10mathrm~mL of 2mathrm~M~NaOH solution is added to 20mathrm~mL of 1mathrm~M~HCl solution kept in a beaker. Now, 10mathrm~mL of this mixture is poured into a volumetric flask of 100mathrm~mL containing 2 moles of mathrmHCl and made the volume upto the mark with distilled water. The solution in this flask is :

Solution & Explanation

### Related Formula Number of millimoles (n) is given by: n = M times V_mathrmmL Molarity (M) of a diluted mixture: M = fractextTotal molestextTotal 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 - Millimoles of mathrmHCl = 20mathrm~mL times 1mathrm~M = 20mathrm~mmol Since millimoles are equal, mathrmHCl and mathrmNaOH completely neutralize each other, producing a neutral aqueous salt solution. ### Step 1: Analyze transfer to volumetric flask Taking 10mathrm~mL of this neutralized solution provides no excess mathrmH^+ or mathrmOH^- ions. It is added to a volumetric flask containing 2mathrm~mol of pure mathrmHCl. ### Step 2: Calculate final molarity of mathrmHCl The volume of the flask is made up to 100mathrm~mL = 0.1mathrm~L: M = frac2mathrm~mol0.1mathrm~L = 20mathrm~M Hence, the resulting solution is 20mathrm~M~HCl. ### Pattern Recognition Stoichiometric neutralizations are evaluated by setting up mole/millimole balance charts. Once stoichiometric equivalence (MV_textacid = MV_textbase) 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

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Some Basic Concepts Of Chemistry

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Q35 jee_main_2025_28_jan_evening Concentration Terms
Concentrated nitric acid is labelled as 75\% by mass. The volume in mL of the solution which contains 30mathrm\ g of nitric acid is Given: Density of nitric acid solution is 1.25mathrm\ g/mL
  • A. 45
  • B. 55
  • C. 32
  • D. 40

Solution

### Related Formula Mass percentage definition: \%text w/w = fractextMass of solutetextMass of solution times 100 Density conversion equation: textVolume of solution = fractextMass of solutiontextDensity of solution ### Core Logic A value of 75\%text w/w HNO_3 implies that 75mathrm\ g of pure textHNO_3 is present in 100mathrm\ g of solution. We need to find the volume that provides exactly 30mathrm\ g of pure acid solute. ### Step 1: Calculate Solution Mass and Volume Mass of solution needed for 30mathrm\ g solute: textMass = frac10075 times 30 = 40mathrm\ g Converting mass to volume using solution density (1.25mathrm\ g/mL): textVolume = frac40mathrm\ g1.25mathrm\ g/mL = 32mathrm\ mL ### Pattern Recognition Break concentration steps down clearly: textMass of solute rightarrow textMass of solution rightarrow textVolume of solution. Combining operations: textVolume = fractextMass solute\% times frac100textdensity = frac3075 times frac1001.25 = 0.4 times 80 = 32. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q jee_main_2025_29_jan_morning Properties of Matter and Their Measurement
Choose the correct statements. (A) Weight of a substance is the amount of matter present in it. (B) Mass is the force exerted by gravity on an object. (C) Volume is the amount of space occupied by a substance. (D) Temperatures below 0^circmathrmC are possible in Celsius scale, but in Kelvin scale negative temperature is not possible. (E) Precision refers to the closeness of various measurements for the same quantity.
  • A. (B), (C) and (D) Only
  • B. (A), (B) and (C) Only
  • C. (A), (D) and (E) Only
  • D. (C), (D) and (E) Only

Solution

### Related Formula T_mathrmK = T_^circmathrmC + 273.15 Absolute zero (0text 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) 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 represents complete cessation of molecular motion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q77 jee_main_2024_01_february_morning Titration
Given below are two statements : Statement (I): Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution. Statement (II) : In this titration phenolphthalein can be used as indicator. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. textBoth Statement I and Statement II are correct
  • B. textStatement I is correct but Statement II is incorrect
  • C. textStatement I is incorrect but Statement II is correct.
  • D. textBoth Statement I and Statement II are incorrect.

Solution

### Core Logic Statement (I): Potassium hydrogen phthalate (KHP) is widely used as a primary standard in analytical chemistry for standardizing strong bases like 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 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 Equivalence pH > 7 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
Q89 jee_main_2024_01_february_morning Stoichiometry
Consider the following reaction: 3PbCl_2 + 2(NH_4)_3PO_4 rightarrow Pb_3(PO_4)_2 + 6NH_4Cl If 72 mathrm~mmol of PbCl_2 is mixed with 50 mathrm~mmol of (NH_4)_3PO_4, then amount of Pb_3(PO_4)_2 formed is ... mmol. (nearest integer)
Numerical Answer. Answer: 24 to 24

Solution

### Related Formula textMoles of Product = textMoles of Limiting Reagent times fractextStoichiometry of ProducttextStoichiometry 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. Let's find the limiting reagent (L.R.) by dividing given millimoles by stoichiometric coefficients: For PbCl_2: frac723 = 24 For (NH_4)_3PO_4: frac502 = 25 Since 24 < 25, PbCl_2 is the limiting reagent and will completely consume. ### Step 1: Calculate Product Moles Moles of Pb_3(PO_4)_2 formed depends entirely on PbCl_2. 3 mmol of PbCl_2 produces 1 mmol of Pb_3(PO_4)_2. Therefore, 72 mmol of PbCl_2 will produce: frac13 times 72 = 24 mathrm~mmol of Pb_3(PO_4)_2. ### Pattern Recognition Always identify the Limiting Reagent by taking the ratio n / textcoefficient. The smallest ratio dictates the extent of the reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q85 jee_main_2024_29_january_evening Volumetric Titration and Molarity
If 50text mL of 0.5text M oxalic acid is required to neutralise 25text mL of mathrmNaOH solution, the amount of mathrmNaOH in 50text mL of given mathrmNaOH solution is ________ g.
Numerical Answer. Answer: 4 to 4

Solution

### Related Formula textEquivalents of Acid = textEquivalents 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 ### Core Logic For oxalic acid (textH_2textC_2textO_4), the valence factor (n-factor) is 2. For textNaOH, the n-factor is 1. Substituting the values into the normality equivalence expression: 50 times 0.5 times 2 = 25 times M_textNaOH times 1 50 = 25 times M_textNaOH implies M_textNaOH = 2text M ### Step 1: Mass Isolation To find the mass of textNaOH present in 50text mL of this solution: textMass = textMolarity times textVolume (in L) times textMolar Mass textMass = 2 times left(frac50, 1000right) times 40 = 2 times 0.05 times 40 = 4text g ### Pattern Recognition Remember to use the correct n-factor (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

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