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Vidaara.orgClass 11 · Chemistry
CodeVID-C11-01-CH-01
Some Basic Concepts of Chemistry — Full Chapter Assignment
Chapter: Some Basic Concepts of Chemistry
Topic: All Topics
Maximum Marks: 40
Time: 90 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • All questions are compulsory.
  • Use atomic masses: H=1, C=12, N=14, O=16, Na=23, S=32, Cl=35.5, Ca=40, Cu=63.5.
  • Take N_A = 6.022 x 10^23 and molar gas volume at STP = 22.4 L.
  • Show all working; answers without steps earn no marks in Sections C and D.
Section A (1 mark each) 6 × 1 = 6 marks
1.
One mole of a substance contains:
  • A.$6.022\times10^{22}$ particles
  • B.$6.022\times10^{23}$ particles
  • C.$3.011\times10^{23}$ particles
  • D.$12\times10^{23}$ particles
2.
The empirical formula of $C_4H_8$ is:
  • A.$CH$
  • B.$CH_2$
  • C.$C_2H_4$
  • D.$CH_4$
3.
The molality unit is:
  • A.$mol\,L^{-1}$
  • B.$mol\,kg^{-1}$
  • C.$g\,L^{-1}$
  • D.dimensionless
4.
The reactant left over after a reaction is the:
  • A.limiting reagent
  • B.excess reagent
  • C.catalyst
  • D.product
5.
The molar mass of $CaCO_3$ is:
  • A.$84\,g\,mol^{-1}$
  • B.$100\,g\,mol^{-1}$
  • C.$106\,g\,mol^{-1}$
  • D.$112\,g\,mol^{-1}$
6.
Equal volumes of gases at the same T and P contain equal numbers of molecules — this is the law of:
  • A.Proust
  • B.Dalton
  • C.Avogadro
  • D.Lavoisier
Section B (2 marks) 4 × 2 = 8 marks
7.
Calculate the number of moles in $9\,g$ of water.
8.
Find the mass percent of oxygen in $CO_2$.
9.
Calculate the molarity of a solution with $0.1\,mol$ solute in $250\,mL$.
10.
State the law of conservation of mass.
Section C (3 marks) 2 × 3 = 6 marks
11.
A compound has 26.7% C, 2.2% H and 71.1% O. Find its empirical formula.
12.
Calculate the number of molecules and atoms in $4.25\,g$ of $NH_3$.
Section D (5 marks) 2 × 5 = 10 marks
13.
$56\,g$ of $N_2$ is mixed with $12\,g$ of $H_2$ for $N_2 + 3H_2 \rightarrow 2NH_3$. Find the limiting reagent, mass of $NH_3$ formed and mass of the reactant left in excess.
14.
A $250\,mL$ solution contains $4\,g$ of $NaOH$. (a) Find its molarity. (b) What volume must be diluted to make $500\,mL$ of $0.05\,M$? (c) How many moles of NaOH are in $50\,mL$ of the original solution?

Answer Key

Section A (1 mark each)
  1. (B) $6.022\times10^{23}$ particles
  2. (B) $CH_2$
  3. (B) $mol\,kg^{-1}$
  4. (B) excess reagent
  5. (B) $100\,g\,mol^{-1}$
  6. (C) Avogadro
Section B (2 marks)
  1. n = 9/18 = 0.5 mol.
  2. M = 44; O mass = 32; % O = (32/44) x 100 = 72.7%.
  3. M = 0.1 / 0.25 = 0.4 mol/L.
  4. Mass is neither created nor destroyed in a chemical reaction; the total mass of reactants equals the total mass of products.
Section C (3 marks)
  1. C = 26.7/12 = 2.22; H = 2.2/1 = 2.2; O = 71.1/16 = 4.44. Divide by 2.22: C = 1, H = 1, O = 2. Empirical formula CHO2 (i.e. simplest ratio 1:1:2).
  2. M(NH3) = 17; n = 4.25/17 = 0.25 mol; molecules = 0.25 x 6.022 x 10^23 = 1.506 x 10^23. Each NH3 has 4 atoms, so atoms = 4 x 1.506 x 10^23 = 6.022 x 10^23.
Section D (5 marks)
  1. n(N2) = 56/28 = 2 mol; n(H2) = 12/2 = 6 mol. Per coefficient: N2 = 2/1 = 2, H2 = 6/3 = 2, equal, so reactants are in exact stoichiometric ratio. NH3 = 2 x 2 = 4 mol = 4 x 17 = 68 g; no reactant left over.
  2. (a) moles = 4/40 = 0.1 mol; V = 0.25 L; M = 0.1/0.25 = 0.4 M. (b) M1V1 = M2V2: 0.4 x V1 = 0.05 x 500, V1 = 62.5 mL. (c) moles in 50 mL = 0.4 x 0.05 = 0.02 mol.
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