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JEE Main Chemistry Test- 2 - JEE MCQ


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25 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main Chemistry Test- 2

JEE Main Chemistry Test- 2 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Main Chemistry Test- 2 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Chemistry Test- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Chemistry Test- 2 below.
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JEE Main Chemistry Test- 2 - Question 1

Which of the following alkenes will react fastest with  under catalytic hydrogenation conditions

JEE Main Chemistry Test- 2 - Question 2

The heat of hydrogenation of 1-hexene is 126 . When a second double bond is introduced in the molecule, the heat of hydrogenation of the resulting compound is 230  . The resulting compound (diene) is

Detailed Solution for JEE Main Chemistry Test- 2 - Question 2
The Diene thus formed is more stable than the initial compound (and rest of the options) due to conjugation (a conjugated diene). Hence the heat of hydrogenation is more as the product will be a thermodynamically controlled one ie a TCP.
JEE Main Chemistry Test- 2 - Question 3

Heat of hydrogenation of benzene is 51 kcal/mol and its resonance energy is 36 kcal/mol. Then the heats of hydrogenation of cyclohexadiene and cyclohexene per mole respectively are

JEE Main Chemistry Test- 2 - Question 4
 is adsorbed on palladium surface.It is a case of
JEE Main Chemistry Test- 2 - Question 5

The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is

JEE Main Chemistry Test- 2 - Question 6
When 0.532 g of benzene (B.P.  C) is burnt in a constant volume system with an excess of oxygen, 22.3 KJ of heat is given out.  for the combustion process is given by :
JEE Main Chemistry Test- 2 - Question 7

The heat released when  & HCl neutralize is :

JEE Main Chemistry Test- 2 - Question 8
The heat of formation of  is 380 kcals/mole and that of  is 195 kcals/mole. The heat (in kcals/mole) of the thermite reaction is
JEE Main Chemistry Test- 2 - Question 9
Enthalpy of Rhombic sulphur of C is
JEE Main Chemistry Test- 2 - Question 10
In the gaseous equilibrium

the formation of  will be favored by
JEE Main Chemistry Test- 2 - Question 11

The  for the decomposition of  (if its degree of dissociation under one atomic pressure is 90%) is

Detailed Solution for JEE Main Chemistry Test- 2 - Question 11
So2cl2---so2+cl2
1. 0. 0
1-.90. .90. .90
total moles at equill. = 1.90
xso2 =.90/1.90 xcl2=.90/1.90 xso2cl2= .10/1.90

kp = pso2 *pcl2 /pso2cl2
kp = .90*.90/1.90*.10 = 4.26
JEE Main Chemistry Test- 2 - Question 12
The pH of  HCl is
JEE Main Chemistry Test- 2 - Question 13
If the maximum concentration of  in water is 0.01M at 298K, its maximum concentration in 0.1M NaCl will be
Detailed Solution for JEE Main Chemistry Test- 2 - Question 13
Pbcl2 = pb +2 + 2 cl - S S 2s 
Ksp = s * (2s)^2 
Ksp = 0.01 * (4 * 10^-4) 
Ksp = 4*10^-6 
Let solubility in Nacl be s’ Pbcl2 = pb + 2cl S’ S’ 2s’ + 0.1 
Neglecting 2s’ Ksp remains same 4*10^-6 = s’ * (0.1) ^2 
By solving this S’ = 4 * 10^-4
JEE Main Chemistry Test- 2 - Question 14

In a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is doubled, then the equilibrium constant will

JEE Main Chemistry Test- 2 - Question 15
Ammonium hydrogen sulphide is contained in a closed vessel at 313 K when total pressure at equilibrium is found to be 0.8 atm. The value of Kp for the reaction
    
Detailed Solution for JEE Main Chemistry Test- 2 - Question 15
If dissociation of NH4HS is x then at equilibrium,

2x=0.8 ( Using PV= nRT you can equate partial pressure to total pressure.)
so, x= 0.4 atm.
Then Kp= [P]²= (0.4)²=0.16
JEE Main Chemistry Test- 2 - Question 16
Reactivity of hydrogen atoms attached to different carbon atoms in alkanes has the order
JEE Main Chemistry Test- 2 - Question 17
Reaction of ethene with  in  gives
JEE Main Chemistry Test- 2 - Question 18
The major product obtained when isobutane is treated with chlorine in the presence of light is
JEE Main Chemistry Test- 2 - Question 19
The compound contains atoms X,Y,Z. The oxidation number of X is 2, Y is 5 and Z is –2, a possible formula of the compound is
JEE Main Chemistry Test- 2 - Question 20

 acts as

*Answer can only contain numeric values
JEE Main Chemistry Test- 2 - Question 21

Enthalpy of neutralization of H3PO3 acid is –106.68 KJ/mole using NaOH. If enthalpy of neutralization of HCl by NaOH is –55.84 KJ/mole. Calculate ΔHionisation of H3PO3 into its ions. (in KJ)


Detailed Solution for JEE Main Chemistry Test- 2 - Question 21

H3PO3 → 2H+ + HPO3–2

2H+ + 2OH → 2H2O
ΔH = –55.84 × 2 = –116.68
Now
–106.68 = ΔHion – 55.84 × 2
ΔHion = 5KJ/mol

*Answer can only contain numeric values
JEE Main Chemistry Test- 2 - Question 22

An aqueous solution of 6.3 g of oxalic acid dihydrate is made upto 250 mL. The volume of0.1 N NaOH required to completely neutralise 10 mL of this solution is :


Detailed Solution for JEE Main Chemistry Test- 2 - Question 22

*Answer can only contain numeric values
JEE Main Chemistry Test- 2 - Question 23

100 mL of 0.6 M acetic acid is shaken with 2 g activated carbon. The final concentration of the solution after adsorption is 0.5 M. What is the amount of acetic acid adsorbed per gram of carbon.


Detailed Solution for JEE Main Chemistry Test- 2 - Question 23

*Answer can only contain numeric values
JEE Main Chemistry Test- 2 - Question 24

What will be the resultant pH when 150 mL of an aqueous solution of HCl (pH = 2.0) is mixed with 350 mL of an aqueous solution of NaOH (pH = 12.0)?


Detailed Solution for JEE Main Chemistry Test- 2 - Question 24

*Answer can only contain numeric values
JEE Main Chemistry Test- 2 - Question 25

How many grams of NH4Cl should be dissolved per litre of solution to have a pH of 5.13 ? Kb for NH3 is 1.8 × 10–5.


Detailed Solution for JEE Main Chemistry Test- 2 - Question 25

NH4Cl  is a salt of strong acid and weak base for solutions of such salts.
pH = 1/2 [pK– log C – pKb]
⇒ 10.26 = 14 – log C – 4.74
⇒ log C = 9.26 – 10.26 = –1.0
∴ C = 10–1 M
[NH4Cl] = 10–1 M
= 10–1 × 53.5 gL–1
= 5.35 gL–1

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