500+ New IIT JEE Main Chemistry MCQ with Solutions

If you are looking for IIT JEE Main Chemistry Multiple Choice Question and Answers(Solutions) Then you are at right place. Here you will find the New Pattern IIT JEE Mains Chemistry Multiple Choice Question with the updated version of syllabus of IIT JEE as per your requirements.

Q1.A sample of protein was analyed for metal content and analysed for metal content and
analysisi revealed that it contained magnesium and titanium in equal amounts, by mass. If
these are the only metallic species present in the protein and contains 0.016% metal, the
minimum possible molar mass of the portein is [Mg=24, Ti = 48]

(A) 600000 (B) 150000 (C) 300000 (D) 1200000

A
Explain: Both the metals are 0.008% by mass and no. of metal atoms in each molecule must be integer.

Q2.1 mol of N2 and 4 mol of H2 are allowed to react in a vessel to form NH3 gas as only product
and after reaction, water is added. Aqueous solution required 1 mol of HCl for complete
reaction. Mol fraction of H2 in the gas mixture after reaction is


(A) 1/6 (B) 5/6 (C) 1/3 (D) 1.0

B
Explain : Only 1 mole of NH3 gas is formed Soluiton requires 1 mole HCl and hence, only 1 mole of NH3 is formed. HCl + NH3 > NH4Cl 1 mole 1 mole For NH3 formation

Q3.pH of 0.05 M H2SO4 solution will be

(A) 0.005 (B) 2 (C) 1 (D) 0.01

B
Explain : 0.005 M H2SO4 =0.01 MH+ = 10–2 MH+,

Q4.Which of the following does not make any change in pH when added to 10 mi dilute HCl

(A) 5 ml pure water (B) 20 ml pure water (C) 10 ml HCl (D) Same 20 ml dilute HCl

B
Explain : Buffer solution is formed. So the pH will not change.

Q5.The pH of a soft drink is 3.82. Its hydrogen ion concentration will be

(A) 1.96×10–2 mol/l (B) 1.96×10–3 mol/l (C)1.5×10–4 mol/ l (D) 1.96×10–1 mol /l

C
Explain : pH=.82 = – log [H+] .: [H+] = 1.5×10–4 mole/litre.

Q6.A monoprotic acid in a 0.1 M solution ionizes to 0.0001%. Its ionization constant is

(A) 1.0×10–3 (B) 1.0×10–6 (C) 1.0×10–8 (D) 1.0×10–10

D
Monobasic acid [H+] = 0.1 M : a2 K C : K=a2×C= 2 0.0001 0.1 100 K=10–13.

Q7.In the reaction A+2B ⇄ 2C, if 2 moles of A.3.0 moles of B and 2.0 moles of C are placed in
a 2.0 lit. flask flask and the equilibrium concentration of C is 0.5 mole/ l. The equilibrium
constant (KC) for the reaction is

(A) 0.073 (B) 0.147 (C) 0.05 (D) 0.026

C

Q8.For which state of matter, the coefficient of cubic expansion is independent to chemical composition ?

(A) solid (B) liquid (C) gas (D) all

C
Explain : Express charle’s Law in ºC For thermal expansion of any substance : Vt = V0(1+t) From Charle’s law application for all the gases : Vt = V0  The coefficient of cubic expansion for all the gases, g = 273 1 /ºC = constant

Q9.4.5 moles each of hydrogen and iodine headed in a sealed ten litre vessel. At equilibrium, 3
moles of HI were found. The equilibrium constant for H2(g) + I 2(g) ⇄ 2HI(g) is

(A) 1 (B) 10 (C) 5 (D) 0.33

Ans: C
Explain : It is weak electrolyte because it is slightly ionized.

Q10.In a chemical equilibrium A+B ⇄ C+D, when one mole each of the two reactants are mixed,
0.6 mole each of the products are formed. The equilibrium constant calculated is

(A) 1 (B) 0.36 (C) 2.25 (D) 4/9

Ans : c
Explain : Gaseous molecules are travelling randomly in in all direction. As all the molecular collisions are perfectly elastic.


Leave a comment