### Related Formula Molarity conversion relation matrix: textMoles = textMolarity (M) times textVolume (L)$\text{Moles} = \text{Molarity (M)} \times \text{Volume (L)}$ textMass = textMoles times textMolar Mass$\text{Mass} = \text{Moles} \times \text{Molar Mass}$ ### Core Logic Let's perform molar quantities verification row-by-row: * Molar mass properties: mathrmCaCO_3 = 100mathrm~g/mol$\mathrm{CaCO_3} = 100\mathrm{~g/mol}$, mathrmCaCl_2 = 40 + (35.5 times 2) = 111mathrm~g/mol$\mathrm{CaCl_2} = 40 + (35.5 \times 2) = 111\mathrm{~g/mol}$. * Initial chemical moles calculated: textMoles of mathrmCaCO_3 = frac1000100 = 10mathrm~mol$\text{Moles of } \mathrm{CaCO_3} = \frac{1000}{100} = 10\mathrm{~mol}$ textMoles of mathrmHCl = 0.76 times frac2501000 = 0.19mathrm~mol$\text{Moles of } \mathrm{HCl} = 0.76 \times \frac{250}{1000} = 0.19\mathrm{~mol}$ * Determine the limiting reactant via stoichiometric ratios: mathrmHCl$\mathrm{HCl}$ acts as the Limiting Reagent (L.R.) because its proportional structural requirement is much smaller. * Moles of product mathrmCaCl_2$\mathrm{CaCl_2}$ formed based on L.R. configuration: textMoles of mathrmCaCl_2 = frac0.192 = 0.095mathrm~mol$\text{Moles of } \mathrm{CaCl_2} = \frac{0.19}{2} = 0.095\mathrm{~mol}$ textMass of mathrmCaCl_2 = 0.095 times 111 = 10.545mathrm~g$\text{Mass of } \mathrm{CaCl_2} = 0.095 \times 111 = 10.545\mathrm{~g}$ ### Pattern Recognition Always compare the available moles divided by the respective stoichiometric coefficients to quickly find the Limiting Reagent: 10/1 gg 0.19/2$10/1 \gg 0.19/2$. This trick saves execution seconds during complex numeric problems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula Number of millimoles (n$n$) is given by: n = M times V_mathrmmL$n = M \times V_{\mathrm{mL}}$ Molarity (M$M$) of a diluted mixture: M = fractextTotal molestextTotal Volume in Liters$M = \frac{\text{Total moles}}{\text{Total Volume in Liters}}$ ### Core Logic Evaluate the first mixing step to determine the net acid-base state: - Millimoles of mathrmNaOH = 10mathrm~mL times 2mathrm~M = 20mathrm~mmol$\mathrm{NaOH} = 10\mathrm{~mL} \times 2\mathrm{~M} = 20\mathrm{~mmol}$ - Millimoles of mathrmHCl = 20mathrm~mL times 1mathrm~M = 20mathrm~mmol$\mathrm{HCl} = 20\mathrm{~mL} \times 1\mathrm{~M} = 20\mathrm{~mmol}$ Since millimoles are equal, mathrmHCl$\mathrm{HCl}$ and mathrmNaOH$\mathrm{NaOH}$ completely neutralize each other, producing a neutral aqueous salt solution. ### Step 1: Analyze transfer to volumetric flask Taking 10mathrm~mL$10\mathrm{~mL}$ of this neutralized solution provides no excess mathrmH^+$\mathrm{H}^+$ or mathrmOH^-$\mathrm{OH}^-$ ions. It is added to a volumetric flask containing 2mathrm~mol$2\mathrm{~mol}$ of pure mathrmHCl$\mathrm{HCl}$. ### Step 2: Calculate final molarity of mathrmHCl$\mathrm{HCl}$ The volume of the flask is made up to 100mathrm~mL = 0.1mathrm~L$100\mathrm{~mL} = 0.1\mathrm{~L}$: M = frac2mathrm~mol0.1mathrm~L = 20mathrm~M$M = \frac{2\mathrm{~mol}}{0.1\mathrm{~L}} = 20\mathrm{~M}$ Hence, the resulting solution is 20mathrm~M~HCl$20\mathrm{~M~HCl}$. ### Pattern Recognition Stoichiometric neutralizations are evaluated by setting up mole/millimole balance charts. Once stoichiometric equivalence (MV_textacid = MV_textbase$MV_{\text{acid}} = MV_{\text{base}}$) is reached, any sub-aliquot of that solution remains completely neutral. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry Class 11 Chemistry: Ionic Equilibrium

### Related Formula The balanced chemical equation for the displacement reaction is: mathrmMg(s) + 2mathrmHCl(aq) rightarrow mathrmMgCl_2mathrm(aq) + mathrmH_2mathrm(g)$\mathrm{Mg(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{MgCl}_2\mathrm{(aq)} + \mathrm{H}_2\mathrm{(g)}$ At STP, 1 mole of any ideal gas occupies a volume of 22.4mathrm~L = 22400mathrm~mL$22.4\mathrm{~L} = 22400\mathrm{~mL}$. ### Core Logic From the stoichiometry of the reaction: - 1 mole of mathrmMg$\mathrm{Mg}$ (24mathrm~g$24\mathrm{~g}$) produces 1 mole of mathrmH_2$\mathrm{H}_2$ (22400mathrm~mL$22400\mathrm{~mL}$ at STP). ### Step 1: Calculate moles of mathrmH_2$\mathrm{H}_2$ gas produced n_mathrmH_2 = frac220mathrm~mL22400mathrm~mL/mol approx 9.8214 times 10^-3mathrm~mol$n_{\mathrm{H}_2} = \frac{220\mathrm{~mL}}{22400\mathrm{~mL/mol}} \approx 9.8214 \times 10^{-3}\mathrm{~mol}$ ### Step 2: Calculate mass of Magnesium required Since the molar ratio of $ ### Step 2: Calculate mass of Magnesium required Since the molar ratio of $\mathrm{Mg} to $ to $\mathrm{H}_2 is $ is $1:1: $: $n_mathrmMg = 9.8214 times 10^-3mathrm~mol$n_{\mathrm{Mg}} = 9.8214 \times 10^{-3}\mathrm{~mol}$ $ $textMass of Mg = 9.8214 times 10^-3mathrm~mol times 24mathrm~g/mol$\text{Mass of Mg} = 9.8214 \times 10^{-3}\mathrm{~mol} \times 24\mathrm{~g/mol}$ $ $textMass of Mg approx 0.2357mathrm~g = 235.7mathrm~mg approx 236mathrm~mg$\text{Mass of Mg} \approx 0.2357\mathrm{~g} = 235.7\mathrm{~mg} \approx 236\mathrm{~mg}$ This matches Option (3). ### Pattern Recognition Always keep a close eye on unit prefixes in options. A mass of $ This matches Option (3). ### Pattern Recognition Always keep a close eye on unit prefixes in options. A mass of $0.2357\mathrm{~g} corresponds to $ corresponds to $235.7\mathrm{~mg}, which rounds directly to $, which rounds directly to $236\mathrm{~mg}, whereas $, whereas $235.7\mathrm{~g}$ is off by a factor of 1000. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### 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