JEE Main Previous Year Questions (2021): Some Basic Concepts of Chemistry

[JEE Main MCQs]

Q1: The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H₃PO₃ solution and 100 mL of 2 M H₃PO₂ solution, respectively, are :
(a) 100 mL and 50 mL
(b) 100 mL and 200 mL
(c) 100 mL and 100 mL
(d) 50 mL and 50 mL
Ans: (b)
Millimoles of H₃PO₃ = M × V = 1 × 50 = 50
For 1 millimole of H₃PO₃, we require 2 millimoles of NaOH.
For 50 millimole of H₃PO₃, we require (2 × 50) = 100 millimoles of NaOH.
Millimoles of NaOH = M × V = 100
1 × V = 100
V = 100 mL
Millimoles of H₃PO₂ = M × V = 2 × 100 = 200
For 1 millimole of H₃PO₂, we require 1 millimoles of NaOH.
For 200 millimoles of H₃PO₂, we require 200 millimoles of NaOH.
So, volume of NaOH = 200 mL

Q2: Complete combustion of 1.80 g of an oxygen-containing compound (C_xH_yO_z) gave 2.64 g of CO₂ and 1.08 g of H₂O. The percentage of oxygen in the organic compound is :
(a) 51.63
(b) 50.33
(c) 63.53
(d) 53.33
Ans: (d)
C_xH_yO_z + O₂ → xCO₂ + Y/2 H₂O
2.64 g of CO2 contains 0.72 g C.
1.08 g of H2O contains 0.12 g H.
∴ Mass of oxygen present = 1.80 - (0.72 +0.12) = 0.96 g
% of O = 0.96/1.80 × 100 = 53.33 %

[JEE Main Numericals]

Q3: The number of atoms in 8g of sodium is x × 10²³. The value of x is ____________. (Nearest integer) [Given : NA = 6.02 × 10²³ mol ^{- 1} 
Atomic mass of Na = 23.0 u]
Ans: 2
Given, mass of Na = 8g Molar mass of Na = 23 gmol^-1 

Number of atoms = 8 × 6.022/23 × 10²³ 
Number of atoms = 2.09 × 10²³ 
x ≈ 2
Hence, answer is 2.

Q4: If 80 g of copper sulphate CuSO₄ . 5H₂O is dissolved in deionised water to make 5L of solution. The concentration of the copper sulphate solution is x × 10 ^{- 3} mol L ^{- 1}. The value of x is _____________. [Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]
Ans: 64
Given, mass of CuSO₄ . 5H₂O = 80 g
The concentration of copper sulphate solution is x × 10^-3 mol/L.
 .
Molar mass of CuSO₄ . 5H₂O = 63.54 + 32 + 16 × 4
= 5 × 18 = 249.54 g/mol

Volume of solution = 5 L
From Eq. (i),
Molarity = 0.3205/5 = 64.11 × 10 ^{- 3} mol/L
∴ x = 64.11
or x ≈ 64
Hence, answer is 64.

Q5: Sodium oxide reacts with water to produce sodium hydroxide. 20.0 g of sodium oxide is dissolved in 500 mL of water. Neglecting the change in volume, the concentration of the resulting NaOH solution is ______________ × 10^-1 M. (Nearest integer)
[Atomic mass : Na = 23.0, O = 16.0, H = 1.0]
Ans: 13
Na₂O + H₂O → 2NaOH
20/62 moles
Moles of NaOH formed = 20/62 × 2
= 1.29 M = 13 × 10^-1 M (Nearest integer)

Q6: The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H₂C₂O₄.2H₂O) in 250 mL of water in mol L^-1 is x × 10^-2. The value of x is _____________. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]
Ans: 20

x = 20

Q7: 100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x × 10 ^{- 2}. The value of x is ____________. (Nearest integer) [Atomic weight : H = 1.008; C = 12.00; O = 16.00]
Ans: 19
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
t = 0    2.27 mole    31.25 mol
t = ∞ 0    19.9 mol    6.81 mol    9.08 mol
mole fraction of CO₂ in the final reaction mixture (heterogenous)

= 0.1902 = 19.02 × 10^-2 
⇒ 19

Q8: In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ______________. (Nearest integer)
[Atomic mass : Ag = 108, Br = 80]
Ans: 40
= 10^-3 mol
⇒ n_Br = n_AgBr = 0.001 mol
⇒ mass_Br = (0.001 × 80) gm = 0.08 gm
⇒ mass% =  = 40 %

Q9: 100 mL of Na₃PO₄ solution contains 3.45 g of sodium. The molarity of the solution is _____________ × 10^-2 mol L^-1. (Nearest integer) [Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]
Ans: 50
Molarity of Na₃PO₄ Solution =

= 0.5 = 50 × 10^-2

Q10: An aqueous KCl solution of density 1.20 g mL^-1 has a molality of 3.30 mol kg^-1. The molarity of the solution in mol L^-1 is ____________ (Nearest integer) [Molar mass of KCl = 74.5]
Ans: 3
1000 kg solvent has 3.3 moles of KCl
1000 kg solvent → 3.3 × 74.5 gm KCl → 245.85
Weight of solution = 1245.85 gm
Volume of solution = 1245.85/1.2 ml
So molarity =  × 1000 = 3.17

Q11: 2SO₂(g) + O₂(g) → 2SO₃(g)
The above reaction is carried out in a vessel starting with partial pressure PSO₂ = 250 m bar, PO₂ = 750 m bar and PSO₃ = 0 bar. When the reaction is complete, the total pressure in the reaction vessel is _______ m bar. (Round off of the nearest integer).
Ans: 875
2SO₂(g) + O₂(g) → 2SO₃(g)

∴ Final total pressure = 625 + 250 = 875 m bar

Q12: The density of NaOH solution is 1.2 g cm^-3. The molality of this solution is _____________ m. (Round off to the Nearest Integer)
[Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of H2O : 1.0 g cm^-3
Ans: 5
Consider 1 litre solution
mass of solution = (1.2 × 1000)g = 1200 gm
Neglecting volume of NaOH
Mass of water = 1000 gm
⇒ Mass of NaOH = (1200 - 1000)gm = 200 gm
⇒ Moles of NaOH = 200g/50g/mol = 5 mol
⇒ molality = 5mol/1kg = 5 m

Q13: The number of significant figures in 0.00340 is ____________.
Ans: 3
Number of significant figures = 3

Q14: When 10 mL of an aqueous solution of Fe²+ ions was titrated in the presence of dil H₂SO₄ using diphenylamine indicator, 15 mL of 0.02 M solution of K₂Cr₂O₇ was required to get the end point. The molarity of the solution containing Fe²⁺ ions is x  10^-2 M. The value of x is __________. (Nearest integer)
Ans: 18
Milli-equivalents of Fe²⁺ = milli-equivalents of K₂Cr₂O₇
M × 10 × 1 = 0.02 × 15 × 6
⇒ M = 0.18 = 18 × 10^-2 M

Q15: Consider the complete combustion of butane, the amount of butane utilized to produce 72.0 g of water is ___________ × 10^-1 g. (in nearest integer)
Ans: 464

Moles of H₂O = 72/18 = 4
Moles of C₄H₁0 used = 1/5 × 4
Wight of C₄H₁₀ used = 4/5 × 58
= 46.4 gm = 464 × 10^-1 g

Q16: If the concentration of glucose (C₆H₁₂O₆) in blood is 0.72 g L^-1, the molarity of glucose in blood is ____________ × 10^-3 M. (Nearest integer) [Given: Atomic mass of C = 12, H = 1, O = 16 u]
Ans: 4


Q17: 4g equimolar mixture of NaOH and Na₂CO₃ contains x g of NaOH and y g of Na₂CO₃. The value of x is ____________ g. (Nearest integer)
Ans: 1
Mass of NaOH = x
Moles of NaOH = x/40
Mass of Na₂CO₃ = y
Moles of Na₂CO₃ = y/100
x/40 = y/106
x + y = 4
x = 1.1, y = 2.9
x = 1.1 ≈ 1 (nearest integer)

Q18: 250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ___________ × 10²¹. (Nearest integer)
(N_A = 6.022 × 10²³)
Ans: 226
We know that, number of moles = V_L × molarity and number of millimoles = V_mL × molarity
So, millimoles of NaOH = 250 × 0.5 = 125
Millimoles of HCl = 500 × 1 = 500
Now, reaction is

125 millimoles of NaOH reacts with 125 millimoles of HCl. So, millimoles of HCl left = 375
Moles of HCl = 375 × 10^-3 
Number of HCl molecules
= Avogadro's constant (N_A) × moles of HCl
= 6.022 × 10²³ × 375 × 10^-3
= 225.8 × 10²¹ = 226 × 10²¹ 
Therefore, the answer is 226.

Q19: An average person needs about 10000 kJ energy per day. The amount of glucose (molar mass = 180.0 g mol^-1) needed to meet this energy requirement is ____________ g.
(Use : Δ_CH(glucose) = - 2700 kJ mol^-1)
Ans: 667
1 mole glucose give 2700 kJ energy,
so mole of glucose needed for 104 kJ energy
= 10000/2700 = 3.703 moles
Weight of glucose = 3.703 × 180 g/moles
= 666.666
≈ 667 g
Hence, the amount of glucose required is 667 g.

Q20: 10.0 mL of Na₂CO₃ solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings :
4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL
Based on these readings, and convention of titrimetric estimation the concentration of Na₂CO₃ solution is ___________ mM.
(Round off to the Nearest Integer).
Ans: 50
From the given value of HCl, it is clear that most appropriate volume of HCl used is 5 ml because it occurs most number of times.
Na₂CO₃ + 2 HCl → 2 NaCl + CO₂ + H₂O
n factor of Na₂CO₃ = 2
n factor of HCl = 1
equivalent of Na₂CO₃ = equivalent of HCl

= 0.05 M
= 50 × 10^-3 M
= 50 mM

Q21: __________ grams of 3-Hydroxy propanal (MW = 74) must be dehydrated to produce 7.8 g of acrolein (MW = 56) (C₃H₄O) if the percentage yield is 64. (Round off to the Nearest Integer).
[Given: Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u ]
Ans: 16

∴ Mass of acrolein = [x/74 × 0.64 ] × 56 = 78
⇒ x = 16.1 gm ≃ 16gm

Q22: A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n × 10^-1, when n = __________. (Round off to the Nearest Integer).
(Given: Atomic masses: C : 12.0 u, H: 1.0 u, N: 14.0 u, Br: 80.0 u]
Ans: 3
Number of moles of benzyl trimethyl
ammonium bromide formed = 23/230 = 0.1
∴ No. of moles of bromomethane consumed
= 3 × 0.1
= 3 × 10^-1

Q23: Complete combustion of 3g of ethane gives x × 10²² molecules of water. The value of x is __________. (Round off to the Nearest Integer). [Use : N_A = 6.023 × 10²³; Atomic masses in u : C : 12.0; O : 16.0; H : 1.0]
Ans: 18

From 1 mol C₂H₆ we get 3 mol H₂O
∴ From 0.1 mol C₂H₆ we get = 3 × 0.1 mol H₂O
∴ Moles of H₂O = 0.3
= 0.3 × 6.023 × 10²³ molecules
= 1.8069 × 10²³ molecules
= 18.069 × 10²² molecules

Q24:

Consider the above reaction. The percentage yield of amide product is __________. (Round off to the Nearest Integer).
(Given: Atomic mass: C : 12.0 u, H : 1.0 u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)
Ans: 77

Stoichiometric moles of amide = 10^-3 mol
Actual weight of amide = 10^-3 × 273 = 0.273 g
% yield = 0.210/0.273 × 100
= 76.9%
≃ 77%

Q25: 15 mL of aqueous solution of Fe²⁺ in acidic medium completely reacted with 20 mL of 0.03 M aqueous . The molarity of the Fe²⁺ solution is __________ × 10^-2 M. (Round off to the Nearest Integer).
Ans: 24

Valance Factor of Cr₂O₇^2- = 6
Valance Factor of Fe²⁺ = 1
mili eq. of Cr₂O₇^2- = mili eq. of Fe²⁺ 
6(0.03 × 20) = 1(M × 15)
M = 6(0.03)4/3
M = 0.24 M
M = 24 × 10^-2 M

Q26: The mole fraction of a solute in a 100 molal aqueous solution is ___________ × 10^-2. (Round off to the Nearest Integer). [Given : Atomic masses : H : 1.0 u, O : 16.0 u ]
Ans: 64
Let weight of H₂O = 1000 g
Moles of solute = 100
(mole)H₂O = 1000/18
Mole fraction of solute =  

x_solute = 64 × 10^-2

Q27: When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, _________ × 10^-5 moles of lead sulphate precipitate out. (Round off to the Nearest Integer).
Ans: 525

For 3 moles of Pb(NO₃)₂ , we require 1 mole of Cr₂(SO₄)₃ 
For 5.25 moles of Pb(NO₃)₂, we require 1/3 × 5.25 mole of Cr₂(SO₄)₃ = 1.75 moles
But we have 2.4 moles. So, Cr₂(SO₄)3 is excess reagent and Pb(NO₃)₂ is limiting reagent, (LR)
Moles of PbSO₄ formed = moles of Pb(NO₃)₂ consumed = 5.25 m mol = 525 × 10^-5 moles

Moles of PbSO₄ formed = moles of Pb(NO₃)₂ consumed
= 5.25 m mol = 525 × 10^-5 moles

Q28: Complete combustion of 750 g of an organic compound provides 420 g of CO₂ and 210 g of H₂O. The percentage composition of carbon and hydrogen in organic compound is 15.3 and ___________ respectively. (Round off to the Nearest Integer).
Ans: 3
18 gm H₂O ⇒ 2 gm H₂ 
210 gm ⇒ 2/18 × 210
= 23.33 gm H₂ 
So, % H₂ = 23.33/750 × 100 = 3.11% ≈ 3%

Q29:

If the above equation is balanced with integer coefficients, the value of c is ___________. (Round off to the Nearest Integer).
Ans: 16


Q30: A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm^-3. The molarity of the solution is ____________ mol dm^-3. (Round off to the Nearest Integer). [Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]
Ans: 9

⇒ 12285 - 364 M = 1000 M
⇒ 1364 M = 12285
⇒ M = 92

Q31:  The NaNO₃ weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na⁺ per mL is _________ g. (Rounded off to the nearest integer)
[Given: Atomic weight in g mol^-1 - Na : 23; N : 14; O : 16]
Ans: 13
Na⁺ = 70 mg/mL
WNa⁺ in 50 mL solution
= 70 × 50 mg
= 3500 mg
= 3.5 gm
Moles of Na⁺ in 50 mL solution = 3.5/23
Moles of NaNO₃ = moles of Na⁺ 
= 3.5/23 mol
Mass of NaNO₃ = 3.5/23 × 85 = 12.934
≃ 13 gm

Q32: The number of significant figures in 50000.020 × 10^-3 is _____________.
Ans: 8
10^-3 has no role in significant digits. Here in 50000.020, Number of significant figure = 8 as all zeroes between non zero digits are counted as significant digits and also the last zero is also significant digit as zeroes at the end or right of the number is significant only if they are present at the right side of the decimal.

Q33: In basic medium CrO₄^-2 oxidises  S₂O₃^-2 to form SO₄^2- and itself changes into Cr(OH)₄^- . The volume of 0.154 M CrO₄^-2 required to react with 40 mL of 0.25 M S₂O₃^2- is ....... mL.
Ans: 173

Applying mole-mole analysisApplying mole-mole analysis 

v = 173 mL

Q34: The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O₂ for complete oxidation and produces 4 times its volume of CO₂ is C_xH_y. The value of y is _____________.
Ans: 8
Combustion reaction:

Suppose, volume of C_xH_y is V and volume of O₂ is 6 times greater than C_xH_y = 6V
then volume of xCO₂ ⇒ V_x = 4 V
x = 4
 
Put value of x = 4 in Eq. (i) we get,
4 + y/4 = 6 ⇒ y = 8

Q35: 1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is __________ × 10^-2.
Ans: 243

Given, weight = 18.6 g
Here, 1 mole of aniline gives 1 mole of acetanilide
∴ mole of aniline = mole of acetanilide

But efficiency of reaction is 90% only.
Hence, mass of acetanilide produced = 2.70 × 90/100g = 2.43g = 243 × 10²
x = 243

Q36: 4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is x × 10^-1. The value of x is _______. (Rounded off to the nearest integer)
Ans: 2
Given, weight of compound A = 4.5 g
Molecular weight of compound A = 90 g/mol
Volume of solution (in mL) = 250 mL
Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).

Hence, x × 10^-1 μ
x = 2