500+ New IIT JEE Main Chemistry MCQ with Solutions MCQ Village

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.

Q11.The following equilibrium exists in aqueous solution CH3COOH ⇄ CH3COOH— + H+. If dilute
HCl is added without a change in temperature then the

(A) Concentration of CH3COO– will increase
(B) Concentration of CH3COO– will decrease
(C) Equilibrium constant will increase
(D) Equilibrium constant will decrease

Explain: When adding HCl in CH3COOH solution the concentration of H+ is increased. So reaction is proceed in reverse direction and the concentration of CH3COO– is decreased.

Q12.Reaction in which yield of product will increases with increases in pressure is

(A) H2(g) + I2(g) ⇄ 2HI(g (B) H2O(g) + CO(g) ⇄ CO2(g)+H2(g)
(C) H2O(g) + C(s) ⇄ CO(g) + H2(g) (D) CO(g)+ 3H2(g) ⇄ CH4(g) + H2O(g)

Explain: In reaction CO+3H2 CH4+H2O Volume is decreasing in forward direction so on increasing pressure the yield pf product will increase.

Q13.The distance between tow nearest neighbors in body centered cubic lattice of axial length, l, is

(A) l (B) (√3/2)I (C) (√2/2)I (D) none

Explain: For BCC :√3l = 4r BCC crystal : √3l = 4r  Distance between two nearest neighbour, 2r = √3l/2I

Q14. Let the height of hop unit cell is ‘h’. The height of tetrahedral voids from the bases i

(A) h/2 (B) h/2,2h/3 (C) h/8,7h/8 (D) h/4,3h/4

Explain: between A & B layers, there are octahedral voids and exctly in between a layer and octahedral voides, there are tetrahedral voides.

Q15.3.0 molal NaOH solution has a density of 1.110 g/ml. The molarity of the solution is

(A) 3.0504 (B) 3.64 (C) 3.05 (D) 2.9732

Ans. : D

Q16.Boiling point of chloroform was raised by 0.323 K, when 0.5143 g of anthracence was dissolved in its its 35 g. Molecular mass of ant hracence is (Kb for CHCl3 = 3.9 k-kg-mol–1)

(A) 79.42 g/mol (B) 132.32 g/mol (C) 177.42 g/mol (D) 242.32 g/mol

Ans. : C
Explain: Here: Tb = 0.323K w=0.5143g weight of Anthracence. W= 35g weight of chloroform Kb= Molal elevation constant (3.9 K–Kg/ mol ) M= b K W 100 35 0.323 0.5143    =177.42g/mol

Q17.A solution containing 30 gms of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mm Hg at 25° C. Further 18 gms of water is then added to the solution. the resulting solution has a vapour pressure of 22.15 mm Hg at 25°C, Calculate the molecular weight of the solute

(A) 74.2 (B) 75.6 (C) 70.3 (D) 78.7

Ans. :C

Q18.With 63 gm of oxalic acid how many litres of N 10 solution can be prepared

(A) 100 litre (B) 10 litre (C) 1 litre (D) 1000 litre

Ans. :B

Q19.The heat evolved on combustion of 1 gm of starch, (C6H10O5), into CO2(g) and H2O(l) is 4.18 Kcal. What is the standard enthalpy of formation of 1 gm of starch ? Heat of formation of CO2(g) and H2O(l) are –94.05 and –68.32 Kcal/mol.

(A) –2.82 Kcal (B) –0.71 Kcal (C) – 1.41 Kcal (D) – 8.46 Kcal

Ans. :C
Explain : First calculate CH per mol and then convert it inot per gm. The required thermochemical equation is 6x C(s)+5x H2 (g)+ 2 5x O2(g)  ( C 6H10O5)x; fH= ? Now, fH= C C(s) C H2 (g) 6x   H  5x   H  – [ H ] C (C6H10O5)x  = 6x × (–94.05)+5x × (–68.32) –(– 4.18) × 162x = – 228.74 x Kcal/mol = 162x – 228.74x = – 1.41 Kcal/gm

Q20.The bond enthalpies of C–C=C and bonds are are 348, 610 and 835 kJ/mol, respectively at
298K and 1 bar. The enthalpy of polymeristion permole mole of 2-Butyne at 298K and 1 bar, as
shown below, is
n CH3–C = C—CH3 (g) → –(CH2–CH=CH–CH2)n –(g)

(A) –123 kJ (B) –132 kJ (C) – 139 kJ (D) – 37 kJ

Ans. :A
Explain : On cheavage of one -bond, one -bond is formed in the polymerisation reactions. H=B.E.C–C  bond = (B.E.CC–B.E.C=C) – (B.E.C–C) = (835–610) – 348 = – 123 kJ/Mol

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