The amount of calcium oxide produced on heating 150mathrm~kg$150\mathrm{~kg}$ limestone (75% pure) is _________________ kg. (Nearest integer) Given : Molar mass (in g mol^-1$^{-1}$ ) of Ca-40, O-16, C-12

### Related Formula P_i = X_i cdot P_texttotal = fracn_in_texttotal cdot P_texttotal$$P_i = X_i \cdot P_{\text{total}} = \frac{n_i}{n_{\text{total}}} \cdot P_{\text{total}}$$ Ratio of partial pressures: fracP_AP_B = fracn_An_B$$\frac{P_A}{P_B} = \frac{n_A}{n_B}$$ ### Core Logic Assume a sample of dry air with total mass = 100 text g$= 100 \text{ g}$: - Mass of mathrmN_2 = 70.0 text g$\mathrm{N}_2 = 70.0 \text{ g}$ - Mass of mathrmO_2 = 27.0 text g$\mathrm{O}_2 = 27.0 \text{ g}$ - Mass of mathrmAr = 3.0 text g$\mathrm{Ar} = 3.0 \text{ g}$ Now, convert masses to moles: n_mathrmN_2 = frac70.028 = 2.5 text moles$$n_{\mathrm{N}_2} = \frac{70.0}{28} = 2.5 \text{ moles}$$ n_mathrmO_2 = frac27.032 = 0.84375 text moles$$n_{\mathrm{O}_2} = \frac{27.0}{32} = 0.84375 \text{ moles}$$ n_mathrmAr = frac3.040 = 0.075 text moles$$n_{\mathrm{Ar}} = \frac{3.0}{40} = 0.075 \text{ moles}$$ Calculate ratios: (i) Ratio of partial pressure of nitrogen to oxygen: fracP_mathrmN_2P_mathrmO_2 = fracn_mathrmN_2n_mathrmO_2 = frac2.50.84375 approx 2.96$$\frac{P_{\mathrm{N}_2}}{P_{\mathrm{O}_2}} = \frac{n_{\mathrm{N}_2}}{n_{\mathrm{O}_2}} = \frac{2.5}{0.84375} \approx 2.96$$ (ii) Ratio of partial pressure of oxygen to argon: fracP_mathrmO_2P_mathrmAr = fracn_mathrmO_2n_mathrmAr = frac0.843750.075 approx 11.25 approx 11.2$$\frac{P_{\mathrm{O}_2}}{P_{\mathrm{Ar}}} = \frac{n_{\mathrm{O}_2}}{n_{\mathrm{Ar}}} = \frac{0.84375}{0.075} \approx 11.25 \approx 11.2$$ ### Pattern Recognition Since total pressure cancels out in a ratio of partial pressures, we only need to calculate the mole ratio directly from the given mass percentages divided by their respective molar masses. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry Class 11 Physics: Kinetic Theory of Gases

### Related Formula Precipitation reaction stoichiometry: textNaI(aq) + textAgNO_3text(aq) longrightarrow textAgI(s) + textNaNO_3text(aq)$$\text{NaI(aq)} + \text{AgNO}_3\text{(aq)} \longrightarrow \text{AgI(s)} + \text{NaNO}_3\text{(aq)}$$ Molarity calculation formula: M = fractextMoles of solute (NaI)textVolume of solution in Liters (L)$$M = \frac{\text{Moles of solute (NaI)}}{\text{Volume of solution in Liters (L)}}$$ ### Execution Step 1: Determine the molar mass of the Silver Iodide (textAgI$\text{AgI}$) precipitate: textMolar Mass of AgI = 108 + 127 = 235 text g mol^-1$$\text{Molar Mass of AgI} = 108 + 127 = 235 \text{ g mol}^{-1}$$ Step 2: Calculate the moles of textAgI$\text{AgI}$ precipitated: textMoles of AgI = frac4.74 text g235 text g mol^-1 approx 0.02017 text mol$$\text{Moles of AgI} = \frac{4.74 \text{ g}}{235 \text{ g mol}^{-1}} \approx 0.02017 \text{ mol}$$ Step 3: Apply the 1:1 reaction stoichiometry to find the moles of textNaI$\text{NaI}$: textMoles of NaI = textMoles of AgI = 0.02017 text mol$$\text{Moles of NaI} = \text{Moles of AgI} = 0.02017 \text{ mol}$$ Step 4: Compute the molarity of the solution, converting 20 text mL$20 \text{ mL}$ to 0.020 text L$0.020 \text{ L}$: textMolarity [NaI] = frac0.02017 text mol0.020 text L = 1.0085 text M$$\text{Molarity [NaI]} = \frac{0.02017 \text{ mol}}{0.020 \text{ L}} = 1.0085 \text{ M}$$ Rounding to the nearest integer value gives **1**. ### Pattern Recognition Precipitation reactions involving silver halides follow a strict 1:1 mole ratio between the halide source and the silver precipitate. Converting mass into moles and dividing by the volume in liters quickly yields the molarity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### 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