JEE Mains 2024 Chemistry Chapter 2: Solutions PYQs with Detailed Answers

JEE Mains 2024 Chemistry Chapter 2: Solutions PYQs with Detailed Answers

Struggling with the “Solutions” chapter in Chemistry? This post offers JEE Mains 2024 Chemistry Chapter 2 Previous Year Questions (PYQs) with clear, step-by-step solutions to help you sharpen your concepts, improve accuracy, and boost exam confidence. 🚀

1. The vapor pressure of pure benzene and methyl benzene at 27 degree Celsius is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapor phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is _________\times 10-2 (nearest integer).Jee main 2024

Answer. 23

$JEE Mains 2024 Chemistry Chapter 2 Solutions handwritten solution for vapor pressure and mole fraction calculation using Raoult's Law.$

2. A solution of 10g of an electrolyte AB 2 in 100g of water balls at 100.52 degree Celsius of ionization of electrolyte(α) ___________\times 10-2 [Given: molar mass of AB 2 =200g/mol , Kp(molal boiling point elevation constant ) =0.52KKg/mol, boiling point of water 100 degree Celsius, AB 2 ionises as AB 2+ -----> A 2+ +2B -]Jee main 2024Answer. 53. When 'x' \times 10-2 ml methanol (molar mass=32g , density=0.792g/cm 3) is added to 100ml .Water (density 1g/cm 3), the following diagram is obtained. x= ________ (nearest integer). Jee main 2024 Answer. 543 4. Considering acetic acid dissociates in water, its dissociation constant is  If 6.25 \times 10-5. If 5ml of acetic acid is dissolved in 1 litre water, the solution will freeze at -x \times 10-2 degree celcius provided pure water freezes at 0 degree celcius. x= _________. (Nearest integer)Kf(water)= 1.86KKg/moldensity of acetic acid is 1.2g/ml molar mass of water = 18g molar mass of acetic acid = 60g  density of water =  1g/cm 3Acetic acid dissociates as CH3COOH  ⇌ C H 3 C O O H^{-} + H + Answer. 195. An artificial cell is made by encapsulating 0.2M glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a  solution of NaCl at 300K is ________\times 10-1 bar. (nearest integer). Jee main 2024Given : R=0.083Lbar/molKAssume complete dissociation of NaClAnswer. 256. 2.7kg of each of water and acetic acid are mixed. The freezing point of the solution will be -x degree celcius . Consider the acetic acid does not dimerise in water, nor dissociates in water. x= ________ (nearest integer) Jee main 2024 Given: Molar mass of water  = 18g  acetic acid = 60gKf of water= 1.86KKg/molKf of acetic acid = 3.90KKg/molfreezing point:  H2O = 273K  ,  acetic acid = 290KAnswer. 317. 2.5g of a non-volatile, non-electrolyte is dissolved in 100g of water at 25 degree celcius .The solution showed a boiling point elevation by 2 degree celcius . Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _________mm of hg . (nearest integer) Jee main 2024Given: 1 atm pressure = 760 mm of hg , molar mass of water = 18gAnswer. 7078. Mass of ethylene glycol (antifreeze) to be added to 18.6kg of water to protect the freezing point at -24 degree celcius is ________ kg . (Molar mass in g/mol for ethylene glycol 62 ,Kf of water = 1.86KKg/mol ) Jee main 2024Answer. 159. The osmotic pressure of a dilute solution is 7\times10 ⁵ Pa at 273K. Osmotic pressure of the same solutionat 283K is  _________\times 104 N/m Jee main 2024Answer. 7310. 0.05M CuSO4 when treated with 0.01M K2Cr2O7 gives green colour solution of Cu2Cr207. The two solutions are seperated as shown below: [SPM: Semi permeable membrane]. Jee main 2024Due to osmosis: (a) Molarity of K2Cr2O7 solution is lowered . (b) Green colour formation observed on side Y. (c) Molarity of CuSO4 solution is lowered. (d) Green colour formation observed on side X.Answer. (c) Molarity of CuSO4 solution is lowered.Osmosis leads to the net movement of water molecule from low osmotic pressure to high osmotic pressure. Since the iM value of 0.05M CuSO4 solution is higher hence it has higher osmotic pressure so the water moves towards CuSO4 solution leads to drop of its molarity. The solute molecules do not cross S.P.M. via osmosis.11. We have three aqueous solutions of NaCl labelled as A,B and C with concentration 0.1M , 0.01M and 0.001M respectively. The value of vant' hoff factor(i) for the solutions will be in the order:Jee main 2024(a) iA<iB<iC (b) iA<iB<iC (c) iA>iB>iC (d) iA=iB=iCAnswer. (b) iA<iB<iC12. Identify the mixture that shows positive deviations from Raoult's law. Jee main 2024(a) (CH3)2CO + CS2 (b) (CH3)2CO + C6H5NH2 (c) CHCl3 + C6H6 (d) CH3Cl + (CH3)2COAnswer. (a)(CH3)2CO + CS213. The solution from the following with highest depression in freezing point/ lowering freezing point is. Jee main 2024(a) 180 gram of Acetic Acid dissolved in Benzene (b) 180 gram of Acetic Acid dissolved in water (c) 180 gram of benzoic acid dissolved in Benzene (d) 180 gram of glucose dissolved in waterAnswer. (b) 180 gram of Acetic Acid dissolved in water 📌 Explanation: Acetic acid ionizes in water, increasing the number of particles in solution, which leads to greater depression in freezing point . In the other options, either no ionization or association (like dimerization) occurs. 🧠 More particles = more freezing point lowering!14. What happens to freezing point of benzene when small quantity of naphthalene is added to benzene. Jee main 2024(a) Increases (b) decreases (c) remains unchanged (d) first decreases and then increasesAnswer. (b) decreases 📌 Explanation: On adding naphthalene (a non-volatile solute) to benzene (solvent), a solution forms, and due to colligative properties, freezing point decreases ❄️📉. 🧪 This is called freezing point depression, a classic example of colligative effect!15. A solution of two miscible liquids show negative division from Raolt's law will have: Jee main 2024(a) Increased vapour pressure, increased boiling point (b) increased vapour pressure, decreased boiling point (c) decreased vapour pressure, decreased boiling point (d) decreased vapour pressure, increased boiling pointAnswer. (d) decreased vapour pressure, increased boiling point 📌 Explanation: In negative deviation from Raoult's Law: Strong intermolecular attractions 💞 between components lead to Lower vapour pressure ⬇️ than expected Which results in higher boiling point 🌡️⬆️ 🧪 This is a classic example of non-ideal solutions showing stronger interactions!