0.082$ bar $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$

Answer

(962)
Sol. Initial pressure of $\mathrm{N}_{2} \mathrm{O}_{5}$
$
=\frac{\frac{37.8}{108} \times 0.082 \times 500}{1}=14.35 \mathrm{bar}
$
$
2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightleftharpoons 2 \mathrm{~N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}
$
$\mathrm{t}=0 \quad 14.35$
$\mathrm{t}=\mathrm{eq} \quad 14.35-2 \mathrm{P} \quad 2 \mathrm{P} \quad \mathrm{P}$
$\mathrm{P}_{\text {Total }}$ at eqb $=14.35+\mathrm{P}=18.65$
$\mathrm{P}=4.3$
$\mathrm{P}_{\mathrm{N}_{2} \mathrm{O}_{5}}=5.75 \mathrm{bar}$
$\mathrm{P}_{\mathrm{N}_{2} \mathrm{O}_{4}}=8.6$ bar
$\mathrm{P}_{\mathrm{O}_{2}}=4.3 \mathrm{bar}$
$\mathrm{k}_{\mathrm{p}}=\frac{(8.6)^{2} \times(4.3)}{(5.75)^{2}}=9.619=\mathrm{x} \times 10^{-2}$
$x=961.9 \approx 962$

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JEE SECTION - B [CHEMISTY - NUMERIC]

SECTION - B [CHEMISTY - NUMERIC]

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