1
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ?
[Molar mass of NaHCO3 = 84 g mol–1]
A
33.6
B
0.84
C
8.4
D
16.8
2
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
An organic compound is estimated through Duma's method and was found to evolve 6 moles of CO2. 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is :
A
C6H8N
B
C6H8N2
C
C12H8N
D
C12H8N2
3
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below :

Mx+ (aq)/M(s) Au3+(aq)/Au(s) Ag+(aq)/Ag(s) Fe3+(aq)/Fe2+ (aq) Fe2+(aq)/Fe(s)
E0Mx+/M/(V) 1.40 0.80 0.77 $$-$$0.44


If $$E_{z{n^{2 + }}/zn}^0$$ = $$-$$ 0.76 V, which cathode will give maximum value of Eocell per electron transferred?
A
Ag+/Ag
B
Fe3+/Fe2+
C
Au3+/Au
D
Fe2+/Fe
4
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The major product of the following reaction is :

JEE Main 2019 (Online) 11th January Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 149 English
A
JEE Main 2019 (Online) 11th January Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 149 English Option 1
B
JEE Main 2019 (Online) 11th January Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 149 English Option 2
C
JEE Main 2019 (Online) 11th January Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 149 English Option 3
D
JEE Main 2019 (Online) 11th January Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 149 English Option 4

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