Solutions Chemistry Online Mock Test: NEET PYQBy admin / March 12, 2025 Class 12 Chemistry Online Free Mock Test 5/5 - (1 vote)0% 9 Table of Contents ToggleReport a questionReport a questionReport a questionReport a questionReport a question What’s wrong with this question?You cannot submit an empty report. Please add some details. Solutions Chemistry Online Mock Test: NEET PYQ 2024-21NEET = National Eligibility-cum-Entrance Test (for admission to undergraduate medical education in all medical institutions)No of Questions: 10 MCQs Time: 20 MinutesPracticing questions from a variety of competitive exams is really valuable. For instance, working on these questions can still sharpen your problem-solving abilities and boost your confidence. Many topics overlap, so broader practice often leads to stronger overall performance 1 / 10Mass of glucose (C₆H₁₂O₆) required to be dissolved to prepare one litre of its solution which is isotonic with 15 g L⁻¹ solution of urea (NH₂CONH₂) is (Given: Molar mass in g mol⁻¹ C:12, H:1, O:16, N:14) [NEET 2024 Rel] A. 55 g B. 15 g C. 30 g D. 45 g 2 / 10The density of the solution is 2.15 g mL⁻¹, then mass of 2.5 mL solution in correct significant figures is: [NEET Re-2022] A. 53.75 g B. 5375 × 10⁻³ g C. 5.4 g D. 5.38 g Solution: In case of multiplication and division, the final result should be reported as having the same number of significant digits as the number with least number of significant digits. 3 / 10The following solutions were prepared by dissolving 10 g of glucose (C₆H₁₂O₆) in 250 mL of water (P₁), 10 g of urea (CH₄N₂O) in 250 mL of water (P₂) and 10 g of sucrose (C₁₂H₂₂O₁₁) in 250 mL of water (P₃). The right option for the decreasing order of osmotic pressure of these solutions is: [NEET 2021] A. P₂ , P₁ , P₃ B. P₁ , P₂ , P₃ C. P₂ , P₃ , P₁ D. P₃ , P₁ , P₂ Solution:Osmotic pressure (π) = iCRT where C is molar concentration of the solutionWith increase in molar concentration of solution osmotic pressure increases.Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.Order of molar mass of solute decreases asSucrose > Glucose > UreaSo, correct order of osmotic pressure of solutions: P3<P1<P2 4 / 10Arrange the following compounds in increasing order of their solubilities in chloroform: [NEET 2024 Rel] NaCl, CH₃OH, cyclohexane, CH₃CNA. NaCl < CH₃CN < CH₃OH < CyclohexaneB. CH₃OH < CH₃CN < NaCl < CyclohexaneC. NaCl < CH₃OH < CH₃CN < CyclohexaneD. Cyclohexane < CH₃CN < CH₃OH < NaCl A B C D 👉 Short Explanation: 👉 Chloroform (CHCl₃) is non-polar / slightly polar solventRule: “Like dissolves like”NaCl → ionic ⇒ least soluble ❌CH₃CN → polar ⇒ low solubilityCH₃OH → polar but has some compatibility ⇒ moderateCyclohexane → non-polar ⇒ most soluble ✔👉 Increasing order: NaCl<CH3CN<CH3OH<Cyclohexane✔ Correct Answer: (A) 5 / 10Gas: KH / K.barAr → 40.3, CO₂ → 1.67, HCHO → 1.83×10−5, CH₄ → 0.413Where KH is Henry's Law constant in water. The order of their solubility in water is: [NEET Re-2022]A. HCHO < CH₄ < CO₂ < ArB. Ar < CO₂ < CH₄ < HCHOC. CO₂ < Ar < CH₄ < HCHOD. HCHO < CO₂ < CH₄ < Ar A B C D Solution: ✔ Correct Answer: (B) 6 / 10The Henry's law constant (Kₕ) values of three gases (A, B, C) in water are 145, 2 × 10⁻⁵ and 35 kbar, respectively. The solubility of these gases in water follow the order(decreasing order): [NEET 2024] B , A , C B , C , A A , C , B A , B , C 7 / 10During the preparation of Mohr's salt solution (Ferrous ammonium sulphate), which of the following acid is added to prevent hydrolysis of Fe²⁺ ion? [NEET 2024] A. dilute hydrochloric acid B. concentrated sulphuric acid C. dilute nitric acid D. dilute sulphuric acid 👉 Short Explanation:👉 Fe²⁺ ions undergo hydrolysis in water forming Fe(OH)₂. To prevent this, solution is kept slightly acidic.✔ Dilute H₂SO₄ is used because:Provides acidic mediumDoes NOT oxidize Fe²⁺ (unlike HNO₃)✔ Correct Answer: (D) dilute sulphuric acid 8 / 10In one molal solution that contains 0.5 mole of a solute, there is [NEET-2022] A. 500 mL of solvent B. 500 g of solvent C. 100 mL of solvent D. 1000 g of solvent Solution: Molality is the moles of solute dissolved per kg of solvent therefore 500g, 1 molal solution contains 0.5 of solute, as 9 / 10Which amongst the following aqueous solution of electrolytes will have minimum elevation in boiling point? Choose the correct option :- [NEET 2023 mpr] A. 0.05 M NaCl B. 0.1 M KCl C. 0.1 M MgSO₄ D. 1 M NaCl 👉 Short ExplanationElevation in boiling point:ΔTb=iKbm👉 Depends on van’t Hoff factor (i) × concentration (m)✔ Correct Answer: (A) 0.05 M NaCl 10 / 10The plot of osmotic pressure (π) vs concentration (mol L⁻¹) for a solution gives a straight line with slope 25.73 L bar mol⁻¹. The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol⁻¹ K⁻¹) [NEET 2024] A. 37°C B. 310°C C. 25.73°C D. 12.05°C Your score isThe average score is 33% 0% Restart quiz ………………………………………………….0% 0 Report a question What’s wrong with this question?You cannot submit an empty report. Please add some details. Solutions Chemistry Online Mock Test: NEET PYQ 2021-17NEET = National Eligibility-cum-Entrance Test (for admission to undergraduate medical education in all medical institutions)No of Questions: 10 MCQs Time: 20 MinutesPracticing questions from a variety of competitive exams is really valuable. For instance, working on these questions can still sharpen your problem-solving abilities and boost your confidence. Many topics overlap, so broader practice often leads to stronger overall performance 1 / 10The mixture which shows positive deviation from Raoult's law is (2020) A. Benzene + Toluene B. Acetone + Chloroform C. Chloroethane + Bromoethane D. Ethanol + Acetone 👉 Short Explanation👉 Positive deviation occurs when:A–B interactions are weaker than A–A or B–BVapour pressure increases✔ Ethanol + Acetone:Breaks strong H-bonding in ethanolForms weaker interactions ⇒ positive deviation❌ Others:Benzene + Toluene → ideal solutionAcetone + Chloroform → strong H-bonding ⇒ negative deviationChloroethane + Bromoethane → nearly ideal✔ Correct Answer: (D) 2 / 10If molality of the dilute solution is doubled, the value of molal depression constant (Kf) will be (NEET 2017) A. halved B. tripled C. unchanged D. doubled 👉 Short Explanation👉 Kf (molal depression constant) is a property of the solvent onlyIt does NOT depend on concentration (molality)It remains constant for a given solvent✔ Even if molality is doubled → Kf remains same✔ Correct Answer: (C) unchanged 3 / 10In water saturated air, the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is (Odisha NEET 2019) A. 1.18 atm B. 1.76 atm C. 1.176 atm D. 0.98 atm 4 / 10The freezing point depression constant (Kf) of benzene is 5.12 K kg mol⁻¹. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places): (2020) A. 0.80 K B. 0.40 K C. 0.60 K D. 0.20 K 5.12 × 0.078 = 0.399K = 0.40K 5 / 10The following solutions were prepared by dissolving 10 g of glucose (C₆H₁₂O₆) in 250 mL of water (P₁), 10 g of urea (CH₄N₂O) in 250 mL of water (P₂) and 10 g of sucrose (C₁₂H₂₂O₁₁) in 250 mL of water (P₃). The right option for the decreasing order of osmotic pressure of these solutions is: [NEET 2021] A. P₂ , P₁ , P₃ B. P₁ , P₂ , P₃ C. P₂ , P₃ , P₁ D. P₃ , P₁ , P₂ Solution:Osmotic pressure (π) = iCRT where C is molar concentration of the solutionWith increase in molar concentration of solution osmotic pressure increases.Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.Order of molar mass of solute decreases asSucrose > Glucose > UreaSo, correct order of osmotic pressure of solutions: P3<P1<P2 6 / 10The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3 : 2 is: [At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume ideal solution] [NEET 2021] A. 160 mm of Hg B. 168 mm of Hg C. 336 mm of Hg D. 350 mm of Hg 7 / 10For an ideal solution, the correct option is (NEET 2019) A. ΔG𝑚𝑖𝑥 = 0 at constant T and P B. ΔS𝑚𝑖𝑥 = 0 at constant T and P C. ΔV𝑚𝑖𝑥 =/= 0 at constant T and P D. Δ𝐻𝑚𝑖𝑥 = 0 at constant T and P 👉 Short ExplanationFor an ideal solution:✔ ΔHmix=0 → no heat absorbed or evolved✔ ΔVmix=0 → no volume change✔ ΔSmix>0 → randomness increases✔ ΔGmix<0 → spontaneous mixing👉 Only correct statement: ΔHmix=0✔ Correct Answer: (D) 8 / 10Which of the following is dependent on temperature? (NEET 2017) A. Molarity B. Mole fraction C. Weight percentage D. Molality 👉 Short Explanation👉 Molarity (M) depends on volume of solution, and volume changes with temperature✔ So, molarity changes with temperature❌ Others:Mole fraction → based on moles (no volume)Weight % → based on massMolality → moles per kg solvent👉 These are temperature independent✔ Correct Answer: (A) Molarity 9 / 10The mixture that forms maximum boiling azeotrope is (NEET 2019) A. heptane + octane B. water + nitric acid C. ethanol + water D. acetone + carbon disulphide 👉 Short Explanation👉 Maximum boiling azeotrope forms when:Solution shows negative deviation from Raoult’s lawStrong intermolecular attractions → lower vapour pressure → higher boiling point✔ Water + nitric acid shows strong interactions ⇒ negative deviation❌ Ethanol + water → positive deviation (minimum boiling azeotrope)✔ Correct Answer: (B) 10 / 10Which of the following statements is correct regarding a solution of two components A and B exhibiting positive deviation from ideal behaviour? (Odisha NEET 2019) A. Intermolecular attractive forces between A–A and B–B are stronger than those between A–B B. ΔHmix =0 at constant T and P C. ΔVmix =0 at constant T and P D. Intermolecular attractive forces between A–A and B–B are equal to those between A–B 👉 Short Explanation👉 Positive deviation from Raoult’s law occurs when:A–B interactions are weaker than A–A and B–BMolecules escape easily ⇒ vapour pressure increases✔ So statement (A) is correct❌ (B), (C), (D) → true for ideal solution, not positive deviation✔ Correct Answer: (A) Your score isThe average score is 0% 0% Restart quiz ………………………………………………….0% 0 Report a question What’s wrong with this question?You cannot submit an empty report. Please add some details. Solutions Chemistry Online Mock Test: NEET PYQ 2016-13NEET = National Eligibility-cum-Entrance Test (for admission to undergraduate medical education in all medical institutions)No of Questions: 10 MCQs Time: 20 MinutesPracticing questions from a variety of competitive exams is really valuable. For instance, working on these questions can still sharpen your problem-solving abilities and boost your confidence. Many topics overlap, so broader practice often leads to stronger overall performance 1 / 10Which of the following statements about the composition of the vapour over an ideal 1:1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25°C. (Given, vapour pressure data at 25°C, benzene = 12.8 kPa, toluene = 3.85 kPa) (NEET-I 2016) A. The vapour will contain equal amounts of benzene and toluene. B. Not enough information is given to make a prediction. C. The vapour will contain a higher percentage of benzene. D. The vapour will contain a higher percentage of toluene. 👉 Short Explanation 👉 In ideal solution (Raoult’s law): Pi=xiP ✔ Benzene has higher vapour pressure (12.8 kPa) than toluene (3.85 kPa) 👉 More volatile component dominates vapour phase ✔ So vapour contains more benzene ✔ Correct Answer: (C) 2 / 10How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO₃? The concentrated acid is 70% HNO₃ (2013 NEET) A. 70.0 g conc. HNO₃ B. 54.0 g conc. HNO₃ C. 45.0 g conc. HNO₃ D. 90.0 g conc. HNO₃ 3 / 10Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression? (2014) A. KCl B. C₆H₁₂O₆ C. Al₂(SO₄)₃ D. K₂SO₄ 👉 Short Explanation Freezing point depression: ΔTf=iKfm 👉 All have same molality ⇒ depends on van’t Hoff factor (i)KCl → 2 ions ⇒ i=2Glucose → non-electrolyte ⇒ i=1K₂SO₄ → 3 ions ⇒ i=3Al₂(SO₄)₃ → 5 ions ⇒ i=5👉 Maximum particles ⇒ maximum depression ✔ Correct Answer: (C) Al₂(SO₄)₃ 4 / 10Which of them is not equal to zero for an ideal solution? (2015 Cancelled) ΔV ΔP ΔH ΔS 👉 Short Explanation For an ideal solution: ✔ ΔHmix=0 ✔ ΔVmix=0 ✔ ΔP=0 (obeys Raoult’s law) ❌ ΔSmix≠0 Entropy increases due to mixing 👉 So the only non-zero quantity is: ΔSmix ✔ Correct Answer: (D) 5 / 10Which one of the following electrolytes has the same value of van’t Hoff factor (i) as that of Al₂(SO₄)₃ (if all are 100% ionised)? (2015 Cancelled) A. Al(NO₃)₃ B. K₄[Fe(CN)₆] C. K₂SO₄ D. K₃[Fe(CN)₆] 👉 Short Explanation👉 For Al₂(SO₄)₃:Al2(SO4)3 → 2Al3+ + 3SO42-👉 Total ions = 5 ⇒ i = 5Now check options:Al(NO₃)₃ → 4 ions (i = 4) ❌K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻ ⇒ 5 ions (i = 5) ✔K₂SO₄ → 3 ions (i = 3) ❌K₃[Fe(CN)₆] → 4 ions (i = 4) ❌✔ Correct Answer: (B) 6 / 10At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be (NEET-I 2016) A. 102°C B. 103°C C. 101°C D. 100°C 7 / 10The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is (NEET-II 2016) A. 0 B. 1 C. 2 D. 3 8 / 10What is the mole fraction of the solute in a 1.00 m aqueous solution? (2015) 1 1.77 0.0177 0.177 9 / 10Which one of the following is incorrect for ideal solution? (NEET-II 2016) A B C D 10 / 10The boiling point of 0.2 mol kg⁻¹ solution of X in water is greater than equimolar solution of Y in water. Which one of the following statements is true in this case? (2015) A. Molecular mass of X is less than the molecular mass of Y B. Y is undergoing dissociation in water while X undergoes no change C. X is undergoing dissociation in water D. Molecular mass of X is greater than the molecular mass of Y 👉 Short Explanation Boiling point elevation: ΔTb=iKbm 👉 Both solutions have same molality ⇒ difference depends on van’t Hoff factor (i) ✔ Higher boiling point ⇒ higher i 👉 So, X must produce more particles ⇒ dissociation occurs ✔ Therefore, X undergoes dissociation ✔ Correct Answer: (C) Your score isThe average score is 0% 0% Restart quiz ………………………………………………….0% 0 Report a question What’s wrong with this question?You cannot submit an empty report. Please add some details. Solutions Chemistry Online Mock Test: NEET PYQ 2012-05NEET = National Eligibility-cum-Entrance Test (for admission to undergraduate medical education in all medical institutions)No of Questions: 06 MCQs Time: 20 MinutesPracticing questions from a variety of competitive exams is really valuable. For instance, working on these questions can still sharpen your problem-solving abilities and boost your confidence. Many topics overlap, so broader practice often leads to stronger overall performance 1 / 6PA and PB are the vapour pressure of pure liquid components A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be (2012) A. PA+xA(PB−PA) B. PA+xA(PA−PB) C. PB+xA(PB−PA) D. PB+xA(PA−PB) 2 / 6200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10⁻3 bar. The molar mass of protein will be (R = 0.083 L bar mol⁻¹ K⁻¹) [2011] A. 51022 g mol⁻¹ B. 122044 g mol⁻¹ C. 31011 g mol⁻¹ D. 61038 g mol⁻¹ 3 / 6The van’t Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively (2011) A. less than one and greater than one B. less than one and less than one C. greater than one and less than one D. greater than one and greater than one 👉 Short Explanation: From the value of van’t Hoff factor i it is possible to determine the degree of dissociation or association. In case of dissociation, i is greater than 1 and in case of association i is less than 1 4 / 6A 0.1 molal aqueous solution of a weak acid is 30% ionized. If KfK_fKf for water is 1.86 °C m⁻¹, the freezing point of the solution will be (2011) A. −0.18°C B. −0.54°C C. −0.36°C D. −0.24°C 5 / 6Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 25°C are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at same temperature will be (Molecular mass of CHCl₃ = 119.5 u and CH₂Cl₂ = 85 u) (2012 Mains) A. 173.9 mm Hg B. 90.63 mm Hg C. 347.9 mm Hg D. 285.5 mm Hg 6 / 6Mole fraction of the solute in a 1.00 molal aqueous solution is (2011) A. 0.1770 B. 0.0177 C. 0.0344 D. 1.7700 Your score isThe average score is 0% 0% Restart quiz Post Views: 114
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