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#concentration

22 public questions tagged with this topic.

CHEMISTRY

What is the molality of a solution containing 8 g of NaOH in 400 g of water? (Molar mass of NaOH = 40 g/mol)

Given: What is the molality of a solution containing 8 g of NaOH in 400 g of water? (Molar mass of NaOH = 40 g/mol) These values define the system as per NCERT data. Formula: Moles = 8 / 40 = 0.2 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Mass of solvent = 0.4 kg. Molality = 0.2 / 0.4 = 0.5 m. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

What is the mass percentage of a solution formed by mixing 120 g of a 20% (w/w) solution with 180 g of a 10% (w/w) solut

Given: What is the mass percentage of a solution formed by mixing 120 g of a 20% (w/w) solution with 180 g of a 10% (w/w) solution? Formula: Mass of solute from first = 0.20 × 120 = 24 g. Substitution & Calculation: Mass of solute from second = 0.10 × 180 = 18 g . Total solute = 24 + 18 = 42 g. Total mass = 120 + 180 = 300 g. Mass % = 42/300 × 100 = 14% . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

A solution has a density of 1.22 g/mL and contains 35% by mass Hâ‚‚SOâ‚„. What is its molarity? (Molar mass of Hâ‚‚SOâ‚„

Given: A solution has a density of 1.22 g/mL and contains 35% by mass H₂SO₄. What is its molarity? (Molar mass of H₂SO₄ = 98 g/mol) These values define the system as per NCERT data. Formula: H₂SO₄ = 35 g. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Assume 100 g solution. . Moles = 35 / 98 ≈ 0.3571 mol. Volume = 100 / 1.22 ≈ 81.97 mL = 0.08197 L. Molarity = 0.3571 / 0.08197 ≈ 4.36 M ≈ 4.4 M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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Plant tissue culture 1

Macronutrients in MS medium are required at concentration:

In plant tissue culture classification, macronutrients are defined as elements required at concentration greater than 0.5 millimolar, reflecting their roles as building blocks and osmotic regulators. They include nitrogen incorporated into amino acids, nucleotides, chlorophyll porphyrin and alkaloids; potassium acting as cofactor for over 40 enzymes, osmoregulator for stomatal opening; calcium essential for middle lamella calcium pectate crosslinking cell walls and as secondary messenger binding calmodulin during stress signaling; magnesium central atom of chlorophyll and bridge for ATP Mg complexes needed by kinases; phosphorus in nucleic acids phospholipids and high-energy ATP; sulfur in cysteine, methionine, coenzymes and Fe-S clusters. In MS medium macronutrients occur at tens of millimolar, for example 20.6 mM NH4+, 18.8 mM NO3-, 20 mM K+, 3 mM Ca2+, 1.5 mM Mg2+. Deficiency below 0.5 mM rapidly causes chlorosis necrosis or hyperhydricity. Micronutrients conversely needed below 0.5 mM as catalytic cofactors, highlighting quantitative distinction guiding stock preparation where macro stock prepared at 10X and micro at 100X to prevent precipitation and toxicity.

Ref: Gamborg et al., Exp Cell Res; MS medium macronutrient composition, Plant Cell Culture Manual.

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Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

For the reaction 2A(g) + 2B(g) 3C(g) , Kc = 64 at 500 K. If 2 moles of A and 2 moles of B are placed in a 1 L vessel, wh

Initial: [A] = 2 M , [B] = 2 M , [C] = 0 . Let 3x be moles of C formed, so A and B decrease by 2x . At equilibrium: [A] = 2 - 2x , [B] = 2 - 2x , [C] = 3x . Kc = ([C]³/[A]²[B]²) = ((3x)³/(2 - 2x)² (2 - 2x)²) = (27x³/(2 - 2x)⁴) = 64 , (27x³/(2 - 2x)⁴) = 64 , (3x/2 - 2x) = 4 , 3x = 8 - 8x , 11x = 8 , x ≈ 0.727 , [C] = 3 × 0.727 ≈ 2.18 M .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

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