### 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\%$\%\mathrm{O} = 100 - (14.5 + 64.46 + 1.8) = 100 - 80.76 = 19.24\%$ ### Step 2: Calculate Molar Ratios Divide each mass percentage by its respective atomic weight: - mathrmC: frac14.512 = 1.208$\mathrm{C}: \frac{14.5}{12} = 1.208$ - mathrmCl: frac64.4635.5 = 1.815$\mathrm{Cl}: \frac{64.46}{35.5} = 1.815$ - mathrmH: frac1.81 = 1.800$\mathrm{H}: \frac{1.8}{1} = 1.800$ - mathrmO: frac19.2416 = 1.202$\mathrm{O}: \frac{19.24}{16} = 1.202$ ### Step 3: Find Simple Integer Ratio Divide by the lowest ratio value (1.202$1.202$): - mathrmC: frac1.2081.202 approx 1 rightarrow times 2 = 2$\mathrm{C}: \frac{1.208}{1.202} \approx 1 \rightarrow \times 2 = 2$ - mathrmCl: frac1.8151.202 approx 1.5 rightarrow times 2 = 3$\mathrm{Cl}: \frac{1.815}{1.202} \approx 1.5 \rightarrow \times 2 = 3$ - mathrmH: frac1.8001.202 approx 1.5 rightarrow times 2 = 3$\mathrm{H}: \frac{1.800}{1.202} \approx 1.5 \rightarrow \times 2 = 3$ - mathrmO: frac1.2021.202 = 1 rightarrow times 2 = 2$\mathrm{O}: \frac{1.202}{1.202} = 1 \rightarrow \times 2 = 2$ Thus, the empirical formula is mathrmC_2mathrmH_3mathrmCl_3mathrmO_2$\mathrm{C}_2\mathrm{H}_3\mathrm{Cl}_3\mathrm{O}_2$. ### Step 4: Compute Mass Empirical formula mass calculation: textMass = (2 times 12) + (3 times 1) + (3 times 35.5) + (2 times 16)$\text{Mass} = (2 \times 12) + (3 \times 1) + (3 \times 35.5) + (2 \times 16)$ textMass = 24 + 3 + 106.5 + 32 = 165.5\,mathrmg\,mol^-1$\text{Mass} = 24 + 3 + 106.5 + 32 = 165.5\,\mathrm{g\,mol}^{-1}$ Expressing in the requested format: 165.5 = 1655 times 10^-1 Rightarrow x = 1655$165.5 = 1655 \times 10^{-1} \Rightarrow x = 1655$ ### 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

### Related Formula Molarity formula based on mass percentage (w/w$w/w$) and density (d$d$): textMolarity = frac\%(w/w) times d times 10textMolar Mass of solute$\text{Molarity} = \frac{\%(w/w) \times d \times 10}{\text{Molar Mass of solute}}$ ### Step 1: Substitute Values Given values: \% = 70$\% = 70$, d = 1.25\,mathrmg\,mL^-1$d = 1.25\,\mathrm{g\,mL}^{-1}$, textMolar Mass = 70\,mathrmg\,mol^-1$\text{Molar Mass} = 70\,\mathrm{g\,mol}^{-1}$. textMolarity = frac70 times 1.25 times 1070 = 1.25 times 10 = 12.5\,mathrmM$\text{Molarity} = \frac{70 \times 1.25 \times 10}{70} = 1.25 \times 10 = 12.5\,\mathrm{M}$ Rounding to the nearest integer gives 13 (or 12.5 as written in standard templates; let us provide 13 matching nearest integer constraints). ### Pattern Recognition Sees: Conversion of mass percentage to molarity tracking. Shortcut: Using the classic shortcut formula frac\% times d times 10M$\frac{\% \times d \times 10}{M}$ simplifies the arithmetic immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula The number of atoms in a given mass of an element is calculated using: textNumber of atoms = fractextMass (g)textMolar Mass (g/mol) times N_A$\text{Number of atoms} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \times N_A$ ### Core Logic Since the mass (10^-9text g$10^{-9}\text{ g}$) is identical for all samples, the number of atoms is inversely proportional to the molar mass of the element: textNumber of atoms propto frac1textMolar Mass $\text{Number of atoms} \propto \frac{1}{\text{Molar Mass}} $ ### Step 1: Molar Mass Values Comparison Let us check the molar masses of the listed elements: * textMolar Mass of Po approx 209text g/mol$\text{Molar Mass of Po} \approx 209\text{ g/mol}$ * textMolar Mass of Pr approx 141text g/mol$\text{Molar Mass of Pr} \approx 141\text{ g/mol}$ * textMolar Mass of Pb approx 207text g/mol$\text{Molar Mass of Pb} \approx 207\text{ g/mol}$ * textMolar Mass of Pt approx 195text g/mol$\text{Molar Mass of Pt} \approx 195\text{ g/mol}$ ### Step 2: Conclusion Praseodymium (Pr) has the least molar mass (141text g/mol$141\text{ g/mol}$), meaning it will yield the maximum total number of atoms for the specified mass sample. ### Pattern Recognition Shortcut: Equal mass given ightarrow$ ightarrow$ Lighter atoms mean more atoms per gram. Find the element with the lowest atomic mass value from the choices. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula textppm = fractextMass of solute (g)textTotal mass of solution/mixture (g) times 10^6$\text{ppm} = \frac{\text{Mass of solute (g)}}{\text{Total mass of solution/mixture (g)}} \times 10^6$ ### Core Logic Let the required mass of pure iron be wmathrm~g$w\mathrm{~g}$. The total mass of the wheat mixture is 150mathrm~kg = 150 times 10^3mathrm~g$150\mathrm{~kg} = 150 \times 10^3\mathrm{~g}$. Applying the parts-per-million concentration condition: 12 = fracw150 times 10^3 times 10^6 implies 12 = w times 6.666$12 = \frac{w}{150 \times 10^3} \times 10^6 \implies 12 = w \times 6.666$ w = frac12 times 150 times 10^310^6 = 1.8mathrm~g text of Iron$w = \frac{12 \times 150 \times 10^3}{10^6} = 1.8\mathrm{~g} \text{ of Iron}$ Now, determine the molar mass of the complete green vitriol salt crystal template, mathrmFeSO_4 cdot 7H_2O$\mathrm{FeSO_4 \cdot 7H_2O}$: M = 56 + 32 + (4 times 16) + (7 times 18) = 56 + 32 + 64 + 126 = 278mathrm~g~mol^-1$M = 56 + 32 + (4 \times 16) + (7 \times 18) = 56 + 32 + 64 + 126 = 278\mathrm{~g~mol^{-1}}$ Set up a stoichiometric mass balance proportion to find the total salt mass w_1$w_1$: textMoles of Fe = frac1.856 = fracw_1278$\text{Moles of Fe} = \frac{1.8}{56} = \frac{w_1}{278}$ w_1 = frac1.8 times 27856 = frac500.456 approx 8.935mathrm~g$w_1 = \frac{1.8 \times 278}{56} = \frac{500.4}{56} \approx 8.935\mathrm{~g}$ Rounding off to the nearest integer value gives **9**. ### Pattern Recognition Always convert concentration metrics back to absolute molar mass equivalence values before distributing across full hydrated molecular templates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry