80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :

Solution & Explanation

### Related Formula mathrmC_mathrmxmathrmH_mathrmy(g) + left(mathrmx + fracmathrmy4right)mathrmO_2(g) longrightarrow mathrmxCO_2(g) + fracmathrmy2mathrmH_2mathrmO_(ell) ### Core Logic Let the volume of hydrocarbon be V = 80 mL. Initial volume of O_2 = 264 mL. At 273 K, H_2O is liquid, so its volume is neglected. Volume of CO_2 formed = 80x mL. Volume of O_2 used = 80left(x + fracy4right) mL. Unreacted O_2 = 264 - 80left(x + fracy4right) mL. Total residual volume = V_CO_2 + V_unreacted \ O_2 = 224 mL. 80x + 264 - 80left(x + fracy4right) = 224 264 - frac80y4 = 224 40 = 20y implies y = 2 After treatment with KOH, CO_2 is absorbed. The remaining volume is unreacted O_2, which is 64 mL. 264 - 80left(x + fracy4right) = 64 Substitute y = 2: 264 - 80left(x + frac12right) = 64 264 - 80x - 40 = 64 224 - 80x = 64 80x = 160 implies x = 2 The hydrocarbon is mathrmC_2mathrmH_2. ### Pattern Recognition Volume decrease by KOH indicates the volume of CO_2 produced. V_CO_2 = 224 - 64 = 160 mL. V_HC = 80 mL. So x = frac16080 = 2. Total volume reduction = 264 - 64 = 200 mL (O_2 consumed). O_2 consumed = 80(x + y/4) = 200 implies 2 + y/4 = 2.5 implies y/4 = 0.5 implies y = 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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Some Basic Concepts Of Chemistry

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 4

Q35 jee_main_2025_28_jan_evening Concentration Terms
Concentrated nitric acid is labelled as 75\% by mass. The volume in mL of the solution which contains 30mathrm\ g of nitric acid is Given: Density of nitric acid solution is 1.25mathrm\ g/mL
  • A. 45
  • B. 55
  • C. 32
  • D. 40

Solution

### Related Formula Mass percentage definition: \%text w/w = fractextMass of solutetextMass of solution times 100 Density conversion equation: textVolume of solution = fractextMass of solutiontextDensity of solution ### Core Logic A value of 75\%text w/w HNO_3 implies that 75mathrm\ g of pure textHNO_3 is present in 100mathrm\ g of solution. We need to find the volume that provides exactly 30mathrm\ g of pure acid solute. ### Step 1: Calculate Solution Mass and Volume Mass of solution needed for 30mathrm\ g solute: textMass = frac10075 times 30 = 40mathrm\ g Converting mass to volume using solution density (1.25mathrm\ g/mL): textVolume = frac40mathrm\ g1.25mathrm\ g/mL = 32mathrm\ mL ### Pattern Recognition Break concentration steps down clearly: textMass of solute rightarrow textMass of solution rightarrow textVolume of solution. Combining operations: textVolume = fractextMass solute\% times frac100textdensity = frac3075 times frac1001.25 = 0.4 times 80 = 32. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q jee_main_2025_29_jan_morning Properties of Matter and Their Measurement
Choose the correct statements. (A) Weight of a substance is the amount of matter present in it. (B) Mass is the force exerted by gravity on an object. (C) Volume is the amount of space occupied by a substance. (D) Temperatures below 0^circmathrmC are possible in Celsius scale, but in Kelvin scale negative temperature is not possible. (E) Precision refers to the closeness of various measurements for the same quantity.
  • A. (B), (C) and (D) Only
  • B. (A), (B) and (C) Only
  • C. (A), (D) and (E) Only
  • D. (C), (D) and (E) Only

Solution

### Related Formula T_mathrmK = T_^circmathrmC + 273.15 Absolute zero (0text K) represents the lowest theoretical temperature limit. ### Core Logic Analyzing each statement based on foundational definitions : * (A) & (B) Incorrect: Mass is the actual matter present; weight is the gravitational force exerted on that mass. These definitions are reversed in the statements. * (C) Correct: Volume correctly defines the space occupied by a substance . * (D) Correct: Celsius values can be negative, whereas Kelvin scale strictly defaults to absolute zero (0text K) as minimum . * (E) Correct: Precision measures how close experimental trials lie relative to each other . Therefore, statements (C), (D), and (E) are correct. ### Pattern Recognition Absolute temperature scale (Kelvin) can never possess real negative values because 0text K represents complete cessation of molecular motion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q77 jee_main_2024_01_february_morning Titration
Given below are two statements : Statement (I): Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution. Statement (II) : In this titration phenolphthalein can be used as indicator. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. textBoth Statement I and Statement II are correct
  • B. textStatement I is correct but Statement II is incorrect
  • C. textStatement I is incorrect but Statement II is correct.
  • D. textBoth Statement I and Statement II are incorrect.

Solution

### Core Logic Statement (I): Potassium hydrogen phthalate (KHP) is widely used as a primary standard in analytical chemistry for standardizing strong bases like NaOH. This is because it is highly pure, non-hygroscopic, stable, and has a relatively high molar mass, making its concentration reliable and stable over time. Statement (II): KHP is a weak acid and NaOH is a strong base. The titration of a weak acid with a strong base yields an equivalence point in the weakly basic range (pH > 7). Phenolphthalein changes colour in the pH range 8.3 to 10.0, making it the perfect indicator for this titration. ### Step 1: Evaluate Statements Statement I is correct. Statement II is correct. ### Pattern Recognition Weak Acid vs Strong Base rightarrow Equivalence pH > 7 rightarrow Phenolphthalein is the indicator of choice. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium Class 11 Chemistry: Some Basic Concepts of Chemistry
Q89 jee_main_2024_01_february_morning Stoichiometry
Consider the following reaction: 3PbCl_2 + 2(NH_4)_3PO_4 rightarrow Pb_3(PO_4)_2 + 6NH_4Cl If 72 mathrm~mmol of PbCl_2 is mixed with 50 mathrm~mmol of (NH_4)_3PO_4, then amount of Pb_3(PO_4)_2 formed is ... mmol. (nearest integer)
Numerical Answer. Answer: 24 to 24

Solution

### Related Formula textMoles of Product = textMoles of Limiting Reagent times fractextStoichiometry of ProducttextStoichiometry of Limiting Reagent ### Core Logic From the balanced chemical equation: 3 text moles of PbCl_2 text react with 2 text moles of (NH_4)_3PO_4. Let's find the limiting reagent (L.R.) by dividing given millimoles by stoichiometric coefficients: For PbCl_2: frac723 = 24 For (NH_4)_3PO_4: frac502 = 25 Since 24 < 25, PbCl_2 is the limiting reagent and will completely consume. ### Step 1: Calculate Product Moles Moles of Pb_3(PO_4)_2 formed depends entirely on PbCl_2. 3 mmol of PbCl_2 produces 1 mmol of Pb_3(PO_4)_2. Therefore, 72 mmol of PbCl_2 will produce: frac13 times 72 = 24 mathrm~mmol of Pb_3(PO_4)_2. ### Pattern Recognition Always identify the Limiting Reagent by taking the ratio n / textcoefficient. The smallest ratio dictates the extent of the reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q85 jee_main_2024_29_january_evening Volumetric Titration and Molarity
If 50text mL of 0.5text M oxalic acid is required to neutralise 25text mL of mathrmNaOH solution, the amount of mathrmNaOH in 50text mL of given mathrmNaOH solution is ________ g.
Numerical Answer. Answer: 4 to 4

Solution

### Related Formula textEquivalents of Acid = textEquivalents of Base N_1 V_1 = N_2 V_2 implies (M_1 times n_1) times V_1 = (M_2 times n_2) times V_2 ### Core Logic For oxalic acid (textH_2textC_2textO_4), the valence factor (n-factor) is 2. For textNaOH, the n-factor is 1. Substituting the values into the normality equivalence expression: 50 times 0.5 times 2 = 25 times M_textNaOH times 1 50 = 25 times M_textNaOH implies M_textNaOH = 2text M ### Step 1: Mass Isolation To find the mass of textNaOH present in 50text mL of this solution: textMass = textMolarity times textVolume (in L) times textMolar Mass textMass = 2 times left(frac50, 1000right) times 40 = 2 times 0.05 times 40 = 4text g ### Pattern Recognition Remember to use the correct n-factor (2) for dibasic oxalic acid during equivalence matching to avoid calculation errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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