Mass of magnesium required to produce 220mathrm~mL$220\mathrm{~mL}$ of hydrogen gas at STP on reaction with excess of dil. HCl is : Given: Molar mass of Mg is 24mathrm~g~mol^-1$24\mathrm{~g~mol}^{-1}$ .

### Related Formula Mass percentage definition: \%text w/w = fractextMass of solutetextMass of solution times 100$$\%\text{ w/w} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$$ Density conversion equation: textVolume of solution = fractextMass of solutiontextDensity of solution$$\text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density of solution}}$$ ### Core Logic A value of 75\%text w/w HNO_3$75\%\text{ w/w HNO}_3$ implies that 75mathrm\ g$75\mathrm{\ g}$ of pure textHNO_3$\text{HNO}_3$ is present in 100mathrm\ g$100\mathrm{\ g}$ of solution. We need to find the volume that provides exactly 30mathrm\ g$30\mathrm{\ g}$ of pure acid solute. ### Step 1: Calculate Solution Mass and Volume Mass of solution needed for 30mathrm\ g$30\mathrm{\ g}$ solute: textMass = frac10075 times 30 = 40mathrm\ g$$\text{Mass} = \frac{100}{75} \times 30 = 40\mathrm{\ g}$$ Converting mass to volume using solution density (1.25mathrm\ g/mL$1.25\mathrm{\ g/mL}$): textVolume = frac40mathrm\ g1.25mathrm\ g/mL = 32mathrm\ mL$$\text{Volume} = \frac{40\mathrm{\ g}}{1.25\mathrm{\ g/mL}} = 32\mathrm{\ mL}$$ ### Pattern Recognition Break concentration steps down clearly: textMass of solute rightarrow textMass of solution rightarrow textVolume of solution$\text{Mass of solute} \rightarrow \text{Mass of solution} \rightarrow \text{Volume of solution}$. Combining operations: textVolume = fractextMass solute\% times frac100textdensity = frac3075 times frac1001.25 = 0.4 times 80 = 32$\text{Volume} = \frac{\text{Mass solute}}{\%} \times \frac{100}{\text{density}} = \frac{30}{75} \times \frac{100}{1.25} = 0.4 \times 80 = 32$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula T_mathrmK = T_^circmathrmC + 273.15$$T_{\mathrm{K}} = T_{^{\circ}\mathrm{C}} + 273.15$$ Absolute zero (0text K$0\text{ K}$) represents the lowest theoretical temperature limit. ### Core Logic Analyzing each statement based on foundational definitions : * (A) & (B) Incorrect: Mass is the actual matter present; weight is the gravitational force exerted on that mass. These definitions are reversed in the statements. * (C) Correct: Volume correctly defines the space occupied by a substance . * (D) Correct: Celsius values can be negative, whereas Kelvin scale strictly defaults to absolute zero (0text K$0\text{ K}$) as minimum . * (E) Correct: Precision measures how close experimental trials lie relative to each other . Therefore, statements (C), (D), and (E) are correct. ### Pattern Recognition Absolute temperature scale (Kelvin) can never possess real negative values because 0text K$0\text{ K}$ represents complete cessation of molecular motion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Core Logic Statement (I): Potassium hydrogen phthalate (KHP) is widely used as a primary standard in analytical chemistry for standardizing strong bases like NaOH$NaOH$. This is because it is highly pure, non-hygroscopic, stable, and has a relatively high molar mass, making its concentration reliable and stable over time. Statement (II): KHP is a weak acid and NaOH$NaOH$ is a strong base. The titration of a weak acid with a strong base yields an equivalence point in the weakly basic range (pH > 7). Phenolphthalein changes colour in the pH range 8.3 to 10.0, making it the perfect indicator for this titration. ### Step 1: Evaluate Statements Statement I is correct. Statement II is correct. ### Pattern Recognition Weak Acid vs Strong Base rightarrow$\rightarrow$ Equivalence pH > 7 rightarrow$\rightarrow$ Phenolphthalein is the indicator of choice. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula textMoles of Product = textMoles of Limiting Reagent times fractextStoichiometry of ProducttextStoichiometry of Limiting Reagent $$ \text{Moles of Product} = \text{Moles of Limiting Reagent} \times \frac{\text{Stoichiometry of Product}}{\text{Stoichiometry of Limiting Reagent}} $$ ### Core Logic From the balanced chemical equation: 3 text moles of PbCl_2 text react with 2 text moles of (NH_4)_3PO_4$3 \text{ moles of } PbCl_2 \text{ react with } 2 \text{ moles of } (NH_4)_3PO_4$. Let's find the limiting reagent (L.R.) by dividing given millimoles by stoichiometric coefficients: For PbCl_2$PbCl_2$: frac723 = 24$\frac{72}{3} = 24$ For (NH_4)_3PO_4$(NH_4)_3PO_4$: frac502 = 25$\frac{50}{2} = 25$ Since 24 < 25$24 < 25$, PbCl_2$PbCl_2$ is the limiting reagent and will completely consume. ### Step 1: Calculate Product Moles Moles of Pb_3(PO_4)_2$Pb_3(PO_4)_2$ formed depends entirely on PbCl_2$PbCl_2$. 3 mmol of PbCl_2$PbCl_2$ produces 1 mmol of Pb_3(PO_4)_2$Pb_3(PO_4)_2$. Therefore, 72 mmol of PbCl_2$PbCl_2$ will produce: frac13 times 72 = 24 mathrm~mmol$\frac{1}{3} \times 72 = 24 \mathrm{~mmol}$ of Pb_3(PO_4)_2$Pb_3(PO_4)_2$. ### Pattern Recognition Always identify the Limiting Reagent by taking the ratio n / textcoefficient$n / \text{coefficient}$. The smallest ratio dictates the extent of the reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula textEquivalents of Acid = textEquivalents of Base$$\text{Equivalents of Acid} = \text{Equivalents of Base}$$ N_1 V_1 = N_2 V_2 implies (M_1 times n_1) times V_1 = (M_2 times n_2) times V_2$$N_1 V_1 = N_2 V_2 \implies (M_1 \times n_1) \times V_1 = (M_2 \times n_2) \times V_2$$ ### Core Logic For oxalic acid (textH_2textC_2textO_4$\text{H}_2\text{C}_2\text{O}_4$), the valence factor (n$n$-factor) is 2$2$. For textNaOH$\text{NaOH}$, the n$n$-factor is 1$1$. Substituting the values into the normality equivalence expression: 50 times 0.5 times 2 = 25 times M_textNaOH times 1$$50 \times 0.5 \times 2 = 25 \times M_{\text{NaOH}} \times 1$$ 50 = 25 times M_textNaOH implies M_textNaOH = 2text M$$50 = 25 \times M_{\text{NaOH}} \implies M_{\text{NaOH}} = 2\text{ M}$$ ### Step 1: Mass Isolation To find the mass of textNaOH$\text{NaOH}$ present in 50text mL$50\text{ mL}$ of this solution: textMass = textMolarity times textVolume (in L) times textMolar Mass$$\text{Mass} = \text{Molarity} \times \text{Volume (in L)} \times \text{Molar Mass}$$ textMass = 2 times left(frac50, 1000right) times 40 = 2 times 0.05 times 40 = 4text g$$\text{Mass} = 2 \times \left(\frac{50}, {1000}\right) \times 40 = 2 \times 0.05 \times 40 = 4\text{ g}$$ ### Pattern Recognition Remember to use the correct n$n$-factor (2$2$) for dibasic oxalic acid during equivalence matching to avoid calculation errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry