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Q. If 1 liter of a 2 M solution is diluted to 3 liters, what is the new molarity?

A. 0.67 M
B. 1 M
C. 1.5 M
D. 2 M

Solution

Using the dilution formula M1V1 = M2V2, (2 M)(1 L) = M2(3 L) => M2 = 2/3 = 0.67 M.

Correct Answer: A — 0.67 M

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Q. If 1 mol of NaCl is dissolved in 1 kg of water, how many particles are present in solution?

A. 1
B. 2
C. 3
D. 4

Solution

NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl produces 2 mol of particles in solution.

Correct Answer: B — 2

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Q. If 1 mol of NaCl is dissolved in 1 kg of water, what is the expected van 't Hoff factor (i)?

A. 1
B. 2
C. 3
D. 4

Solution

NaCl dissociates into 2 ions (Na+ and Cl-), so the van 't Hoff factor i = 2.

Correct Answer: B — 2

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Q. If 1 mol of NaCl is dissolved in water, how many particles are present in solution?

A. 1
B. 2
C. 3
D. 4

Solution

NaCl dissociates into 2 ions (Na+ and Cl-), so 1 mol of NaCl results in 2 mol of particles.

Correct Answer: B — 2

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Q. If 1 mole of a gas occupies 22.4 L at STP, how many liters will 0.5 moles occupy?

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 5.6 L

Solution

Volume = moles x volume per mole = 0.5 moles x 22.4 L/mole = 11.2 L.

Correct Answer: A — 11.2 L

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Q. If 1 mole of a gas occupies 22.4 L at STP, how much volume will 0.5 moles occupy?

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 5.6 L

Solution

Volume = moles x volume per mole = 0.5 moles x 22.4 L/mole = 11.2 L.

Correct Answer: A — 11.2 L

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Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the expected change in freezing point?

A. 0.0 °C
B. -1.86 °C
C. -3.72 °C
D. -5.58 °C

Solution

The freezing point depression for 1 mole of a non-electrolyte solute in 1 kg of water is -1.86 °C.

Correct Answer: B — -1.86 °C

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Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the expected freezing point depression?

A. -1.86 °C
B. -3.72 °C
C. -0.52 °C
D. -2.00 °C

Solution

The freezing point depression is calculated using the formula ΔTf = Kf * m, where Kf for water is 1.86 °C kg/mol.

Correct Answer: A — -1.86 °C

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Q. If 1 mole of a non-electrolyte solute is dissolved in 1 kg of water, what is the freezing point depression?

A. 0 °C
B. 1.86 °C
C. 3.72 °C
D. 5.58 °C

Solution

The freezing point depression is calculated using the formula ΔTf = i * Kf * m. For a non-electrolyte, i = 1, Kf for water = 1.86 °C kg/mol, and m = 1 mol/kg gives ΔTf = 1.86 °C.

Correct Answer: B — 1.86 °C

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Q. If 1 mole of a non-volatile solute is dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.512 °C kg/mol)

A. 0.512 °C
B. 1.024 °C
C. 2.048 °C
D. 0.256 °C

Solution

Boiling point elevation = i * Kb * m = 1 * 0.512 * 1 = 0.512 °C.

Correct Answer: B — 1.024 °C

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Q. If 1 mole of an ideal gas occupies 22.4 L at STP, what is the pressure exerted by the gas?

A. 1 atm
B. 2 atm
C. 0.5 atm
D. 4 atm

Solution

At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.4 L at a pressure of 1 atm.

Correct Answer: A — 1 atm

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Q. If 1 mole of an ideal gas occupies 22.4 L at STP, what is the volume occupied by 2 moles at the same conditions?

A. 11.2 L
B. 22.4 L
C. 44.8 L
D. 56.8 L

Solution

According to Avogadro's Law, 2 moles of gas will occupy double the volume, which is 44.8 L at STP.

Correct Answer: C — 44.8 L

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Q. If 1 mole of NaCl is dissolved in 1 kg of water, what is the expected van 't Hoff factor (i)?

A. 1
B. 2
C. 3
D. 4

Solution

NaCl dissociates into 2 ions (Na+ and Cl-), so the van 't Hoff factor (i) is 2.

Correct Answer: B — 2

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Q. If 1 mole of solute is dissolved in 1 liter of solution, what is the concentration in terms of molarity?

A. 1 M
B. 2 M
C. 0.5 M
D. 0.25 M

Solution

Molarity (M) = moles of solute / liters of solution = 1 mole / 1 L = 1 M.

Correct Answer: A — 1 M

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Q. If 10 g of CaCO3 decomposes completely, how many grams of CO2 are produced?

A. 22 g
B. 10 g
C. 44 g
D. 20 g

Solution

10 g of CaCO3 = 0.1 moles. CaCO3 → CaO + CO2, so 0.1 moles of CO2 = 0.1 * 44 g = 4.4 g.

Correct Answer: C — 44 g

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Q. If 10 g of Na reacts with excess Cl2, what is the mass of NaCl produced?

A. 58.5 g
B. 10 g
C. 20 g
D. 30 g

Solution

10 g of Na = 0.43 moles. Na + Cl2 → NaCl, so 0.43 moles of NaCl = 0.43 * 58.5 g = 25.2 g.

Correct Answer: A — 58.5 g

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Q. If 10 grams of calcium carbonate (CaCO3) decomposes, how many grams of calcium oxide (CaO) are produced?

A. 5 g
B. 10 g
C. 8 g
D. 7 g

Solution

The balanced equation is CaCO3 → CaO + CO2. The molar mass of CaCO3 is 100 g/mol and CaO is 56 g/mol. Thus, 10 g of CaCO3 produces (10 g / 100 g/mol) x 56 g/mol = 5.6 g of CaO.

Correct Answer: C — 8 g

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Q. If 10 grams of NaCl are dissolved in water, how many moles of NaCl are present? (Molar mass of NaCl = 58.5 g/mol)

A. 0.17 moles
B. 0.5 moles
C. 1.0 moles
D. 1.5 moles

Solution

Moles of NaCl = mass (g) / molar mass (g/mol) = 10 g / 58.5 g/mol = 0.171 moles.

Correct Answer: A — 0.17 moles

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Q. If 10 grams of NaCl is dissolved in 500 mL of water, what is the mass/volume percent concentration?

A. 1%
B. 2%
C. 5%
D. 10%

Solution

Mass/volume percent = (mass of solute / volume of solution) x 100 = (10 g / 500 mL) x 100 = 2%.

Correct Answer: B — 2%

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Q. If 10 grams of NaCl is dissolved in 500 mL of water, what is the molality of the solution? (Molar mass of NaCl = 58.5 g/mol)

A. 0.34 m
B. 0.17 m
C. 0.85 m
D. 0.50 m

Solution

Molality (m) = moles of solute / kg of solvent. Moles of NaCl = 10 g / 58.5 g/mol = 0.171 moles. Mass of water = 0.5 kg. Molality = 0.171 moles / 0.5 kg = 0.342 m.

Correct Answer: A — 0.34 m

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Q. If 10 grams of NaCl is dissolved in enough water to make 500 mL of solution, what is the molality of the solution? (Molar mass of NaCl = 58.5 g/mol)

A. 0.34 m
B. 0.17 m
C. 0.85 m
D. 0.50 m

Solution

Moles of NaCl = 10 g / 58.5 g/mol = 0.171 moles. Mass of solvent (water) = 0.5 kg. Molality (m) = moles of solute / kg of solvent = 0.171 moles / 0.5 kg = 0.34 m.

Correct Answer: A — 0.34 m

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Q. If 10 grams of NaOH are dissolved in water, how many moles of NaOH are present? (Molar mass of NaOH = 40 g/mol)

A. 0.25 moles
B. 0.5 moles
C. 1 mole
D. 2.5 moles

Solution

To find the number of moles, use the formula: moles = mass/molar mass. Thus, 10 g / 40 g/mol = 0.25 moles.

Correct Answer: B — 0.5 moles

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Q. If 10 grams of NaOH is dissolved in 500 mL of solution, what is the molality of the solution? (Molar mass of NaOH = 40 g/mol)

A. 0.5 m
B. 1 m
C. 2 m
D. 0.25 m

Solution

Moles of NaOH = 10 g / 40 g/mol = 0.25 moles. Mass of solvent (water) = 0.5 kg. Molality (m) = moles of solute / kg of solvent = 0.25 moles / 0.5 kg = 0.5 m.

Correct Answer: B — 1 m

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Q. If 10 grams of NaOH is dissolved in enough water to make 500 mL of solution, what is the molarity of the solution? (Molar mass of NaOH = 40 g/mol)

A. 0.5 M
B. 1 M
C. 2 M
D. 0.25 M

Solution

Moles of NaOH = 10 g / 40 g/mol = 0.25 moles. Molarity = 0.25 moles / 0.5 L = 0.5 M.

Correct Answer: B — 1 M

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Q. If 100 g of glucose (C6H12O6) is dissolved in 1 L of solution, what is the molarity of the solution? (Molar mass of glucose = 180 g/mol)

A. 0.56 M
B. 1.0 M
C. 0.33 M
D. 0.75 M

Solution

Moles of glucose = 100 g / 180 g/mol = 0.556 moles. Molarity = moles of solute / liters of solution = 0.556 moles / 1 L = 0.56 M.

Correct Answer: A — 0.56 M

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Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature?

A. 50°C
B. 25°C
C. 75°C
D. 0°C

Solution

The final temperature will be 50°C due to equal masses and specific heat capacities.

Correct Answer: A — 50°C

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Q. If 100 g of water at 0°C is mixed with 100 g of water at 100°C, what will be the final temperature of the mixture?

A. 50°C
B. 25°C
C. 75°C
D. 0°C

Solution

Using the principle of conservation of energy, the final temperature will be 50°C.

Correct Answer: A — 50°C

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Q. If 100 g of water at 80°C is mixed with 200 g of water at 20°C, what will be the final temperature?

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the principle of conservation of energy, the final temperature can be calculated to be 40°C.

Correct Answer: B — 40°C

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Q. If 100 J of heat is added to a system and 40 J of work is done by the system, what is the change in internal energy?

A. 60 J
B. 40 J
C. 100 J
D. 140 J

Solution

According to the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 100 J - 40 J = 60 J.

Correct Answer: A — 60 J

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Q. If 100 J of heat is added to a system at a constant temperature of 300 K, what is the change in entropy?

A. 0.33 J/K
B. 0.25 J/K
C. 0.5 J/K
D. 0.75 J/K

Solution

The change in entropy ΔS = Q/T = 100 J / 300 K = 0.33 J/K.

Correct Answer: A — 0.33 J/K

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