Quantitative analysis of an organic compound (X) shows following % composition. mathrmC:14.5\% quad mathrmCl:64.46\% quad mathrmH:1.8\% The empirical formula mass of the compound (X) is mathrmx times 10^-1. The value of mathrmx is: (Given molar mass in mathrmg\,mol^-1 of C: 12, H: 1, O: 16, Cl: 35.5)

Numerical Answer Type:
Enter a numerical value Answer: 1655 to 1655 +4 marks

Solution & Explanation

### Step 1: Determine Oxygen Percentage The total percentage must equal 100%. The remaining composition corresponds to Oxygen: \%mathrmO = 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24\% ### Step 2: Calculate Molar Ratios Divide each mass percentage by its respective atomic weight: - mathrmC: frac14.512 = 1.208 - mathrmCl: frac64.4635.5 = 1.815 - mathrmH: frac1.81 = 1.800 - mathrmO: frac19.2416 = 1.202 ### Step 3: Find Simple Integer Ratio Divide by the lowest ratio value (1.202): - mathrmC: frac1.2081.202 approx 1 rightarrow times 2 = 2 - mathrmCl: frac1.8151.202 approx 1.5 rightarrow times 2 = 3 - mathrmH: frac1.8001.202 approx 1.5 rightarrow times 2 = 3 - mathrmO: frac1.2021.202 = 1 rightarrow times 2 = 2 Thus, the empirical formula is mathrmC_2mathrmH_3mathrmCl_3mathrmO_2. ### Step 4: Compute Mass Empirical formula mass calculation: textMass = (2 times 12) + (3 times 1) + (3 times 35.5) + (2 times 16) textMass = 24 + 3 + 106.5 + 32 = 165.5\,mathrmg\,mol^-1 Expressing in the requested format: 165.5 = 1655 times 10^-1 Rightarrow x = 1655 ### Pattern Recognition Sees: Multi-element empirical calculation. Trap: Forgetting to compute Oxygen by missing that the percentages do not sum to 100% initial value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

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Some Basic Concepts Of Chemistry

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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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