MOLE CONCEPT

• One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the C12
• 12gm of C12 contains 6.022 x1023 atoms
• The number 6.022 x1023 is called Avogadro’s constant or Avogadro’s number.
• The mass of one mole of a substance in grams is called its molar mass.
• The molar mass in grams is numerically equal to the atomic/molecular/formula mass in u.(u is the unified mass)

CONVERSION OF MASS & VOLUME TO MOLES

Ex. Which has maximum number of atom?                                                              (2003, 1M)

(a) 24 g of C (12)                                             (b) 56 g of Fe (56)

(c) 27 g of Al (27)                                            (d) 108 g of Ag (108)

Solution: Number of atoms = Number of moles

Numbers of atoms in 24g C= 2412 ×NA=2NA

Number of atoms in 56 g of Fe= 5656NA=NA

Number of atoms in 27 g of Al= 2727NA=NA

Number of atoms in 108 of Ag= 108108NA=NA

Hence, 24g of carbon has the maximum number of atoms.                                      Answer (a)

Ex. The ratio mass of oxygen and nitrogen particular gaseous mixture is 1: 4. The ratio of number of their molecule is                                                                                              (2014 Main)

Solution: no2nN2= (mO2)(MO2)(mN2)(MN2)

Where, mO2  = given mass O2, mN2  = given mass of N2, MO2  = molecular mass of O2, MN2  = molecular mass of N2, nO2  = Number of moles of O2, nN2  = number of moles of N2

Ex.:- How many moles of electron weighs 1 kg?                                               (2002, 3M)

a 6.023×1023                                       b19.108×1031

c 6.0239.108×1054                             d19.108×6.023×108

Solution: Mass of electron = 9.108 x 10-31 kg

9.108 x 10-31 kg = 1.0 electron

∴   1 kg = 19.108 ×10-31  electrons= 10319.108× 16.023 ×1023

∴  19.108×6.023 ×108 mole of electrons                                                                Answer (d)

MOLARITY MOLALITY, MOLE FRACTION AND NORMALITY

• Molarity is the number of moles of solute in per liter of solution. Unit is moles per liter.

Molarity= Gram Mole of solutelitre of solutiuon

• Molality is the number of solute present in 1kg of solvent.

Molarity= Moles of solutewt. of solvent in kg

• Normality= Gram Equivalent of soluteLitre of solution
• Mole fraction of solute= Moles of solute Moles of solute+moles of solvent

Ex.:  3g of activated charcoal was added to 50mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042N. The amount of acetic acid adsorbed (per gram of charcoal) is.                                                      (2015 JEE MAINS)

(a) 18 mg                                            (b) 36 mg

(c) 42 mg                                            (d) 54 mg

Solution. Given, initial strength of acetic acid = 0.06N

Final strength = 0.042 N;        Volume = 50mL

\ Initial millimoles of CH3COOH = 0.06 x 50 =3

Final millimoles of CH3COOH = 0.042 x 50 = 2.1

\ Millimoles of CH3COOH adsorbed = 3 – 2.1 = 0.9 mmol.

= 0.9 x 60mg = 54 mg                                  Answer (d)

Ex.: Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is                                                                                        (2011)

(a) 1.78 M                                         (b) 2.00M

(c) 2.05 M                                         (d) 2.22 M

Solution: Molarity= Moles of solute Volume of Solution(L)

Moles of urea= 12060=2

Weight of solution = weight of solvent + weight of solute

= 1000 +120 = 1120 g

⇒       Volume= 1120g1.15g/mL×11000L/L=0.973L

Ex.: The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl 250 mL of 2 M HCl will be                                                                                                                                      (2013 Main)

(a) 0.875 M                                            (b) 1.00 M

(c) 1.75 M                                              (d) 0.0975 M

Solution: From the formula, Mf= M1V1+M2V2V1+V2

Given, V1 = 750 mL,  M1 = 0.5 M

V2 = 250mL, M2 = 2 M

= 750 × 0.5+250 × 2750+250= 8751000=0.875M                      Answer (a)

Day - 3

PERCENTAGE COMPOSITION

Mass % of an element=mass of that element in the compound ×100molar mass of the compound

Let’s take example of carbon-dioxide (CO2)

Mass % of C=1244 x 100=27.2%

Mass % of O= 3244x 100=72.73%

Ex.:- The most abundant element by mass in the body of a healthy human adult are oxygen (61.4%) carbon (22.9%), hydrogen (10.0%) and nitrogen (2.6%) The weight which a 75 kg person would gain if all Hatoms are replaced by 2 Hatoms are replaced by 2H atoms is       2017 JEE Main)

(a) 15Kg                                                  (b) 37.5 kg

(b) 7.5 Kg                                                (d) 10kg

Solution: Given, abundance of element by mass

Oxygen = 61.4% carbon= 22.9%, hydrogen = 10% and nitrogen = 2.6%

Total weight of person = 75 kg

Mass due to 1H= 75 ×10100=7.5kg

1H atoms are replaced by 2H atoms

Mass due to 2H=7.5 ×2kg

\    Mass gain by person = 7.5 kg                                                                             Answer (c)

Ex.:- Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5% 90% and 5% respectively the atomic mass of Fe is                                                                                                            (2009)

(a) 55. 85                                                          (b) 55.95

(c) 55.75                                                           (d) 56.05

Solution: From the given relative abundance the average weight of Fe can be calculated as

A= 54 × 5 × 56 × 90 ×57 ×5100=55.95                                                            Answer (b)

If we know percent composition we can find empirical formula

An empirical formula represents the simplest whole number ratio of various atoms present in a compound

•  Molecular formula shows the exact number of different types of atoms present in a molecule of a compound.
•  If the mass per cent of various elements present in a compound is known, its empirical formula can be determined.
•  Molecular formula = n x (Empirical formula), where n is a simple number and may have values 1, 2, 3….

Following steps should be followed to determine empirical formula of the compound

1.  Step 1: Conversion of mass per cent of various elements into grams.
2.  Step 2: Convert mass obtained in step1 into number of moles
3.  Step 3: Divide the mole value obtained in step 2 by the smallest mole value (out of the mole value of various elements calculated)
4.  Step 4: In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.
5.  Step 5: Write empirical formula by mentioning the numbers after writing the symbols of respective elements

Example: - In an organic compound contains 37% carbon 50% oxygen and 13% hydrogen what is the empirical formula of compound. If it’s V. D is to find the molecular formula.

Empirical formula CH4O.

Mol. Wt. = VD x 2 = 32

Emp. formula wt = 32

n = M.W/E.W =1

So molecular formula = CH4O