JEE Main Chemistry questions with solutions

Q1. A 100 mL sample of a urea solution contains 6.02 × 10²⁰ molecules of urea. What is the molarity of the solution? (Take Avogadro constant, N_A = 6.02 × 10²³ mol^−1.)

1. 0.02 M
2. 0.01 M
3. 0.001 M
4. 0.1 M

Answer: 0.01 M

6.02e20 / 6.02e23 = 1e-3 mol of urea in 0.100 L, so molarity = 0.001/0.1 = 0.01 M. Stored answer 0.001 M confuses moles with molarity.

Q2. When 4.9 g of H2SO4 reacts with NaCl, the products formed are 6 g of sodium hydrogen sulphate and 1.825 g of HCl. What mass of NaCl was decomposed?

1. 6.921 g
2. 4.65 g
3. 2.925 g
4. 1.4 g

Answer: 2.925 g

H2SO4 (4.9 g) + NaCl = NaHSO4 (6 g) + HCl (1.825 g); by conservation NaCl = 7.825 - 4.9 = 2.925 g (= 0.05 mol). Stored 4.65 g is wrong.

Q3. Which of the following has the least mass?

1. 0.2 mol of hydrogen gas
2. 6.023 × 10²² molecules of nitrogen
3. 0.1 g of silver
4. 0.1 mol of oxygen gas

Answer: 0.1 g of silver

0.2 mol H2 = 0.4 g; 6.023e22 N2 = 0.1 mol = 2.8 g; 0.1 g Ag = 0.1 g; 0.1 mol O2 = 3.2 g. Least is 0.1 g of silver, not the hydrogen.

Q4. What amount of magnesium phosphate, Mg3(PO4)2, in moles is needed to provide 0.25 mole of oxygen atoms?

1. 1.25 × 10^−2
2. 2.5 × 10^−2
3. 0.02
4. 3.125 × 10^−2

Answer: 3.125 × 10^−2

Each Mg3(PO4)2 contains 8 O atoms, so moles = 0.25/8 = 0.03125 = 3.125e-2. Stored 2.5e-2 is wrong.

Q5. A sulphuric acid solution has a concentration of 3.60 M and contains 29% H2SO4 by mass. If the molar mass of H2SO4 is 98 g mol⁻¹, what is the density of the solution in g mL⁻¹?

1. 1.45
2. 1.64
3. 1.88
4. 1.22

Answer: 1.22

Density = molarity x molar mass / (10 x %w/w) = 3.60 x 98 / (10 x 29) = 1.22 g/mL. Stored 1.64 is wrong.

Q6. A sample of gas has a volume of 300 cc at 27°C under a pressure of 620 mm. What will its volume be at 47°C and 640 mm pressure?

1. 260 cc
2. 310 cc
3. 390 cc
4. 450 cc

Answer: 310 cc

The correct option is derived from applying the combined gas law, which states that the volume of a gas is directly proportional to its temperature and inversely proportional to its pressure. By substituting the initial and final conditions into the equation, the calculations yield a final volume of 310 cc.

Q7. Haemoglobin has 0.33% iron by mass. If its molecular mass is about 67200 and the atomic mass of iron is 56, how many iron atoms are present in a single haemoglobin molecule?

1. 6
2. 1
3. 2
4. 4

Answer: 4

The percentage of iron by mass in haemoglobin indicates that for every 100 grams of haemoglobin, there are 0.33 grams of iron. Given the molecular mass of haemoglobin is approximately 67200 g/mol, the mass of iron in one mole of haemoglobin is calculated to be 0.33% of 67200, which equals about 221 grams. Dividing this by the atomic mass of iron (56 g/mol) shows that there are approximately 4 iron atoms in each haemoglobin molecule.

Q8. What volume of 20-volume hydrogen peroxide is needed to produce 5 L of oxygen gas at STP?

1. 250 mL
2. 125 mL
3. 100 mL
4. 50 mL

Answer: 250 mL

To make 5000 mL O2 you need 5000/20 = 250 mL of 20-volume H2O2. Stored 125 mL is wrong.

Q9. For the reaction 2Al(s) + 6HCl(aq) → 2Al³+(aq) + 6Cl^−(aq) + 3H₂(g), which statement is correct?

1. 11.2 L of H₂(g) at STP is formed for each mole of HCl(aq) used up
2. 6 L of HCl(aq) is used for every 3 L of H₂(g) formed
3. 33.6 L of H₂(g) is formed, independent of temperature and pressure, for every mole of Al that reacts
4. 67.2 L of H₂(g) at STP is formed for every mole of Al that reacts

Answer: 11.2 L of H₂(g) at STP is formed for each mole of HCl(aq) used up

2Al->3H2 means 1 mol Al gives 1.5 mol H2 = 33.6 L (not 67.2). The correct true statement is option 0: 6 HCl -> 3 H2, so 1 mol HCl gives 0.5 mol H2 = 11.2 L at STP.

Q10. Commercial concentrated sulphuric acid contains 95% H₂SO₄ by mass. If its density is 1.834 g cm⁻³, what is the molarity of the solution?

1. 17.8 M
2. 12.0 M
3. 10.5 M
4. 15.7 M

Answer: 17.8 M

The correct option is right because the molarity is determined by the concentration of the solute in the solution, which is calculated using the density and mass percentage of the acid, leading to a result of 17.8 M.

Q11. When 50 mL of a 16.9% AgNO₃ solution is combined with 50 mL of a 5.8% NaCl solution, what mass of precipitate is produced? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)

1. 28 g
2. 3.5 g
3. 7 g
4. 14 g

Answer: 7 g

The correct option is 7 g because the reaction between AgNO₃ and NaCl produces AgCl as a precipitate, and stoichiometric calculations based on the concentrations and volumes of the solutions show that 7 g of AgCl is formed.

Q12. A transition metal M gives a volatile chloride whose vapour density is 94.8. If this chloride contains 74.75% chlorine by mass, what is the formula of the metal chloride?

1. MCl₃
2. MCl₂
3. MCl₄
4. MCl₅

Answer: MCl₄

MW = 2 x 94.8 = 189.6; Cl mass = 0.7475 x 189.6 = 141.7 g -> 141.7/35.5 = 4 Cl atoms, so MCl4. Stored MCl2 is wrong.

Q13. On complete combustion, a gaseous hydrocarbon produces 0.72 g of water and 3.08 g of carbon dioxide. What is the empirical formula of this hydrocarbon?

1. C2H4
2. C3H4
3. C6H5
4. C7H8

Answer: C7H8

CO2 3.08/44 = 0.07 mol C; H2O 0.72/18 = 0.04 mol -> 0.08 mol H. C:H = 0.07:0.08 = 7:8 -> C7H8. Stored C3H4 is wrong.

Q14. A sample of washing soda has the following composition by mass: - Na₂CO₃: molecular mass 106.0, mass percentage 84.8 - NaHCO₃: molecular mass 84.0, mass percentage 8.4 - NaCl: molecular mass 58.5, mass percentage 6.8 If 1 kg of this sample is treated completely with excess HCl, the amount of carbon dioxide released will be:

1. 9 mol of CO₂
2. 16 mol of CO₂
3. 17 mol of CO₂
4. 18 mol of CO₂

Answer: 9 mol of CO₂

Na2CO3 848/106 = 8 mol -> 8 CO2; NaHCO3 84/84 = 1 mol -> 1 CO2; NaCl gives none. Total 9 mol CO2. Stored 17 is wrong.

Q15. A nitrogen–hydrogen gaseous compound has hydrogen making up 12.5% of its mass. Its density with respect to hydrogen is 16. What is the molecular formula of this compound?

1. NH₂
2. N₃H
3. NH₃
4. N₂H₄

Answer: N₂H₄

The correct option, N₂H₄, is derived from the given mass percentage of hydrogen and the density ratio, which indicates that the compound contains two nitrogen atoms for every four hydrogen atoms, aligning with the calculated molecular formula.

Q16. When 100 mL of a 20.8% BaCl₂ solution is mixed with 50 mL of a 9.8% H₂SO₄ solution, the mass of BaSO₄ produced will be: (Ba = 137, Cl = 35.5, S = 32, H = 1, O = 16)

1. 23.3 g
2. 11.65 g
3. 30.6 g
4. 33.2 g

Answer: 11.65 g

BaCl2 = 20.8 g/208 = 0.1 mol; H2SO4 = 4.9 g/98 = 0.05 mol (limiting). BaSO4 = 0.05 x 233 = 11.65 g. Stored 30.6 g is wrong.

Q17. A 2 g sample containing only CO and CO₂ is treated with excess I₂O₅. If 2.54 g of I₂ is obtained, what is the mass percentage of CO₂ in the original sample?

1. 35
2. 70
3. 30
4. 60

Answer: 30

I2 = 2.54/254 = 0.01 mol; 5CO -> 1 I2, so CO = 0.05 mol = 1.4 g, CO2 = 0.6 g -> 30% CO2. Stored 70 is wrong.

Q18. A sample of a gas has a mass of 7.5 g and occupies 5.6 L at STP. Which gas is it?

1. N₂O
2. NO
3. CO
4. CO₂

Answer: NO

5.6 L at STP = 0.25 mol; molar mass = 7.5/0.25 = 30 g/mol = NO. Stored CO (28 g/mol) is wrong.

Q19. How much oxygen is needed to completely burn 2.8 kg of ethylene?

1. 2.8 kg
2. 6.4 kg
3. 9.6 kg
4. 96 kg

Answer: 9.6 kg

To completely burn ethylene (C2H4), a specific stoichiometric amount of oxygen is required. The balanced chemical equation shows that for every mole of ethylene, three moles of oxygen are needed, which translates to approximately 9.6 kg of oxygen for 2.8 kg of ethylene.

Q20. A 3 L mixture containing propane (C₃H₈) and butane (C₄H₁₀) is completely burned at 25°C. If the combustion yields 10 L of carbon dioxide, determine the volume ratio of propane to butane in the original mixture.

1. 2:1
2. 1:2
3. 1.5:1.5
4. 0.5:2.5

Answer: 2:1

Let propane=x, butane=y L. x+y=3 and 3x+4y=10 give y=1, x=2, so propane:butane = 2:1. Stored 1:2 is inverted.

Q21. An organic substance is found to have 49.3% carbon and 6.84% hydrogen by mass. If its vapour density is 73, what is its molecular formula?

1. C₃H₆O₂
2. C₄H₁₀O₂
3. C₆H₁₀O₄
4. C₃H₁₀O₂

Answer: C₆H₁₀O₄

C:H:O = 49.3/12 : 6.84/1 : 43.86/16 = 1.5:2.5:1 = 3:5:2 -> empirical C3H5O2 (73); MW 146 doubles it to C6H10O4. Stored C4H10O2 (90) is wrong.

Q22. At what wavenumber does the first line of the hydrogen Balmer series occur in the atomic spectrum?

1. 9R / 400 cm⁻¹
2. 7R / 144 cm⁻¹
3. 3R / cm⁻¹
4. 5R / 36 cm⁻¹

Answer: 5R / 36 cm⁻¹

Wavenumber = R(1/2^2 - 1/3^2) = R(5/36) = 5R/36. Stored 9R/400 is wrong.

Q23. In the hydrogen emission spectrum, a spectral series has its limit at 12186.3 cm⁻¹. This series is identified as

1. Lyman series
2. Balmer series
3. Paschen series
4. Brackett series

Answer: Paschen series

12186.3 cm^-1 = R/9 (R=109677), so n=3 -> Paschen series. Stored Balmer (R/4 = 27419) is wrong.

Q24. Two rapidly moving particles X and Y have de Broglie wavelengths of 1 nm and 4 nm, respectively. If the mass of X is nine times the mass of Y, what is the ratio of their kinetic energies, X: Y?

1. 3: 1
2. 9: 1
3. 5: 12
4. 16: 9

Answer: 16: 9

KE ratio EX/EY = (mY lambdaY^2)/(mX lambdaX^2) = (1 x 16)/(9 x 1) = 16:9. Stored 9:1 is wrong.

Q25. What is the ratio of the magnetic moments of Fe(III) and Co(II)?

1. √7: √3
2. 3: 7
3. √3: √7
4. 3: √7

Answer: √7: √3

The magnetic moment is determined by the number of unpaired electrons in the d-orbitals of the transition metals. Fe(III) has 5 unpaired electrons, leading to a magnetic moment of √35, while Co(II) has 4 unpaired electrons, resulting in a magnetic moment of √24. The ratio of their magnetic moments simplifies to √7: √3.

Q26. Given that Planck’s constant is 6.63 × 10⁻³⁴ J s and the speed of light is 3.0 × 10⁸ m s⁻¹, which of the following is nearest to the wavelength, in nanometres, of light having frequency 8 × 10¹⁵ s⁻¹?

1. 5 × 10⁻¹⁸
2. 4 × 10¹
3. 3 × 10⁷
4. 2 × 10⁻²⁵

Answer: 4 × 10¹

The wavelength of light can be calculated using the formula λ = (c)/(f), where c is the speed of light and f is the frequency. Plugging in the values, we find that the wavelength corresponding to a frequency of 8 × 10¹⁵ s⁻¹ is approximately 3.75 × 10⁻⁷ m, which converts to 375 nm, closest to 4 × 10¹ nm.

Q27. A lithium ion and a proton are each accelerated through the same potential difference. If their de Broglie wavelengths are λ₁ and λ₂, respectively, and m_(Li) = 9mₚ, what is the ratio λ₁: λ₂?

1. 1: 2
2. 1: 4
3. 1: 1
4. 1: 3√(3)

Answer: 1: 3√(3)

The de Broglie wavelength is inversely proportional to the momentum of a particle, which depends on its mass and velocity. Since the lithium ion has a mass 9 times that of the proton and both are accelerated through the same potential difference, the proton will have a higher velocity and thus a shorter wavelength, leading to the ratio of their wavelengths being 1: 3√3.

Q28. The ionisation energy of He⁺ is 19.6 × 10⁻¹⁸ J atom⁻¹. What is the energy of the first stationary state (n = 1) of Li²⁺?

1. 4.41 × 10⁻¹⁶ J atom⁻¹
2. 4.41 × 10⁻¹⁷ J atom⁻¹
3. −2.2 × 10⁻¹⁵ J atom⁻¹
4. 8.82 × 10⁻¹⁷ J atom⁻¹

Answer: 4.41 × 10⁻¹⁷ J atom⁻¹

E(Li2+, n=1) = E(He+, n=1) x (3/2)^2 = 19.6e-18 x 2.25 = 4.41e-17 J/atom. The stored value 4.41e-16 is ten times too large.

Q29. What is the kinetic energy of an electron in the second Bohr orbit of a hydrogen atom, taking a₀ as the Bohr radius?

1. h² / 4π²mₑa₀²
2. h² / 16π²mₑa₀²
3. h² / 32π²mₑa₀²
4. h² / 64π²mₑa₀²

Answer: h² / 32π²mₑa₀²

The kinetic energy of an electron in the Bohr model is derived from its orbital radius and the quantization of angular momentum. In the second orbit, the radius is twice that of the first, leading to the specific factor of 1/32 in the kinetic energy formula.

Q30. The energy of an electron is expressed as E = -2.178 × 10⁻¹⁸ (Z² / n²). For a hydrogen atom, the wavelength of radiation needed to promote an electron from n = 1 to n = 2 is (take h = 6.62 × 10⁻³⁴ Js and c = 3.0 × 10⁸ m/s):

1. 1.214 × 10⁻⁷ m
2. 2.816 × 10⁻⁷ m
3. 6.500 × 10⁻⁷ m
4. 8.500 × 10⁻⁷ m

Answer: 1.214 × 10⁻⁷ m

DeltaE(1->2) = 2.178e-18 x 0.75 = 1.6335e-18 J; lambda = (6.62e-34 x 3e8)/1.6335e-18 = 1.214e-7 m. The stored 2.816e-7 m is wrong.