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Solved Examples  Some Basic Concepts of Chemistry, CBSE, Class 11, Chemistry
Solved Objective
Ex.1 8 litre of H_{2} and 6 litre of Cl_{2} 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)
H_{2} Cl_{2} → 2 HCl
Volume before reaction 8 lit 6 lit 0
Volume after reaction 2 0 12
Volume after reaction
= Volume of H_{2} left Volume of HCl formed = 2 12 = 14 lit
Ex.2 Naturally occurring chlorine is 75.53% Cl^{35} which has an atomic mass of 34.969 amu and 24.47% Cl^{37} 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^{23}gm (B) 16.93 × 10^{23} gm
(C) 17.93 × 10^{23} 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 CO_{2} 
(A) 22 gm (B) 44 gm (C) 88 gm (D) None of these.
Sol. (C)
Q N molecules of CO_{2} has molecular mass = 44.
2N molecules of CO_{2} has molecular mass = 44 x 2 = 88 gm.
Ex.7 How many carbon atoms are present in 0.35 mol of C_{6}H_{12}O_{6} 
(A) 6.023 × 10^{23} carbon atoms (B) 1.26 × 10^{23} carbon atoms
(C) 1.26 × 10^{24} carbon atoms (D) 6.023 × 10^{24} carbon atoms
Sol. (C)
1 mol of C_{6}H_{12}O_{6} has = 6 N atoms of C
0.35 mol of C_{6}H_{12}O_{6} has = 6 × 0.35 N atoms of C
= 2.1 N atoms = 2.1 × 6.023 × 10^{23 } = 1.26 × 10^{24} carbon atoms
Ex.8 How many molecules are in 5.23 gm of glucose (C_{6}H_{12}O_{6}) 
(A) 1.65 × 10^{22} (B) 1.75 × 10^{22} (C) 1.75 × 10^{21 }(D) None of these
Sol. (B)
Q 180 gm glucose has = N molecules
5.23 gm glucose has = = 1.75 × 10^{22} molecules
Ex.9 What is the weight of 3.01 × 10^{23} molecules of ammonia 
(A) 17 gm (B) 8.5 gm (C) 34 gm (D) None of these
Sol. (B)
Q 6.023 × 10^{23} molecules of NH_{3} has weight = 17 gm
3.01 × 10^{23 } molecules of NH_{3} has weight
= = 8.50 gm
Ex.10 How many significant figures are in each of the following numbers 
(a) 4.003 (b) 6.023 × 10^{23} (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 × 10^{19} (B) 2.69 × 10^{19} (C) 1.69 × 10^{19} (D) 2.69 × 10^{19}
Sol. (D)
Q 22.4 litre water vapour at STP has
= 6.023 × 10^{23} 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 × 10^{19} years (B) 19.098 years
(C) 19.098 × 10^{9} years (D) None of these
Sol. (C)
Q 10^{6} rupees are spent in 1sec.
6.023 × 10^{23} rupees are spent in = sec
= years , = 19.098 × 10^{9} 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) 10^{4} 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 × 10^{23} = 40 gm
Q 40 gm of element has 1 gm atom.
40 x 10^{3} gm of element has , = 10^{3} gm atom.
Ex.14 Calculate the number of Cl^{} and Ca^{ 2} ions in 222 g anhydrous CaCl_{2} 
(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 CaCl_{2} = 111 g
Q 111 g CaCl_{2} has = N ions of Ca^{ 2}
222g of CaCl_{2} has
= 2N ions of Ca^{ 2}
Also Q 111 g CaCl_{2} has = 2N ions of Cl^{}
222 g CaCl_{2} has = ions of Cl^{}
= 4N ions of Cl^{} .
Ex.15 The density of O_{2} 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 O_{2} gas occupies volume = 1 litre.
32 gm of O_{2} 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 × 10^{23} atoms of carbon (D) 1.12 litre of O_{2} 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 × 10^{23} atoms of C = = 1.99 gm.
(D) Mass of 1.12 litre of O_{2} 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 KClO_{3} → 2KCl 3O_{2}
Mole for reaction 2 2 3
Q 3 × 22.4 litre O_{2} is formed by 2 mol KClO_{3}
11.2 litre O_{2} is formed by = mol KClO_{3}
Ex.18 Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone (CaCO_{3}).
(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
CaCO_{3} = 95 kg
200 kg impure sample has pure CaCO_{3}
= = 190 kg. CaCO_{3} → CaO CO_{2}
Q 100 kg CaCO_{3} gives CaO = 56 kg.
190 kg CaCO_{3} gives CaO = = 106.4 kg.
Ex.19 The chloride of a metal has the formula MCl_{3}. The formula of its phosphate will be
(A) M_{2}PO_{4} (B) MPO_{4} (C) M_{3}PO_{4} (D) M(PO_{4})_{2}
Sol. (B) AlCl_{3} as it is AlPO_{4}
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 AgNO_{3}) 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 2HNO_{3} → AgNO_{3} NO_{2} H_{2}O
108
AgNO_{3} NaCl → AgCl NaNO_{3}
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 H_{2}SO_{4}
(a) moles (b) Molecules (c) Total H atoms (d) Total O atoms
(e) Total electrons
Sol. Molecular wt of H_{2}SO_{4} = 98
(a) moles = =
(b) Since 1 mole = 6.023 × 10^{23} molecules.
= 6.023 × 10^{23} × molecules = 3.011 × 10^{23} molecules
(c) 1 molecule of H_{2}SO_{4} Contains 2 H atom
3.011 × 10^{23} of H_{2}SO_{4} contain 2 × 3.011 × 10^{23} atoms = 6.023 × 10^{23} atoms
(d) 1 molecules of H_{2}SO_{4} contains 4 O atoms
3.011 × 10^{23} molecular of H_{2}SO_{4} contains = 4 × 3.011 × 10^{23} = 12.044 × 10^{23}
(e) 1 molecule of H_{2}SO_{4} contains 2H atoms 1 S atom 4 O atom
this means 1 molecule of H_{2}SO_{4} Contains (2 16 4 × 8) e^{}
So 3.011 × 10^{23} molecules have 3.011 × 10^{23} × 50 electrons = 1.5055 × 10^{25} 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 × 10^{23} × 3 × 1.6 × 10^{19} , = 8.67 × 10^{4} 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 Na_{2}CO_{3} to convert all Ca into 0.16 g CaCO_{3}. A 0.115 gm sample of compound was carried through a series of reactions until all its S was changed into SO_{4}^{2} and precipitated as 0.344 g of BaSO_{4}. A 0.712 g sample was processed to liberated all of its N as NH_{3} and 0.155 g NH_{3} was obtained. The formula mass was found to be 156. Determine the empirical and molecular formula of the compound.
Sol. Moles of CaCO_{3} = = 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 = CaC_{2}N_{2}S_{2},
Molecular formula wt = 156 , n × 156 = 156 ⇒ n = 1
Hence, molecular formula = CaC_{2}N_{2}S_{2}
Ex.5 A polystyrne having formula Br_{3}C_{6}H_{3}(C_{3} H_{8})_{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 CuSO_{4}.5H_{2}O and MgSO_{4}. 7H_{2}O was heated until all the water was drivenoff. if 5.0 g of mixture gave 3 g of anhydrous salts, what was the percentage by mass of CuSO_{4}.5H_{2}O in the original mixture ?
Sol. Let the mixture contain x g CuSO_{4}.5H_{2}O
⇒ = 3 ⇒ x = 3.56
⇒ Mass percentage of CuSO_{4}. 5H_{2}O = = 71.25 % Ans.
Ex .8 367.5 gm KClO_{3} (M = 122.5) when heated, How many litre of oxygen gas is proudced at S.T.P.
Sol. KClO_{3} → KCl O_{2}
Applying POAC on O, moles of O in KClO_{3} = moles of O in O_{2}
3 × moles of KClO_{3} = 2 × moles of O_{2}
3 × = 2 × n, n = ×
Volume of O_{2} 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 H_{2}PtCl_{6}→ BH_{2}PtCl_{6} → 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 BH_{2}PtCl_{6} = moles of Pt atoms in the product
1 × moles of BH_{2}PtCl_{6} = moles of Pt in the product
mol. wt. of BH_{2}PtCl_{6} = 532
From the formula BH_{2}PtCl_{6}, we get
mol. wt. of B = mol. wt. of BH_{2}PtCl_{6}  mol. wt. of H_{2}PtCl_{6}
= 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_{x}H_{y}O_{z} O_{2} → xCO_{2}H_{2}O
10 ml after explosion volume of gas = 90 ml
90 = volume of CO_{2} gas volume of unreacted O_{2}
on treatment with KOH solution volume reduces by 20 ml. This means the volume of CO_{2} = 20 ml
the volume of unreacted O_{2} = 70 ml
volume of reacted O_{2} = 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 C_{2}H_{6}O Ans.
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