Solved Examples: Some Basic Concepts of Chemistry | Additional Study Material for JEE PDF Download

Solved Examples - Some Basic Concepts of Chemistry, CBSE, Class 11, Chemistry 

Solved Objective 

Ex.1 8 litre of H₂ and 6 litre of Cl₂ are allowed to react to maximum possible extent. Find out the final volume of reaction mixture. Suppose P and T remains constant throughout the course of reaction -

(A) 7 litre (B) 14 litre (C) 2 litre (D) None of these.

Sol. (B)

H₂ Cl₂ → 2 HCl

Volume before reaction 8 lit 6 lit 0

Volume after reaction 2 0 12

Volume after reaction

= Volume of H₂ left Volume of HCl formed = 2 12 = 14 lit

Ex.2 Naturally occurring chlorine is 75.53% Cl³⁵ which has an atomic mass of 34.969 amu and 24.47% Cl³⁷ which has a mass of 36.966 amu. Calculate the average atomic mass of chlorine-

(A) 35.5 amu (B) 36.5 amu (C) 71 amu (D) 72 amu

Sol. (A)

Average atomic mass

=  =  = 35.5 amu.

Ex.3 Calculate the mass in gm of 2g atom of Mg-

(A) 12 gm (B) 24 gm (C) 6 gm (D) None of these.

Sol. (D)

Q 1 gm atom of Mg has mass = 24 gm

2 gm atom of Mg has mass = 24 x 2 = 48 gm.

Ex.4 In 5 g atom of Ag (At. wt. of Ag = 108), calculate the weight of one atom of Ag -

(A) 17.93 × 10^-23gm (B) 16.93 × 10^-23 gm

(C) 17.93 × 10²³ gm (D) 36 × 10^-23 gm

Sol. (A)

Q N atoms of Ag weigh 108 gm

1 atom of Ag weigh =  =  = 17.93 × 10^-23 gm.

Ex.5 In 5g atom of Ag (at. wt. = 108), calculate the no. of atoms of Ag -

(A) 1 N (B) 3N (C) 5 N (D) 7 N.

Sol. (C)

Q 1 gm atom of Ag has atoms = N

5 gm atom of Ag has atoms = 5N.

Ex.6 Calculate the mass in gm of 2N molecules of CO₂ -

(A) 22 gm (B) 44 gm (C) 88 gm (D) None of these.

Sol. (C)

Q N molecules of CO₂ has molecular mass = 44.

2N molecules of CO₂ has molecular mass = 44 x 2 = 88 gm.

 
Ex.7 How many carbon atoms are present in 0.35 mol of C₆H₁₂O₆ -

(A) 6.023 × 10²³ carbon atoms (B) 1.26 × 10²³ carbon atoms

(C) 1.26 × 10²⁴ carbon atoms (D) 6.023 × 10²⁴ carbon atoms

Sol. (C)

1 mol of C₆H₁₂O₆ has = 6 N atoms of C

0.35 mol of C₆H₁₂O₆ has = 6 × 0.35 N atoms of C

= 2.1 N atoms = 2.1 × 6.023 × 10²³  = 1.26 × 10²⁴ carbon atoms

Ex.8 How many molecules are in 5.23 gm of glucose (C₆H₁₂O₆) -

(A) 1.65 × 10²² (B) 1.75 × 10²² (C) 1.75 × 10²¹ (D) None of these

Sol. (B)

Q 180 gm glucose has = N molecules

5.23 gm glucose has = = 1.75 × 10²² molecules

Ex.9 What is the weight of 3.01 × 10²³ molecules of ammonia -

(A) 17 gm (B) 8.5 gm (C) 34 gm (D) None of these

Sol. (B)

Q 6.023 × 10²³ molecules of NH₃ has weight = 17 gm

3.01 × 10²³  molecules of NH₃ has weight

= = 8.50 gm

Ex.10 How many significant figures are in each of the following numbers -

(a) 4.003 (b) 6.023 × 10²³ (c) 5000

(A) 3, 4, 1 (B) 4, 3, 2 (C) 4, 4, 4 (D) 3, 4, 3

 Sol. (C) 

Ex.11 How many molecules are present in one ml of water vapours at STP -

(A) 1.69 × 10¹⁹ (B) 2.69 × 10^-19 (C) 1.69 × 10^-19 (D) 2.69 × 10¹⁹

 Sol. (D)

Q 22.4 litre water vapour at STP has

= 6.023 × 10²³ molecules

1 × 10^-3 litre water vapours at STP has

= × 10^-3 = 2.69 × 10 ¹⁹ 

Ex.12 How many years it would take to spend Avogadro's number of rupees at the rate of 1 million rupees in one second -

(A) 19.098 × 10¹⁹ years (B) 19.098 years

(C) 19.098 × 10⁹ years (D) None of these

Sol. (C)

Q 10⁶ rupees are spent in 1sec.

6.023 × 10²³ rupees are spent in = sec

=  years , = 19.098 × 10⁹ year

Ex.13 An atom of an element weighs 6.644 × 10^-23 g. Calculate g atoms of element in 40 kg-

(A) 10 gm atom (B) 100 gm atom (C) 1000 gm atom (D) 10⁴ gm atom

Sol. (C)

Q weight of 1 atom of element

= 6.644 × 10^-23 gm

weight of 'N' atoms of element

= 6.644 × 10^-23 × 6.023 × 10²³ = 40 gm

Q 40 gm of element has 1 gm atom.

40 x 10³ gm of element has  , = 10³ gm atom.

Ex.14 Calculate the number of Cl^- and Ca ² ions in 222 g anhydrous CaCl₂ -

(A) 2N ions of Ca ² 4 N ions of Cl^- (B) 2N ions of Cl^- & 4N ions of Ca ²

(C) 1N ions of Ca ² & 1N ions of Cl^- (D) None of these.

Sol. (A)

Q mol. wt. of CaCl₂ = 111 g

Q 111 g CaCl₂ has = N ions of Ca ²

222g of CaCl₂ has

= 2N ions of Ca ²

Also Q 111 g CaCl₂ has = 2N ions of Cl^-

222 g CaCl₂ has =  ions of Cl^-

= 4N ions of Cl^- .

Ex.15 The density of O₂ at NTP is 1.429g / litre. Calculate the standard molar volume of gas-

(A) 22.4 lit. (B) 11.2 lit (C) 33.6 lit (D) 5.6 lit.

Sol. (A)

Q 1.429 gm of O₂ gas occupies volume = 1 litre.

32 gm of O₂ gas occupies =  ,= 22.4 litre/mol.

Ex.16 Which of the following will weigh maximum amount-

(A) 40 g iron (B) 1.2 g atom of N

(C) 1 × 10²³ atoms of carbon (D) 1.12 litre of O₂ at STP

Sol. (A)

(A) Mass of iron=40g (B) Mass of 1.2 g atom of N = 14 × 1.2 = 16.8 gm

(D) Mass of 1 × 10²³ atoms of C =  = 1.99 gm.

(D) Mass of 1.12 litre of O₂ at STP =  = 1.6 g

Ex.17 How many moles of potassium chlorate to be heated to produce 11.2 litre oxygen -

(A) mol (B) mol (C) mol (D) mol.

Sol. (B)

2 KClO₃ → 2KCl 3O₂

Mole for reaction 2 2 3

Q 3 × 22.4 litre O₂ is formed by 2 mol KClO₃

11.2 litre O₂ is formed by = mol KClO₃

Ex.18 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone (CaCO₃).

(A) 104.4 kg (B) 105.4 kg (C) 212.8 kg (D) 106.4 kg

Sol. (D)

Q 100 kg impure sample has pure

CaCO₃ = 95 kg

200 kg impure sample has pure CaCO₃

= = 190 kg. CaCO₃ → CaO CO₂

Q 100 kg CaCO₃ gives CaO = 56 kg.

190 kg CaCO₃ gives CaO = = 106.4 kg.

Ex.19 The chloride of a metal has the formula MCl₃. The formula of its phosphate will be-

(A) M₂PO₄ (B) MPO₄ (C) M₃PO₄ (D) M(PO₄)₂

Sol. (B) AlCl₃ as it is AlPO₄

Ex.20 A silver coin weighing 11.34 g was dissolved in nitric acid. When sodium chloride was added to the solution all the silver (present as AgNO₃) was precipitated as silver chloride. The weight of the precipitated silver chloride was 14.35 g. Calculate the percentage of silver in the coin

(A) 4.8 % (B) 95.2% (C) 90 % (D) 80%

Sol. (B)

Ag 2HNO₃ → AgNO₃ NO₂ H₂O

108

AgNO₃ NaCl → AgCl NaNO₃

143.5

143.5 gm of silver chloride would be precipitated by 108 g of silver.

or 14.35 g of silver chloride would be precipitated 10.8 g of silver.

Q 11.34 g of silver coin contain 10.8 g of pure silver.

100 g of silver coin contain × 100 = 95.2 %.

Solved Subjective 

Ex.1 Calculate the following for 49 gm of H₂SO₄

(a) moles (b) Molecules (c) Total H atoms (d) Total O atoms

(e) Total electrons

Sol. Molecular wt of H₂SO₄ = 98

(a) moles =  =

(b) Since 1 mole = 6.023 × 10²³ molecules.

 = 6.023 × 10²³ × molecules = 3.011 × 10²³ molecules

(c) 1 molecule of H₂SO₄ Contains 2 H atom

3.011 × 10²³ of H₂SO₄ contain 2 × 3.011 × 10²³ atoms = 6.023 × 10²³ atoms

(d) 1 molecules of H₂SO₄ contains 4 O atoms

3.011 × 10²³ molecular of H₂SO₄ contains = 4 × 3.011 × 10²³ = 12.044 × 10²³

(e) 1 molecule of H₂SO₄ contains 2H atoms 1 S atom 4 O atom

this means 1 molecule of H₂SO₄ Contains (2 16 4 × 8) e^-

So 3.011 × 10²³ molecules have 3.011 × 10²³ × 50 electrons = 1.5055 × 10²⁵ e^-

Ex.2 Calculate the total ions & charge present in 4.2 gm of N^-3

Sol. mole =  =  = 0.3

total no of ions = 0.3 × N_A ions

total charge = 0.3 N_A × 3 × 1.6 × 10^-19

= 0.3 × 6.023 × 10²³ × 3 × 1.6 × 10^-19 , = 8.67 × 10⁴ C Ans. 

Ex.3 Find the total number of iron atom present in 224 amu iron.

Sol. Since 56 amu = 1 atom
therefore 224 amu =  × 224 = 4 atom Ans. 

Ex.4 A compound containing Ca, C, N and S was subjected to quantitative analysis and formula mass determination. A 0.25 g of this compound was mixed with Na₂CO₃ to convert all Ca into 0.16 g CaCO₃. A 0.115 gm sample of compound was carried through a series of reactions until all its S was changed into SO₄^2- and precipitated as 0.344 g of BaSO₄. A 0.712 g sample was processed to liberated all of its N as NH₃ and 0.155 g NH₃ was obtained. The formula mass was found to be 156. Determine the empirical and molecular formula of the compound.

Sol. Moles of CaCO₃ =  = Moles of Ca

Wt of Ca =  × 40

Mass % of Ca =

Similarly Mass % of S =

Similarly Mass % of N = = 17.9

⇒ Mass % of C = 15.48

Now :

Elements Ca S N C

Mass % 25.6 41 17.9 15.48

Mol ratio 0.64 1.28 1.28 1.29

Simple ratio 1 2 2 2

Empirical formula = CaC₂N₂S₂,

Molecular formula wt = 156 , n × 156 = 156 ⇒ n = 1

Hence, molecular formula = CaC₂N₂S₂

Ex.5 A polystyrne having formula Br₃C₆H₃(C₃ H₈)_n found to contain 10.46% of bromine by weight. Find the value of n. (At. wt. Br = 80)

 Sol. Let the wt of compound is 100 gm & molecular wt is M

Then moles of compound =

Moles of Br = × 3

wt of Br = × 3 × 80 = 10.46

M = 2294.45 = 240 75 44 n , Hence n = 45 Ans. 

Ex.6 A sample of clay was partially dried and then analysed to 50% silica and 7% water. The original clay contained 12% water. Find the percentage of silica in the original sample.

 Sol. In the partially dried clay the total percentage of silica water = 57%. The rest of 43% must be some impurity. Therefore the ratio of wts. of silica to impurity = . This would be true in the original sample of silica.

The total percentage of silica impurity in the original sample is 88. If x is the percentage of silica, ; x = 47.3% Ans.

Ex.7 A mixture of CuSO₄.5H₂O and MgSO₄. 7H₂O was heated until all the water was driven-off. if 5.0 g of mixture gave 3 g of anhydrous salts, what was the percentage by mass of CuSO₄.5H₂O in the original mixture ?

 Sol. Let the mixture contain x g CuSO₄.5H₂O

⇒ = 3 ⇒ x = 3.56

⇒ Mass percentage of CuSO₄. 5H₂O = = 71.25 % Ans. 

Ex .8 367.5 gm KClO₃ (M = 122.5) when heated, How many litre of oxygen gas is proudced at S.T.P.

 Sol. KClO₃ → KCl O₂

Applying POAC on O, moles of O in KClO₃ = moles of O in O₂

3 × moles of KClO₃ = 2 × moles of O₂

3 × = 2 × n, n = ×

 Volume of O₂ gas at S.T.P = moles × 22.4

=  = 9 × 11.2 = 100.8 lit Ans. 

Ex.9 0.532 g of the chloroplatinate of a diacid base on ignition left 0.195 g of residue of Pt. Calculate molecular weight of the base (Pt = 195)

 Sol. Suppose the diacid base is B.

B H₂PtCl₆→ BH₂PtCl₆ → Pt

diacid acid chloroplatinate

base 0.532 g 0.195 g

Since Pt atoms are conserved, applying POAC for Pt atoms,

moles of Pt atoms in BH₂PtCl₆ = moles of Pt atoms in the product

1 × moles of BH₂PtCl₆ = moles of Pt in the product

mol. wt. of BH₂PtCl₆ = 532

From the formula BH₂PtCl₆, we get

mol. wt. of B = mol. wt. of BH₂PtCl₆ - mol. wt. of H₂PtCl₆

= 532 - 410 = 122. Ans. 

Ex.10 10 mL of a gaseous organic compound containing. C, H and O only was mixed with 100 mL of oxygen and exploded under conditions which allowed the water formed to condense. The volume of the gas after explosion was 90 mL. On treatment with potash solution, a further contraction of 20 mL in volume was observed. Given that the vapour density of the compound is 23, deduce the molecular formula. All volume measurements were carried out under the same conditions.

 Sol. C_xH_yO_z O₂ → xCO₂H₂O

10 ml after explosion volume of gas = 90 ml

90 = volume of CO₂ gas volume of unreacted O₂

on treatment with KOH solution volume reduces by 20 ml. This means the volume of CO₂ = 20 ml

the volume of unreacted O₂ = 70 ml

volume of reacted O₂ = 30 ml

V.D of compoud = 23

molecular wt 12x y 16z = 46 ...(1)

from equation we can write

, x  -  = 3

4x y - 2z = 12 ...(2)

& 10x = 20 ⇒ x = 2

from eq. (1) & (2) ; z = 1 & y = 6; Hence C₂H₆O Ans.