1
MHT CET (PCB) 2025 9th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

What is the total number of moles of ' C ' atoms and ' H ' atoms respectively present in n mole molecule represented by following structure?

MHT CET (PCB) 2025 9th April Evening Shift Chemistry - Hydrocarbons Question 1 English
A

$5 n$ and $8 n$

B

$4 n$ and $7 n$

C

$4 n$ and $6 n$

D

5 n and 7 n

2
MHT CET (PCB) 2025 9th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

Which from following reactions performs highest + ve work at $25^{\circ} \mathrm{C}$ ?

A

$\mathrm{CH}_{4(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \longrightarrow \mathrm{CH}_3 \mathrm{Cl}_{(\mathrm{g})}+\mathrm{HCl}_{(\mathrm{g})}$

B

$3 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{N}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})}$

C

$\mathrm{C}_2 \mathrm{H}_{2(\mathrm{~g})}+\frac{5}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(l)}$

D

$\quad 2 \mathrm{C}_2 \mathrm{H}_{6(\mathrm{~g})}+7 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 4 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$

3
MHT CET (PCB) 2025 9th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

Calculate the number of unit cells present in 0.6 g metal if the product of density and unit cell volume is $3.6 \times 10^{-22} \mathrm{~g}$.

A

$1.666 \times 10^{21}$

B

$3.333 \times 10^{21}$

C

$2.499 \times 10^{21}$

D

$4.833 \times 10^{21}$

4
MHT CET (PCB) 2025 9th April Evening Shift
MCQ (Single Correct Answer)
+1
-0

Identify the product formed in the following reaction.

$$ \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CHO} \xrightarrow[\text { ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {i) } \mathrm{LiAlH}_4} \text { Product } $$
A

$\quad \mathrm{CH}_3-\left(\mathrm{CH}_2\right)_3-\mathrm{CH}_2-\mathrm{OH}$

B

$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{OH}$

C

$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}$

D

$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{OH}$