1
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$$ \begin{aligned} &\text { Find the compound ' } \mathrm{A} \text { ' from the following reaction sequences. }\\ &\mathrm{A} \xrightarrow{\text { aqua-regia }} \mathrm{B} \xrightarrow[\text { (2) } \mathrm{AcOH}]{\text { (1) } \mathrm{KNO}_2 \mid \mathrm{NH}_4 \mathrm{OH}} \text { yellow ppt } \end{aligned} $$

A
NiS
B
MnS
C
ZnS
D
CoS
2
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$$\begin{aligned} & \mathrm{S}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+2 x \mathrm{kcal} \\ & \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g})+y \mathrm{kcal} \end{aligned}$$

The heat of formation of $\mathrm{SO}_2(\mathrm{~g})$ is given by :

A
$2 x+y$ kcal
B
$\frac{2 x}{y} \mathrm{~kcal}$
C
$y-2 x \mathrm{~kcal}$
D
$x+y \mathrm{~kcal}$
3
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The structure of the major product formed in the following reaction is :

JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English

A
JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English Option 1
B
JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English Option 2
C
JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English Option 3
D
JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English Option 4
4
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The elemental composition of a compound is $54.2 \% \mathrm{C}, 9.2 \% \mathrm{H}$ and $36.6 \% \mathrm{O}$. If the molar mass of the compound is $132 \mathrm{~g} \mathrm{~mol}^{-1}$, the molecular formula of the compound is : [Given : The relative atomic mass of $\mathrm{C}: \mathrm{H}: \mathrm{O}=12: 1: 16$ ]

A
$\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_2$
B
$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
C
$\mathrm{C}_4 \mathrm{H}_9 \mathrm{O}_3$
D
$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3$
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