1
JEE Main 2022 (Online) 25th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Match List I with List II:

List I
(molecule)
List II
(hybridization ; shape)
(A) XeO$$_3$$ (I) sp$$^3$$d ; linear
(B) XeF$$_2$$ (II) sp$$^3$$ ; pyramidal
(C) XeOF$$_4$$ (III) sp$$^3$$d$$^3$$ ; distorted octahedral
(D) XeF$$_6$$ (IV) sp$$^3$$d$$^2$$ ; square pyramidal

Choose the correct answer from the options given below:

A
A-II, B-I, C-IV, D-III
B
A-II, B-IV, C-III, D-I
C
A-IV, B-II, C-III, D-I
D
A-IV, B-II, C-I, D-III
2
JEE Main 2022 (Online) 25th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is

A
1 : 4
B
1 : 0.25
C
1 : 0.20
D
1 : 5
3
JEE Main 2022 (Online) 25th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$${K_{{a_1}}}$$, $${K_{{a_2}}}$$ and $${K_{{a_3}}}$$ are the respective ionization constants for the following reactions (a), (b) and (c).

(a) $${H_2}{C_2}{O_4} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + H{C_2}O_4^ - $$

(b) $$H{C_2}O_4^ - \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + {C_2}O_4^{2 - }$$

(c) $${H_2}{C_2}O_4^{} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2{H^ + } + {C_2}O_4^{2 - }$$

The relationship between $${K_{{a_1}}}$$, $${K_{{a_2}}}$$ and $${K_{{a_3}}}$$ is given as :

A
$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$+$$ $${K_{{a_2}}}$$
B
$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$-$$ $${K_{{a_2}}}$$
C
$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$/$$ $${K_{{a_2}}}$$
D
$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$\times$$ $${K_{{a_2}}}$$
4
JEE Main 2022 (Online) 25th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is $${\Lambda _{m1}}$$ and that of 20 moles another identical cell heaving 80 mL NaCl solution is $${\Lambda _{m2}}$$. The conductivities exhibited by these two cells are same. The relationship between $${\Lambda _{m2}}$$ and $${\Lambda _{m1}}$$ is

A
$${\Lambda _{m2}}$$ = 2$${\Lambda _{m1}}$$
B
$${\Lambda _{m2}}$$ = $${\Lambda _{m1}}$$ / 2
C
$${\Lambda _{m2}}$$ = $${\Lambda _{m1}}$$
D
$${\Lambda _{m2}}$$ = 4$${\Lambda _{m1}}$$
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