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 5

Q83 jee_main_2024_27_jan_morning Stoichiometry
Mass of methane required to produce 22text g of textCO_2 after complete combustion is textquadquad g. (Given Molar mass in textg mol^-1: textC=12.0, textH=1.0, textO=16.0)
Numerical Answer. Answer: 8 to 8

Solution

### Related Formula Balanced combustion chemical equation: textCH_4 + 2textO_2 rightarrow textCO_2 + 2textH_2textO textMoles = fractextMasstextMolar Mass ### Step 1: Determine moles of product generated textMolar Mass of CO_2 = 12 + (2 times 16) = 44text g mol^-1 textMoles of CO_2 text produced = frac2244 = 0.5text moles ### Step 2: Relate to input mass via stoichiometry metrics From the balanced equation, 1text mole of CH_4 produces 1text mole of CO_2. textRequired Moles of CH_4 = 0.5text moles textMolar Mass of CH_4 = 12 + (4 times 1) = 16text g mol^-1 textMass of CH_4 = 0.5 times 16 = 8text g ### Pattern Recognition 22text g of CO_2 is exactly half a mole. By stoichiometry ratios, half a mole of methane is needed, which translates to 8text g. ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q88 jee_main_2024_30_jan_morning Mole Concept
0.05text cm thick coating of silver is deposited on a plate of 0.05text m^2 area. The number of silver atoms deposited on plate are ________ times 10^23. (At mass Ag=108, d=7.9text g cm^-3)
Numerical Answer. Answer: 11 to 11

Solution

### Related Formula textVolume = textArea times textThickness textMass = textDensity times textVolume textMoles = fractextMasstextMolar Mass textNumber of Atoms = textMoles times N_A ### Step 1: Calculate Volume of Coating Area = 0.05 text m^2 = 0.05 times 10^4 text cm^2 = 500 text cm^2 Thickness = 0.05 text cm Volume = 500 text cm^2 times 0.05 text cm = 25 text cm^3 ### Step 2: Calculate Mass and Moles Mass = Volume times Density = 25 text cm^3 times 7.9 text g/cm^3 = 197.5 text g Moles of Ag = frac197.5108 = 1.8287 text moles ### Step 3: Calculate Number of Atoms textNumber of Atoms = 1.8287 times 6.022 times 10^23 = 11.01 times 10^23 Rounding to nearest integer, we get 11. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry Class 12 Chemistry: Electrochemistry
Q62 jee_main_2024_31_jan_evening Stoichiometry and Calculations
A sample of CaCO_3 and MgCO_3 weighed 2.21text g is ignited to constant weight of 1.152text g. The composition of mixture is: (Given molar mass in mathrmg\,mol^-1 mathrmCaCO_3:100, MgCO_3:84)
  • A. 1.187mathrm~gmathrm~CaCO_3 + 1.023mathrm~gmathrm~MgCO_3
  • B. 1.023mathrm~gmathrm~CaCO_3 + 1.023mathrm~gmathrm~MgCO_3
  • C. 1.187mathrm~gmathrm~CaCO_3 + 1.187mathrm~gmathrm~MgCO_3
  • D. 1.023mathrm~gmathrm~CaCO_3 + 1.187mathrm~gmathrm~MgCO_3

Solution

### Related Formula mathrmCaCO_3(s) xrightarrowDelta mathrmCaO(s) + mathrmCO_2(g) mathrmMgCO_3(s) xrightarrowDelta mathrmMgO(s) + mathrmCO_2(g) ### Core Logic Let the weight of mathrmCaCO_3 be xtext g. Then, the weight of mathrmMgCO_3 = (2.21 - x)text g. Moles of mathrmCaCO_3 decomposed = Moles of mathrmCaO formed. fracx100 = textMoles of CaO formed textWeight of CaO formed = fracx100 times 56 Moles of mathrmMgCO_3 decomposed = Moles of mathrmMgO formed. frac(2.21 - x)84 = textMoles of MgO formed textWeight of MgO formed = frac2.21 - x84 times 40 ### Step 1: Setting up the Equation The total weight of the residue (mathrmCaO + mathrmMgO) is given as 1.152text g. frac2.21 - x84 times 40 + fracx100 times 56 = 1.152 ### Step 2: Solving for x frac88.4 - 40x84 + 0.56x = 1.152 1.0523 - 0.4761x + 0.56x = 1.152 0.0839x = 0.0997 x = 1.188text g So, weight of mathrmCaCO_3 approx 1.187text g (accounting for rounding) Weight of mathrmMgCO_3 = 2.21 - 1.188 = 1.022text g approx 1.023text g. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q83 jee_main_2024_31_jan_morning Stoichiometry
Number of moles of methane required to produce 22g CO_2(g) after combustion is x times 10^-2 moles. The value of x is
Numerical Answer. Answer: 50 to 50

Solution

### Step 1: Stoichiometric Equation CH_4(g) + 2O_2(g) rightarrow CO_2(g) + 2H_2O_(l) 1 mole of CH_4 produces 1 mole of CO_2. ### Step 2: Moles Calculation Molar mass of CO_2 = 12 + 2(16) = 44 text g/mol n_CO_2 = fractextMasstextMolar mass = frac2244 = 0.5 text moles Since 1 mole of CH_4 produces 1 mole of CO_2, the moles of CH_4 required is 0.5 moles. ### Step 3: Finding x 0.5 text moles = 50 times 10^-2 text moles x = 50 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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