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JEE SECTION - B [CHEMISTY - NUMERIC]

JEE Main 3-April-2025 Paper - Shift 1 · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 5 questions.

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SECTION - B [CHEMISTY - NUMERIC]

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5 questions · self-marked practice — reveal the answer and mark yourself.

Question 24 Marks
Given :
$
\begin{array}{l}
\Delta H_{sub}^{\ominus}[C(\text { graphite })]=710 kJ mol^{-1} \\
\Delta_{C-H} H^{\ominus}=414 kJ mol^{-1} \\
\Delta_{H-H} H^{\ominus}=436 kJ mol^{-1} \\
\Delta_{C-C} H^{\ominus}=611 kJ mol^{-1}
\end{array}
$
The $\Delta H _{ f }^{\ominus}$ for $CH _2= CH _2$ is __________ $kJ mol ^{-1}$
(nearest integer value)
Answer
(25)
Image
$\begin{aligned} & {\left[\Delta H _{ f }^{\circ}\right]_{ C _2 H _{ tov }}=2 \times\left[\Delta H _{ sut }^{\circ}\right]_{ C _{(0)}}+2 \times \Delta H _{ H - H }^{\circ}-1 \times \Delta H _{ C - C }^{ o }-4 \times \Delta H _{ C - H }^{\circ} } \\ \Rightarrow \quad & {\left[\Delta H _{ f }^{ o }\right]_{ C _2 H _{4(B)}}=(2 \times 710)+(2 \times 436)-611-4 \times 414 } \\ \Rightarrow \quad & {\left[\Delta H _{ f }^{ o }\right]_{ C _2 H _{4( P )}}=25 kJ / mol }\end{aligned}$
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Question 34 Marks
The number of optical isomers exhibited by the iron complex (A) obtained from the following reaction is __________
$
FeCl_3+KOH+H_2 C_2 O_4 \rightarrow A
$
Answer
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Question 44 Marks
0.5 g of an organic compound on combustion gave 1.46 g of $CO _2$ and 0.9 g of $H _2 O$. The percentage of carbon in the compound is __________. (Nearest integer)
[Given : Molar mass (in $g mol ^{-1}$ ) $C : 12, H : 1$, O : 16]
Answer
(80)
Organic $\rightarrow CO _2$
Compound
Applying POAC on 'C'
(mole) of 'C' in compound $= n _{ CO _2} \times 1$
So mass of 'C' in compound
$
=\frac{1.46}{44} \times 12
$
So, $\%$ of 'C' in compound $=\frac{1.46}{44} \times \frac{12}{0.5} \times 100$ $
= 79.63
$
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Question 54 Marks
During estimation of nitrogen by Dumas' method of compound $X (0.42 g)$ :
Image
__________ mL of $N _2$ gas will be liberated at STP. (nearest integer)
(Given molar mass in $g~mol ^{-1}: C : 12, H : 1$, $N : 14)$
Answer
(111)
M.wt. of given compound $=86$
Image
Applying POAC on ' N '
$
\begin{array}{l}
n_{X} \times 2=n_{N_2} \times 2 \\
\frac{0.42}{86}=n_{N_2} \\
\Rightarrow(\text { Volume })_{N_2} \text { at } STP=\frac{0.42}{86} \times 22.4 L \\
=0.1108 L=110.8 m \ell
\end{array}
$
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