Download No. of Moles = Mass ÷ Molar mass - CLASS 11-B

Class 11 - Chemistry - CH1 - Mole Concept
Q1: Define mole
Answer: A mole (or mol) is defined as the amount of substance which contains equal number of particles
(atoms / molecules / ions) as there are atoms in exactly 12.000g of carbon-12.
One mole of carbon-12 atom has a mass of exactly 12.000 grams and contains 6.02 × 10 23 atoms.
A mole is just a number like a dozen. A dozen equals to 12 eggs , a gross of Pencil equals to 144 Pencil.
Similarly, mol is equal to 6.022 × 1023(Avagadro constant). Mol is also known as chemist dozen.
The value 6.022 × 1023 is known as Avogadro Constant (NA), after the Italian scientist who first
recognized the importance of the mass/number relationship
Q2: Why was there need to use this number called mole?
Answer: Atoms and molecules are extremely small in size and their numbers in even a small amount
of any substance is really very large. To handle such large numbers, a unit of similar magnitude is
required.
Q3: 1 mol of chlorine atom contains
(a) 6.022 × 1023 atoms
(b) one atom
(c) 35.5 g of Cl
(d) All of the above.
Answer: Both (a) and (b) are correct. 1 mol of Cl atom = 36.5 g of Cl (molar mass)
= 6.02 × 1023 atoms.
Q4: 1.0 mole of Chlorine molecule (Cl2) contains
(i) how many number of molecules.
(ii) how many number of atoms.
(iii) how much it weighs.
Answer: (i) 1.0 mol of Cl2 contains 6.022 × 1023molecules.
(ii) One molecule of Cl2 contains 2 atoms of Cl. ∴ Cl2 contains 2 × 6.022 × 1023 atoms
i.e. 12.44 × 1023 atoms
(iii) Molar mass of Cl is 35.5 gm/mol. 1.0 mol of Cl2 weighs = 2 × 35.5 = 71.1 g
Mole in terms of number,
1 Mole of particle = 6.022 × 1023 particles
Q5: Which of the following is correct option?
1.0 mole of NH3 (ammonia) contains ...
(a) 6.022 × 1023 molecules
(b) 4 mol of atoms
(c) 1 mol of Nitrogen atoms
(d) 3 × 6.022 × 1023 of H atoms
Answer: All of the above options are correct.
Q6: What is atomic mass unit (amu)?
Answer: Atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12
atom. One atomic mass unit also called one Dalton.
Mass of one mole of C-12 atoms = 12 g = mass of 6.022 × 1023 C-12 atoms
1 amu = 1g per mol = 1/NA = 1/ (6.022 × 1023) = 1.66 × 10-24g.
Q7: What is the mass of one 12C atom? Express it in grams as well as in amu.
Answer: one mole of C-12 atoms = 12 g = mass of 6.022 × 1023 C-12 atoms
∴ mass of 1 C-12 atom = 12g ÷ 6.022 × 1023 = 1.994 × 10-23g
or = 12g ÷ 6.022 × 1023 = 12g × 1.66 × 10-24g = 12 amu.
Q8: What is molar mass?
Answer: The mass of one mole of an element or one mole of compound is referred as molar mass.
It is expressed as g mol-1.
Example:
molar mass of Mg = 24 g mol-1.
molar mass of methane (CH4) = (12 + 4) g mol-1 = 16 g mol-1.
Q9: What is Gram atomic mass or molar mass of an element?
Answer: Gram atomic mass or molar mass of an element is mass of 1 mol of atoms or atomic mass
expressed in grams. For example, atomic mass of Mg = 24u, therefore, molar mass of Mg is 24 grams
per mol. Molar mass of an element is also called one gram atom.
Q10: What is Gram molecular mass or molar mass of molecular substance?
Answer: Gram molecular mass or molar mass of a molecular substance is the mass of 1 mol of molecules
or molecular mass expressed is grams. For example, molecular mass of H 2O is 18u (2u + 16u), therefore,
molar mass of H2O is 18 g mol-1.
Mole in terms of mass,
No. of Moles = Mass ÷ Molar mass
Q11: In 14g of N2(nitrogen gas), calculate
(i) number of moles (take molar mass of 28 g mol-1)
(ii) number of molecules
(iii) number of atoms
Answer:
(i) No. of moles = 14g ÷ 28 g mol-1 = 0.5 moles
(ii) No. of molecules = (number of moles) × NA = 0.5 × 6.022 × 1023 = 3.011 × 1023molecules
(iii) One molecule of N2 contains 2 N atoms.
No. of atoms = 2 × 3.011 × 1023 = 6.022 × 1023 atoms = 1 NA atoms.
Q12: One gram molecule H2O contains how many moles of molecules and number of molecules?
Answer: 1 gram molecule H2O = 1 mol of molecule H2O = 6.022 × 1023 molecules of H2O
Q13: How many moles of molecule are present in 15g of C2H4 (ethylene). Given atomic mass of C =
12.0 amu and of H = 1.0 amu.
Answer: Molecular mass of C2H4 = 2 × 12.0 amu + 4 × 1.0 amu = 24 + 4 = 28amu
∴ Molar mass of C2H4 = 28 g mol-1
No. of moles of molecule C2H4 = mass ÷molar mass = 15 ÷ 28 = 0.536 mol.
Q14: Calculate number of bromide ions in 3 moles of mercury(II) bromide.
Answer: Mercury(II) bromide (HgBr2) in ionic form in an aqueous solution is
HgBr2 → Hg+2 + 2Br-1
∴ 1 molecule of HgBr2 contains 2 Br-1 ions.
⇒ 1 mol of HgBr2 contains 2 mol Br-1 ions.
∴ 3 molecule of HgBr2 contains (2×3) = 6 Br-1 ions. = 6 × 6.022 × 1023 = 1.807 × 1024ions
Q15: What is Molar Volume?
Answer: One mole of any gas contains 22.4 liters at N.T.P or S.T.P. (0°C and 1atm/760 mm of Hg), the
volume is called molar volume of standard molar volume. For molecular gases (e.g. CH 4), it is
expressed as gram molecular volume and for atomic gases (e.g. He gas), it is expressed as gram atomic
volume.
e.g. One mol of CH4 = 22.4 litres at STP gram molecular volume = 1 gram molecule of CH4
= 6.022 × 1023 molecules of CH4 = 16g of CH4
Mole in terms of molar volume, at STP
No. of Moles = Volume of gas (dm3) ÷ 22.4 (dm3 mol-1)
Q16: What is the volume in litres at S.T.P. of 3 moles of hydrogen sulfide H 2S gas?
Answer: At S.T.P., 1 mole = 22.4 litres molar volume
3 moles = 3 × 22.4 = 67.21 litres of Hydrogen Sulphide (H 2S).
Q17: A balloon is filled with He gas at N.T.P. The volume of the balloon is 2.24 dm 3, find the
amount of He gas in terms of moles is present.
Answer: At N.T.P., 22.4 dm3 contains 1 mol of Helium gas.
2.24 dm3 will have = 2.24 × 1/ 22.4 = 0.1 mol.
Q18: Calculate the molar mass of Glucose (C6H12O6).
Answer:
Atomic mass of C = 12u
Atomic mass of H = 1 u
Atomic mass of O = 16 u
No. of C atoms glucouse = 6
No. of H atoms glucouse = 12
No. of O atoms glucouse = 6
Molar mass of glucouse = 6 × 12 + 12 × 1 + 6 × 16 = 180g
⇒ 1 mole or 6.022 × 1023 molecules of glucouse (C6H12O6) weighs = 180g
Q19: A 12.0 g sample of gas occupies 19.2 L at STP. What is the molecular weight of this gas?
Answer: According to Avagadro's law, at n.t.p. 1 molar volume of gas = 22.4 L
At n.t.p. 19.2 L gas weighs = 12.0 g
At n.t.p. 22.4 L gas weighs = 12.0 x 22.4 /19.2 = 14.0g / mol
Q20(NCERT): Calculate the molecular mass of the following:
(i) H2O
(ii) CO2
(iii) CH4
Answer:
Molecular mass is the sum of atomic masses of the elements present in a molecule. It is obtained by
multiplying the atomic mass of each element by the number of its atoms and adding them together.
(i) Molecular mass of Water (H2O) = (2× Atomic mass of H) + (1 × Atomic mass of Oxygen)
= 2 × 1.008u + 1 × 16.00u = 2.016 u + 16.00 u = = 18.016 u = 18.02 u
(ii) Molecular mass of CO2 = (1 × Atomic mass of C ) + (2 × Atomic mass of O)
= 1 × 12.011u + 2 × 16.00u = = 12.011 u + 32.00 u = 44.01 u
(iii) Molecular mass of CH4 = (1 × Atomic mass of C ) + (4 × Atomic mass of H)
= 1 × 12.011u + 4 × 1.008u = = 12.011 u + 4.032 u = 16.043 u
Q21(NCERT): How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Answer: 1 mole of CuSO4 contains = 1 mole of Cu + 1 mol of S + 4 moles of O.
Molar mass of CuSO4 = (63.5g) + (32.00g) + 4(16.00g)
= 63.5 + 32.00 + 64.00 = 159.5 g
Since, 1 mole of CuSO4 contains 1 mole of Copper.
⇒ 159.5 g of CuSO4 contains 63.5 g of copper.
100g of CuSO4 will have Cu = (63.5 × 100) / 159.5 = 39.81g
Q22(NCERT): What will be the mass of one 12C atom in g?
Answer: 1 mole of carbon atoms = 6.022 × 1023 atoms of carbon
= 12 g of carbon
Mass of one 12C atom = 12 ÷ 6.022 × 1023 = 1.994 × 10-23 g
Q23 (NCERT): Calculate the number of atoms in each of the following
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He.
Answer:
(i) 1 mole of Ar = 6.022 × 1023 atoms of Ar
52 mol of Ar = 52 × 6.022 × 1023 atoms of Ar
= 3.131 × 1025 atoms of Ar
(ii) 1 atom of He = 4 u of He
⇒ 4 u of He = 1 atom of He
1 u of He atom of He = 1/4 atoms of He
52u of He atom of He = 52 × 1/4 = 13 atoms of He
(iii) 4 g of He = 6.022 × 1023 atoms of He
52 g of He atoms of He = 6.022 × 1023 × 52 / 4 = 7.8286 × 1024 atoms of He
Law of Conservation of Mass states that matter can neither be created nor destroyed.
Law of Definite Proportions states that a given compound always contains exactly the same proportion
of elements by weight.
Law of Multiple Proportions states that if two elements can combine to form more than one
compound, the masses of one element that combine with a fixed mass of the other element, are in the
ratio of small whole numbers.
Gay Lussac’s Law of Gaseous Volumes: This law was given by Gay Lussac in 1808. He observed that
when gases combine or are produced in a chemical reaction they do so in a simple ratio by volume
provided all gases are at same temperature and pressure.
Avogadros Law - Avogadro proposed that equal volumes of gases at the same temperature and pressure
should contain equal number of molecules.
Mole Fraction: It is ratio of number of moles of a particular component to the total number of moles of
all the components.
Mole-fraction of solute =
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Molality (m). It is defined as number of moles of solute (B) per 1000 g or 1 kg of solvent.
Molality =
𝑛𝑜 .𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑘𝑔 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Molarity (M). It is expressed as the number of moles of solute per litre of solution.
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
Molarity =𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒𝑠
An empirical formula represents the simplest whole number ratio of various atoms
present in a compound whereas the molecular formula shows the exact number of
different types of atoms present in a molecule of a compound.
Many a time, the reactions are carried out when the reactants are not present in the
amounts as required by a balanced chemical reaction. In such situations, one reactant is in
excess over the other. The reactant which is present in the lesser amount gets consumed
after sometime and after that no further reaction takes place whatever be the amount of
the other reactant present. Hence, the reactant which gets consumed, limits the amount of
product formed and is, therefore, called the limiting reagent.
Worksheets in progress----