JEE Main 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.