1
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The major product formed in the following reaction is :
JEE Main 2018 (Online) 15th April Evening Slot Chemistry - Hydrocarbons Question 149 English
A
JEE Main 2018 (Online) 15th April Evening Slot Chemistry - Hydrocarbons Question 149 English Option 1
B
JEE Main 2018 (Online) 15th April Evening Slot Chemistry - Hydrocarbons Question 149 English Option 2
C
JEE Main 2018 (Online) 15th April Evening Slot Chemistry - Hydrocarbons Question 149 English Option 3
D
JEE Main 2018 (Online) 15th April Evening Slot Chemistry - Hydrocarbons Question 149 English Option 4
2
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are Mx and My, respectively where Mx = $${3 \over 4}$$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :
A
$${4 \over 3}$$
B
$${3 \over 4}$$
C
$${1 \over 2}$$
D
$${1 \over 4}$$
3
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For a first order reaction, A $$ \to $$ P, t1/2 (half-life) is 10 days The time required for $${1 \over 4}$$th conversion of A (in days) is : (ln 2 = 0.693,    ln 3 = 1.1)
A
5
B
3.2
C
4.1
D
2.5
4
JEE Main 2018 (Online) 15th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
At a certain temperature in a $$5$$ $$L$$ vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,
         CO + Cl2 $$\rightleftharpoons$$ COCl2
At equilibrium, if one mole of CO is present then equilibrium constant (Kc) for the reaction is :
A
2
B
2.5
C
3
D
4

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