Class 11 Chemistry Chapter 1 Worksheets – Some Basic Concepts of Chemistry (NCERT Based)

Leave a Comment / By /
Class 11 Chemistry Chapter 1 Worksheet + Solutions

Some Basic Concepts of Chemistry Worksheet | मोल अवधारणा, मोलरता/मोललता, अनुभविक सूत्र, लिमिटिंग अभिकर्मक

यह printable worksheet (CBSE + NCERT pattern) में MCQ, Assertion–Reason, Fill in the blanks, VSA, SA और Numericals शामिल हैं। नीचे answers + stepwise solutions भी दिए गए हैं।

Quick Revision (1-minute)

  • 1 mol = 6.022×1023 particles (NA).
  • n = m/M (m in g, M in g/mol)
  • N = n × NA
  • Molarity (M) = moles / volume (L) → temperature dependent
  • Molality (m) = moles / mass of solvent (kg) → temperature independent
  • Mass % = (mass solute / mass solution) × 100
  • Mole fraction (χ) = moles component / total moles (sum = 1)
  • Limiting reagent पूरी तरह consume होता है और product decide करता है
  • Empirical formula = simplest whole-number ratio
  • Molecular formula = (empirical) × n, where n = molar mass / empirical mass
Common Mistakes: units miss, mL↔L confusion, limiting reagent ignore, sig-fig rounding गलत।

Section A: MCQs (15)

  1. 1) 1 mol CO2 में total atoms कितने होंगे?
  2. 2) 11.2 L O2 (STP) में moles कितने?
  3. 3) 18 g H2O में molecules की संख्या?
  4. 4) Molarity (M) depend करती है:
  5. 5) Molality (m) depend करती है:
  6. 6) 1 mol NaCl में total ions कितने?
  7. 7) Empirical formula बताता है:
  8. 8) 0.1 mol H2SO4 में oxygen atoms कितने?
  9. 9) 2.0 M solution का मतलब:
  10. 10) 5 g H2 में moles (M=2 g/mol):
  11. 11) 0.00450 में significant figures कितने?
  12. 12) 3.20×102 में sig figs कितने?
  13. 13) Mole fraction का total (all components) होता है:
  14. 14) 1 mole particles का value:
  15. 15) Dilution formula:

Section B: Assertion–Reason (5)

Directions: (A) A & R true, R correct explanation; (B) both true but R not explanation; (C) A true R false; (D) A false R true.

  1. 1) A: Molarity temperature के साथ change होती है.
    R: Volume temperature से change होता है.
  2. 2) A: Molality temperature से independent होती है.
    R: Molality mass of solvent पर based होती है.
  3. 3) A: Empirical formula CO2 का CO होता है.
    R: Empirical formula simplest ratio देता है.
  4. 4) A: Mole fraction unitless होता है.
    R: Mole fraction moles/moles ratio है.
  5. 5) A: Limiting reagent product amount decide करता है.
    R: Limiting reagent completely consume हो जाता है.

Section C: Fill in the Blanks (10)

  1. Avogadro number =
  2. 1 mol gas (STP) का volume = L
  3. n = m/M में n =
  4. Molarity = moles of solute /
  5. Molality = moles of solute / mass of solvent in
  6. Mole fraction का sum =
  7. Empirical formula = atoms का ratio
  8. 1 mol H2O में total atoms = mol atoms
  9. Dilution: M1V1 =
  10. 1000 mL = L

Section D: Very Short Answer (8)

  1. Mole (मोल) define करो.
  2. Molar mass क्या होता है?
  3. Empirical formula और molecular formula में difference (1 line).
  4. Molarity और molality में एक main difference.
  5. Limiting reagent क्या होता है?
  6. Significant figures क्या indicate करते हैं?
  7. Mole fraction define करो.
  8. % composition का formula लिखो.

Section E: Short Answer (6)

  1. 4.4 g CO2 में moles और molecules निकालो.
  2. 0.25 mol NaOH का mass निकालो.
  3. 10 g CaCO3 में Ca atoms कितने?
  4. 250 mL 0.5 M NaCl solution में NaCl का mass (M=58.5).
  5. 2.0 g H2 + 16 g O2 react करते हैं, limiting reagent identify करो.
  6. 2.50 g compound में C=40%, H=6.7%, O=53.3% (approx). Empirical formula.
Tip: Solution में हमेशा units लिखो और volume को L में convert करो (mL ÷ 1000).

Section F: Numericals Corner (12)

  1. 5.85 g NaCl में moles, formula units और total ions.
  2. 9 g water में oxygen atoms कितने?
  3. 0.1 mol Al2(SO4)3 में total ions कितने?
  4. 2.5 M HCl का 200 mL solution बनाने के लिए moles of HCl?
  5. 500 mL 1.0 M glucose solution में glucose mass (M=180).
  6. 25% (w/w) NaOH solution में 200 g solution में NaOH mass.
  7. 0.2 mol urea (NH2CONH2) में N atoms कितने?
  8. 2.0 L 0.1 M H2SO4 में moles of H2SO4 और total H+ ions (complete dissociation assume).
  9. 0.5 m NaCl solution में 1 kg water के लिए moles of NaCl.
  10. 20% (w/v) glucose का meaning + 250 mL में glucose mass.
  11. 0.004560 को 3 significant figures में round करो.
  12. Empirical formula CH2 है, molar mass 56 g/mol. Molecular formula find करो.

Answer Key + Stepwise Solutions

MCQ Answer Key

1-C2-B3-A4-B5-C 6-B7-B8-A9-B10-A 11-B12-B13-B14-B15-B

Assertion–Reason Answers

1-A, 2-A, 3-D, 4-A, 5-A

Fill in the Blanks Answers

  1. 6.022×1023
  2. 22.4
  3. moles
  4. volume of solution (L)
  5. kg
  6. 1
  7. simplest
  8. 3
  9. M2V2
  10. 1

Stepwise Solutions (Section E + F)

E1) 4.4 g CO2: moles & molecules

Given: m = 4.4 g, M(CO2) = 44 g/mol

n = m/M = 4.4/44 = 0.10 mol

N = n × NA = 0.10 × 6.022×1023 = 6.022×1022 molecules

E2) 0.25 mol NaOH: mass

M(NaOH)=40 g/mol

m = n×M = 0.25×40 = 10 g

E3) 10 g CaCO3: Ca atoms

M(CaCO3) ≈ 100 g/mol

n = 10/100 = 0.10 mol

Ca atoms = 0.10×6.022×1023 = 6.022×1022

E4) 250 mL 0.5 M NaCl: mass

V = 250 mL = 0.250 L

n = M×V = 0.5×0.250 = 0.125 mol

m = 0.125×58.5 = 7.31 g (approx)

E5) 2.0 g H2 + 16 g O2: limiting reagent

Reaction: 2H2 + O2 → 2H2O

n(H2) = 2.0/2 = 1.0 mol

n(O2) = 16/32 = 0.50 mol

0.50 mol O2 के लिए H2 required = 2×0.50 = 1.0 mol

Conclusion: Stoichiometric mixture, no limiting reagent.

E6) C=40%, H=6.7%, O=53.3%: empirical formula

Assume 100 g sample:

C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33

Divide by 3.33 → C:1, H:2, O:1

Empirical formula: CH2O

F1) 5.85 g NaCl: moles, formula units, ions

M(NaCl)=58.5 g/mol

n = 5.85/58.5 = 0.10 mol

Formula units = 0.10×6.022×1023 = 6.022×1022

NaCl → Na+ + Cl− (2 ions)

Total ions = 2×6.022×1022 = 1.204×1023

F2) 9 g H2O: oxygen atoms

n = 9/18 = 0.50 mol

O atoms = 0.50×6.022×1023 = 3.011×1023

F3) 0.1 mol Al2(SO4)3: total ions

Al2(SO4)3 → 2Al3+ + 3SO42− (total 5 ions)

Moles of ions = 0.1×5 = 0.50 mol

Number of ions = 0.50×6.022×1023 = 3.011×1023

F4) 2.5 M HCl, 200 mL: moles

V = 200 mL = 0.200 L

n = M×V = 2.5×0.200 = 0.50 mol

F5) 500 mL 1.0 M glucose: mass

V = 0.500 L

n = 1.0×0.500 = 0.50 mol

M(glucose)=180 g/mol

m = 0.50×180 = 90 g

F6) 25% (w/w) NaOH: 200 g solution me solute mass

NaOH mass = (25/100)×200 = 50 g

F7) 0.2 mol urea: N atoms

Urea has 2 nitrogen atoms.

Moles of N atoms = 0.2×2 = 0.40 mol

Number = 0.40×6.022×1023 = 2.409×1023

F8) 2.0 L 0.1 M H2SO4: H+ ions

n(H2SO4) = 0.1×2.0 = 0.20 mol

H+ moles = 0.20×2 = 0.40 mol

H+ ions = 0.40×6.022×1023 = 2.409×1023

F9) 0.5 m NaCl: 1 kg water me moles

Molality m = n/kg solvent

0.5 = n/1 → n = 0.50 mol

F10) 20% (w/v) glucose: meaning + 250 mL me mass

20% (w/v) = 20 g in 100 mL solution

For 250 mL: (20/100)×250 = 50 g

F11) 0.004560 ko 3 sig figs

3 significant figures में: 0.00456

F12) Empirical CH2, molar mass 56: molecular formula

Empirical mass = 12 + 2 = 14 g/mol

n = 56/14 = 4

Molecular formula = (CH2)4 = C4H8

Chapter 1 FAQs

Molarity solution के volume पर depend करती है, इसलिए temperature change से बदल सकती है। Molality solvent के mass पर depend करती है, इसलिए temperature-independent रहती है।
Molecular formula = Empirical formula × n, जहाँ n = (molar mass / empirical formula mass) होता है।
Limiting reagent पूरी तरह consume हो जाता है और वही product की maximum quantity decide करता है।

Leave a Comment

Shopping Basket
Scroll to Top