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Chemistry (SSC Chemistry)NCTB SSC Chemistry (Classes 9-10)Mole & Molarity (W = SVM/1000)Reaction Enthalpy (ΔH)Oxidation & Titration

SSC Chemistry Equation & Solution Suite

All-in-one NCTB SSC Chemistry quantitative solver covering Graham's Law, Bohr Model, Mole & Molarity, Empirical Formula, Limiting Reactants, Oxidation Numbers, Bond Energy ΔH, pH, and Hydrocarbons.

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SSC Chemistry Unified Solutions SuiteNCTB 9-10

Quantitative Chemistry & Stoichiometry Solver

কেলাস পানি ও মোলার ভরআয়ন-ইলেকট্রন রেডক্স সমতাকরণ
6ম অধ্যায়: Mole & MolarityModule: #mole_molarity
নির্ণয় করতে হবে:
Step-by-Step Derivation
Final Answer:
2.6497g (গ্রাম)
1. Solute Mass Formula
W=S×V×M1000W = \frac{S \times V \times M}{1000}
2. Mass Evaluation
W=0.1 M×250 mL×105.988 g/mol1000=2.6497 gramsW = \frac{0.1\text{ M} \times 250\text{ mL} \times 105.988\text{ g/mol}}{1000} = 2.6497\text{ grams}

Lab Preparation Note: Weigh 2.6497 g of solute on an analytical balance, transfer to a 250 mL volumetric flask, dissolve in distilled water, and dilute to the calibration mark.

NCTB কারিকুলাম ৯-১০ স্ট্যান্ডার্ড সমীকরণChemistry 9-10
Result:
2.6497g (গ্রাম)

Formula & Derivation

Theoretical foundation, dimensional analysis, and governing boundary conditions.

NCTB SSC Chemistry Quantitative Formula Cheat Sheet

Scoring full marks in secondary chemistry requires complete mastery over quantitative formulas across Chapters 2, 3, 6, 7, 8, 9, and 11. Below is the comprehensive standardized formula bank:

Chapter 2: States of Matter (Graham's Law)

• Diffusion rate vs Molar mass: r₁ / r₂ = √(M₂ / M₁)

• Diffusion time vs Molar mass: t₁ / t₂ = √(M₁ / M₂)

• Heavier gases diffuse more slowly than lighter gases (r ∝ 1/√M)

Chapter 3: Structure of Matter (Bohr & Isotopes)

• Average atomic mass: A_avg = Σ (A_i × p_i) / 100

• Bohr angular momentum: mvr = (n × h) / (2π)

• Planck's constant: h = 6.626 × 10⁻³⁴ J·s

• Orbital electron capacity: 2n² (subshells: s=2, p=6, d=10, f=14)

Chapter 6: Mole Concept & Molarity

• Unified mole formula: n = W / M = V / 22.4 = N / (6.023 × 10²³)

• Solution molarity: W = (S × V × M) / 1000

• Percentage composition: %Element = (n × Ar / Mr) × 100%

• Solution dilution: V₁S₁ = V₂S₂

Chapter 6: Empirical Formula & Limiting Reactant

• Atom ratio: ratio = %mass / Ar

• Molecular formula: (Empirical Formula)_n where n = M / M_empirical

• Vapor density relation: Molar mass = 2 × Vapor Density (M = 2D)

• Limiting reactant: min(n_i / stoichiometric_coefficient)

Chapter 7: Chemical Reactions (Oxidation Numbers)

• Neutral compound sum of oxidation numbers = 0

• Polyatomic ion sum of oxidation numbers = ionic charge

• Oxidation = Loss of electrons (LEO)

• Reduction = Gain of electrons (GER)

Chapter 8: Chemistry & Energy (Bond Energy & ΔH)

• Reaction enthalpy: ΔH = Σ B_broken - Σ B_formed

• ΔH < 0 indicates an exothermic reaction

• ΔH > 0 indicates an endothermic reaction

• Core bond energies (kJ/mol): C-H: 414, Cl-Cl: 244, C-Cl: 326, H-Cl: 431, O=O: 498, O-H: 464

Chapter 9: Acid-Base Equilibrium (pH & Titration)

• Hydronium concentration: pH = -log₁₀[H⁺]

• Hydroxide concentration: pOH = -log₁₀[OH⁻]

• Water ion product at 25°C: pH + pOH = 14

• Neutralization titration: (V_A × S_A) / a = (V_B × S_B) / b

Chapter 11: Mineral Resources — Fossils (Hydrocarbons)

• Alkanes: C_n H_(2n+2) (saturated paraffins)

• Alkenes: C_n H_(2n) (double bonded olefins)

• Alkynes: C_n H_(2n-2) (triple bonded acetylenes)

• Methane (CH₄), Ethane (C₂H₆), Propane (C₃H₈), Butane (C₄H₁₀)

Solved Textbook Examples

Three fully worked pedagogical exemplars: standard textbook, advanced edge-case, and authentic past board examination.

Board Exam Special (Molarity & Standard Solution Prep)bg-teal-500/10 text-teal-400 border border-teal-500/20
10.6 grams of sodium carbonate (Na₂CO₃) is dissolved in 250 mL of aqueous solution. What is the molarity of the solution? How many additional grams of Na₂CO₃ must be added to make it a semimolar (0.5 M) solution?
Given Parameters:
  • Solution volume V = 250 mL
  • Dissolved solute mass W = 10.6 g
  • Molecular mass of Na₂CO₃ M = (2 × 23) + 12 + (3 × 16) = 106 g/mol
  • Target semimolar concentration S₂ = 0.5 M
Procedural Solution:
Part 1: Apply standard molarity formula: S = (1000 × W) / (V × M)
S = (1000 × 10.6) / (250 × 106) = 10600 / 26500 = 0.4 mol/L (M) [This is a decimolar solution].
Part 2: Calculate required mass for 0.5 M solution: W₂ = (S₂ × V × M) / 1000
W₂ = (0.5 × 250 × 106) / 1000 = 13250 / 1000 = 13.25 g
Additional solute required = W₂ - W = 13.25 - 10.6 = 2.65 grams.
Final Answer:The initial molarity is 0.4 M, and 2.65 grams of additional Na₂CO₃ is required to achieve a 0.5 M semimolar solution.
Board Exam Special (Limiting Reactant & Theoretical Yield)bg-amber-500/10 text-amber-400 border border-amber-500/20
In a closed vessel, 5 grams of hydrogen gas (H₂) is mixed with 35 grams of oxygen gas (O₂) and sparked to produce water. Identify the limiting reactant and calculate the maximum mass of water (H₂O) formed in grams.
Given Parameters:
  • Balanced equation: 2H₂ + O₂ ⟶ 2H₂O
  • Mass of H₂ supplied = 5 g, M(H₂) = 2.016 g/mol
  • Mass of O₂ supplied = 35 g, M(O₂) = 32.00 g/mol
Procedural Solution:
Step 1: Compute moles of each reactant:
n(H₂) = 5 / 2.016 = 2.480 mol
n(O₂) = 35 / 32.00 = 1.0938 mol
Step 2: Compare stoichiometric requirements (2 mol H₂ : 1 mol O₂):
Moles of H₂ required to react with 1.0938 mol O₂ = 1.0938 × 2 = 2.1876 mol.
Since 2.480 mol H₂ is available (greater than the required 2.1876 mol), H₂ is present in excess and O₂ is completely consumed. Hence, O₂ is the limiting reactant.
Step 3: Calculate theoretical yield of water:
Moles of H₂O formed = 1.0938 × 2 = 2.1876 mol
Theoretical mass of H₂O = 2.1876 mol × 18.015 g/mol = 39.41 grams.
Excess hydrogen remaining = (2.480 - 2.1876) × 2.016 = 0.59 grams.
Final Answer:Oxygen (O₂) is the limiting reactant, and 39.41 grams of water is produced (with 0.59 g of unreacted H₂ remaining).
Board Exam Special (Reaction Enthalpy ΔH from Bond Energies)bg-emerald-500/10 text-emerald-400 border border-emerald-500/20
For the chlorination of methane: CH₄ + Cl₂ ⟶ CH₃Cl + HCl, calculate the enthalpy change of the reaction (ΔH) using NCTB textbook bond energy values. (C-H = 414, Cl-Cl = 244, C-Cl = 326, H-Cl = 431 kJ/mol)
Given Parameters:
  • Bonds broken: 1 × C-H and 1 × Cl-Cl (or total: 4 × C-H, 1 × Cl-Cl)
  • Bonds formed: 1 × C-Cl and 1 × H-Cl (or total: 3 × C-H, 1 × C-Cl, 1 × H-Cl)
  • Bond energies: B(C-H)=414, B(Cl-Cl)=244, B(C-Cl)=326, B(H-Cl)=431 kJ/mol
Procedural Solution:
Step 1: Total energy absorbed to break reactant bonds (B₁):
B₁ = [1 × B(C-H)] + [1 × B(Cl-Cl)] = 414 + 244 = 658 kJ/mol
Step 2: Total energy released upon forming product bonds (B₂):
B₂ = [1 × B(C-Cl)] + [1 × B(H-Cl)] = 326 + 431 = 757 kJ/mol
Step 3: Calculate reaction enthalpy ΔH = B₁ - B₂:
ΔH = 658 - 757 = -99 kJ/mol
Since ΔH < 0, the reaction is exothermic, releasing 99 kJ of thermal energy per mole.
Final Answer:The enthalpy change ΔH = -99 kJ/mol. The reaction is exothermic.
Board Exam Special (Empirical & Molecular Formula Determination)bg-purple-500/10 text-purple-400 border border-purple-500/20
An organic compound contains 40.0% Carbon, 6.67% Hydrogen, and the remaining percentage is Oxygen. If the vapor density of the compound is 30, determine its empirical and molecular formulas.
Given Parameters:
  • %C = 40.0%, %H = 6.67%
  • %O = 100 - (40.0 + 6.67) = 53.33%
  • Vapor density (D) = 30
  • Atomic masses: C = 12.011, H = 1.008, O = 15.999
Procedural Solution:
Step 1: Divide percentages by respective atomic masses:
C = 40.0 / 12 = 3.333
H = 6.67 / 1 = 6.670
O = 53.33 / 16 = 3.333
Step 2: Divide by the smallest quotient (3.333):
C = 3.333 / 3.333 = 1
H = 6.670 / 3.333 ≈ 2
O = 3.333 / 3.333 = 1
Hence, the empirical formula is CH₂O.
Step 3: Empirical formula mass = 12 + (2 × 1) + 16 = 30 g/mol.
Step 4: Molecular mass M = 2 × Vapor Density = 2 × 30 = 60 g/mol.
Multiplier n = 60 / 30 = 2.
Molecular formula = (CH₂O)₂ = C₂H₄O₂ (Ethanoic acid / Acetic acid).
Final Answer:The empirical formula is CH₂O and the molecular formula is C₂H₄O₂.

Casio Scientific Calculator Keystroke GuideExam Ready

Exact button sequences permitted in board and university entrance examination halls:

Casio fx-991ES PLUS / 2nd Edition
The standard board tool for mole, molarity, and buffer pH.
1. Input Exponent (6.023 × 10²³)
6.023x10^x23
Do not type "× 10 ^ 23" manually; always use the dedicated x10^x key.
2. Calculate pH = -log[H+]
(-)log(1.5x10^x(-)3)=
Yields exact pH value without intermediate rounding.

Practical Applications

How this scientific principle drives chemical engineering, aerospace, medicine, and research.

Board Exam Pitfalls & Scoring Recommendations:

  • Volume Units: STP molar volume is 22.4 L. Always verify whether volume is stated in milliliters (mL) or liters (L) before substitution. 1000 mL = 1 L = 1 dm³.
  • Hydrated Salts: When determining the molar mass of hydrated crystals (e.g. CuSO₄·5H₂O), remember to include all water of crystallization (5 × 18.015 = 90.075 g/mol).
  • Limiting Reactant Identification: Never identify the limiting reactant by comparing initial masses directly. You must convert both reactants to moles and normalize by their stoichiometric coefficients.
  • Bond Counting: Sketch structural diagrams to verify the exact number of covalent bonds. For example, O₂ contains an O=O double bond (498 kJ/mol) and CO₂ contains two C=O double bonds.
  • Offline CLI Tool: For offline study or scripting, run our standalone Python CLI (scripts/ssc_chemistry_calculator.py) on your local computer.

Frequently Asked Questions

Answers to common conceptual misconceptions and board examination guidelines.

What is the standard molar gas volume at STP according to NCTB?▼
Under National Curriculum and Textbook Board (NCTB) secondary chemistry guidelines, standard temperature and pressure (STP: 0°C or 273 K and 1 atm) dictates that 1 mole of any ideal gas occupies exactly 22.4 liters (22.4 dm³). This calculator implements n = V / 22.4 for all gas conversions.
How do I handle volume units in the molarity equation?▼
When solution volume is provided in milliliters (mL or cm³), use the standard formula W = (S × V × M) / 1000, where W is solute mass in grams, S is molar concentration in mol/L, V is in mL, and M is molar mass in g/mol. For dilution problems, V₁S₁ = V₂S₂ applies.
What is the thermodynamic significance of positive vs. negative ΔH?▼
Reaction enthalpy is calculated as ΔH = (Energy to break reactant bonds) - (Energy released forming product bonds). A negative ΔH (ΔH < 0) denotes an exothermic reaction releasing heat into surroundings. A positive ΔH (ΔH > 0) denotes an endothermic reaction absorbing heat from surroundings.
How can I calculate complex chemical formulas on a Casio scientific calculator?▼
On Casio fx-991EX (ClassWiz) or fx-991CW, you can enter compound formula calculations directly using parentheses (e.g. 63.55 + 32.07 + 4×16 + 5×(2×1.01 + 16) for CuSO₄·5H₂O) without intermediate rounding errors.