1
JEE Main 2023 (Online) 6th April Evening Shift
+4
-1

The volume of $$0.02 ~\mathrm{M}$$ aqueous $$\mathrm{HBr}$$ required to neutralize $$10.0 \mathrm{~mL}$$ of $$0.01 ~\mathrm{M}$$ aqueous $$\mathrm{Ba}(\mathrm{OH})_{2}$$ is (Assume complete neutralization)

A
7.5 mL
B
5.0 mL
C
10.0 mL
D
2.5 mL
2
JEE Main 2023 (Online) 31st January Evening Shift
+4
-1
When a hydrocarbon A undergoes complete combustion it requires 11 equivalents of oxygen and produces 4 equivalents of water. What is the molecular formula of $A$ ?
A
$\mathrm{C}_{9} \mathrm{H}_{8}$
B
$\mathrm{C}_{5} \mathrm{H}_{8}$
C
$\mathrm{C}_{11} \mathrm{H}_{4}$
D
$\mathrm{C}_{11} \mathrm{H}_{8}$
3
JEE Main 2023 (Online) 30th January Evening Shift
+4
-1
$1 \mathrm{~L}, 0.02 \mathrm{M}$ solution of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right]$ Br is mixed with $1 \mathrm{~L}, 0.02 \mathrm{M}$ solution of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}$. The resulting solution is divided into two equal parts $(\mathrm{X})$ and treated with excess of $\mathrm{AgNO}_{3}$ solution and $\mathrm{BaCl}_{2}$ solution respectively as shown below:

$1 \mathrm{~L}$ Solution $(\mathrm{X})+\mathrm{AgNO}_{3}$ solution (excess) $\longrightarrow \mathrm{Y}$

$1 \mathrm{~L}$ Solution $(\mathrm{X})+\mathrm{BaCl}_{2}$ solution (excess) $\longrightarrow \mathrm{Z}$

The number of moles of $\mathrm{Y}$ and $\mathrm{Z}$ respectively are
A
$0 .01,0.01$
B
$0.01,0.02$
C
$0.02,0.01$
D
$0.02,0.02$
4
JEE Main 2023 (Online) 29th January Evening Shift
+4
-1

When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.5 equivalents of oxygen and produces 3 equivalents of water. What is the molecular formula of A?

A
$$\mathrm{C_9H_6}$$
B
$$\mathrm{C_6H_6}$$
C
$$\mathrm{C_8H_6}$$
D
$$\mathrm{C_9H_9}$$
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