Solved Examples: Some Basic Concepts of Chemistry- 2 | Chemistry Class 11 - NEET PDF Download

Q.1. Out of the following the largest number of atoms are contained in
(a) 11 g of CO₂
(b) 4g of H₂ 
(c) 5 g of NH₃ 
(d) 8 g of SO₂
Ans. (b)
Solution.
(a) CO₂ = 11/44 = 0.25 mole
(b) 4/2 = 2 mole
(c) 5/17 = 2 × N_A largest no. of atoms
(d) 8/64 = 1/8

Q.2. The percentage of nitrogen in urea, (NH₂CONH₂₎ is: 
(a) 38.4
(b) 46.6
(c) 59.1
(d) 61.3
Ans. (b)
Solution.

⇒ 60 gm urea = 28 gN
100 g 28 × 100/60 = 46.6%

Q.3. The empirical formula of a compound is CH₂O, if vapour density is 90. Then the molecular formula is 
(a) CH₂O
(b) C₂H₄O₂ 
(c) C₃H₆O₃ 
(d) C₆H₁₂O₆
Ans. (d)
Solution.
⇒ V. D × = mw
 ⇒ 90 × 2 = 180 gm
Empirical formula weight = 30 gm
Value of n = 180/30 = 6
C₆H₁₂O₆

Q.4. An organic compound having carbon and hydrogen has 80% carbon. The empirical formula of the hydrocarbon is 
(a) CH₄
(b) CH₃
(c) CH₂
(d) CH
Ans. (b)
Solution.
C = 80%, h = 20%
%C = 80/12, %H = 20/1
= 6.66 = 20
= 6.66/6.66 = 20/6.66 = 3 {least Ratio}
1 : 3
C₁H₃ = CH₃

Q.5. A compound contains 90% C and 10% H. The empirical formula of the compound is 
(a) C₈H₁₀
(b) C₁₅ H₃₀
(c) C₃H₄
(d) C₁₅H₃₂
Ans. (c)
Solution. 
%C = 90/12, %H = 10/1
= 7.5/7.5 = 10/7.5 = 1.33
= 1
C₃H₄

Q.6. An organic compound contains C = 50% and H = 9.25%. Its empirical formula is 
(a) C₃H₆
(b) C₃H₇O₂
(c) C₂H₄O
(d) C₄H₈O
Ans. (b)
Solution.
C% = 50/12 = 4.1666/2.546 = 1.636 × 2
%H = 9.25%/1 = 9.25/2.546 = 2.546/2.546 = 3.633 × 2
O% = 40.76/16 = 2.546/2.546 = 1 × 2
Ch₃H₇O₂

Q.7. A metal M having an atomic weight of 197 yields a chloride containing 35.1% chlorine.
The empirical formula of the compound is 
(a) MCl₃ 
(b) MCl
(c) MCl₂ 
(d) MCl₄
Ans. (a)
Solution.
Eq. weight of metal chloride = % of metal/% chloride × 35.5 = 64.9/35.1 × 35.5 = 64.9
Eq = 197/V_F 
= 64.9 = 197/V_f = 197/64.9
V_F = 3
M⁺³  Cl^⊖
MCL₃

Q.8. When 1 gram CaCO₃ is dissolved in excess dilute acid the volume of CO₂ evolved at STP will be 
(a) One litre
(b) 224 ml
(c) 22.4 litre
(d) 2.24 litre
Ans. (b)
Solution.
CaCO₃ ⟶ Cao + CO₂
1 gm  -------- 44 × 10^-2 gm
44 g  -----  22.4 litres
44 × 10^-2 g ------ = 0.224 litre = 224 ml

Q.9. 8 gram of sulphur is completely burnt in a large excess of oxygen, the volume in litres of SO₂ formed as reduced to STP is 
(a) 5.6
(b) 8.0
(c) 11.2
(d) 16.0
Ans. (a)
Solution.
S + O₂ → SO₂
32 g ------- 64 g = 1 mole = 22.4 liter
8g ----- 64 x 8/32 = 16 g
mole = 16/64 = 1/4 mole
5.6 litre value

Q.10. 0.01 mole of iodoform (CHI₃) reacts with Ag powder to produce a gas whose volume at NTP is 
(a) 224 ml
(b) 112 ml
(c) 336 ml
(d) 1120 ml
Ans. (b)
Solution.
2CHl₃ + 6Ag ⟶ H − C ≡ C − H + 6 Agl
2 -------- 1 mole
10^-2 --------- ?
10^-2 × 1/2 = 0.5  × 10^-2
= 5 × 10^-3
∴ 1 mole ------ 22.4 literes
5 × 10^-3 ------- 22.4 × 5 × 10^-3
= 112 × 10^-3
112 ml

Q.11. Molarity of liquid HCl with density equal to 1.17 g/ cc is 
(a) 36.5
(b) 18.25
(c) 32.05
(d) 4.65
Ans. (c)
Solution.
d = M/v
1 cc = 1 ml
1000 ml = 1.17 × 1000 = 1170 g/36.5  = 32.05

Q.12. How many milliliters (mL) of 1 M H₂SO₄ solution are required to neutralize 10 mL of 1 M NaOH solution? 
(a) 2.5 mL
(b) 5.0 mL
(c) 10.0 mL
(d) 20.0 mL
Ans. (b)
Solution.
Neutralization
Acid = Base (milliequivalent)
N₁V₁ = N₂V₂
⇒ 1 × V = 10 × 1 (N = m × VF)
V = 5ml

Q.13. The No. of moles of barium carbonate which contain 1.5 moles of oxygen atoms is 
(a) 1.0 mole
(b) 1.5 mole
(c) 0.5 mole
(d) 2.0 mole
Ans. (c)
Solution.
Baco₃ → 1 mole BaCo₃  
→ 1 mole Ba
→ 1 mole C
→ 3 mole (O)
3 mole oxygen - 1 mole BaCo₃
1.5 mole oxygen - 1.5/3 = 0.5 mole BaC0₃

Q.14. A molecule of Haemoglobin contains 0.33% of iron by weight. The molecular weight of Haemoglobin is 67200. The number of iron atoms (At. wt. = 56) present in one molecule of Haemoglobin is 
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (d)
Solution.
Mass of Fe in 1 mole of  Hb = ?
100 gm Hb – 0.33 gm Fe
1 mole_(hb) = 67200g - 0.33 × 67200/100
= 221.76 gm
Mole of Fe = 221.76/56 = ≅ 3.9 ≅ 4
= 4N_A atom of Fe
Na molecules of Hb – 4N_A atom of Fe
1 mole - 4N_A/N_A
= 4 atoms of Fe

Q.15. 48 gram of Mg contains the same number of atoms as 160 gram of another element. The atomic mass of the element is 
(a) 24
(b) 320
(c) 80
(d) 40
Ans. (c)
Solution.
mole of mg = 48/24 = 2 mole = 2 × N_A
(A) = 160/2 = 80

Q.16. The number of moles of oxygen in one litre of air (21% oxygen by volume) at STP would be 
(a) 0.186 mole
(b) 0.21 mole
(c) 2.10 mole
(d) 0.0093 mole
Ans. (d)
Solution.
In 100 ml air ---------- 21 ml oxygen
1000 ml air ----------- 21 × 1000/100 = 210 ml
22400 ml – 1 mole oxygen
210 ml - 210 × 1/22400 = 0.009375 mole oxygen

Q.17. What is correct for 10 g of CaCO₃? 
(a) It contains 1 g atom of carbon
(b) It contains 0.3 g atoms of oxygen
(c) It contains 12 g of calcium
(d) It refers to 0.1 g equivalent of CaCO₃.
Ans. (b)
Solution.
10/100 = 0.1 mole CaCO₃
1 mole CaCO₃ →1 mole Ca
→1 mole C
→3 mole
∴ 0.1 mole CaCO₃ → 0.1 Ca
→0.1C
→0.3(oxygen)

Q.18. If isotopic distribution of C-12 and C-14 is 98% and 2% respectively, then the number of C-14 atoms in 12 g of carbon is 
(a) 1.032 × 10²² 
(b) 3.01 × 10²²
(c) 5.88 × 10²³
(d) 6.02 × 10²³
Ans. (a)
Solution.
In 100 gram of carbon C-14 is=2 gram
12 gm = 2 × 12/100 = 0.24 gram
Mole of C-14 = 0.24/14 =o.017
Total no. of atoms C-14 = 0.017 × NA
= 0.0017 × 6.023 × 10²³
= 0.102 × 10²³
= 1.02 × 10²² atom

Q.19. The number of moles of water present in 100g of water of 90% purity is 
(a) 3
(b) 18
(c) 5
(d) 5.5
Ans. (c)
Solution.
100% purity = 100/18 mole H₂O
90% = 100 × 90/18 × 100
5 mole of H₂O

Q.20. Which of the gases contains the same number of molecules as that of 16 grams of oxygen 
(a) 16g of O₃ 
(b) 32g of SO₂ 
(c) 16g of SO₂ 
(d) All
Ans. (b)
Solution.
16/16 = 1 mole N_A atom = N_A/2 = molecals
(a) 6/48 = 1/2 × N_A
(b) 32/64 = 1/2
(c) 16/4 = 1/4

Q.21. The correct arrangement of the following in order of increasing mass is
(i) N₂ molecule 
(ii) oxygen atom 
(iii) 1 Avogram 
(iv) 1 × 10^-10 gram atom of copper 
(a) I > II > III > IV
(b) III < II < I < IV
(c) I > III > II > IV
(d) IV > I > II > III
Ans. (d)
Solution.
(a) N₂ mole = 28 gm
(b) O atom = 16 gm
(c) 1/6 × 10 - 23 gm
(d) 63.5
1 mole of Cu = 63.5 gm
1 × 10^-10 mol = 63.5 × 10^-10 gm

Q.22. 0.5 mole of a gas (Mol. Wt. 80) occupies 11.2 litres at STP. The volume occupied by 0.25 mole of a lighter gas (Mol. Wt. 20) at STP will be
(a) 11.2 lit
(b) 5.6 lit
(c) 8 lit
(d) 22.4 litre
Ans. (b)
Solution.
0.5 mole - 11.2 litre
0.25 mole → 11.2 × 0.25/0.5 = 5.6 litre

Q.23. The molecular weight of an unknown substance is found to 24000. If it contains 0.2% Magnesium, then the number of Magnesium atoms that can be present in a molecule of it is 
(a) 1
(b) 2
(c) 4
(d) 10
Ans. (b)
Solution.
100 gm-------2 gm Mg
2400--------2 × 2400/100 = 48 g
Mole of mg = 48/24 = 2 mole
Total atom of Mg = 2 × N_A
NA molecules of unknown sub = 2N_A
0.1 mole = 2 × N_A/N_A = 2

Q.24. The percentage of magnesium in chlorophyll is 2.68%. The number of magnesium atoms in 2 gram of chlorophyll is 
(a) 1.34 × 10²¹ 
(b) 1.34 × 10^–21 
(c) 1.35 × 10^–24 
(d) 1.35 × 10²⁴
Ans. (a)
Solution.
100 g – 2.68 g Mg
2g - 2.68 × 2/100 = weight of Mg
Mole= 2.68 × 2/100 × 24 = 0.2 × 10^-2
Total atom of Mg = 0.22 × 10^-2 × 6.023 × 10²³
= 1.34 × 10²¹

Q.25. Ordinary water contains one part of heavy water per 6000 parts by weight. The number of heavy water molecules present in a drop of water of volume 0.01 ml is 
(a) 2.5 × 10¹⁶ 
(b) 5 × 10¹⁷ 
(c) 5 × 10¹⁶ 
(d) 7.5 × 10¹⁶
Ans. (c)
Solution.
D₂O = Mw = 20
Mole of D₂O = 0.01/20 = 5 × 10^-4 mole D₂O
6000 parts ----5 × 10^-4 mol D₂O
1 part----- 5 × 10^-4/6000 × N_A
= 5 × 10¹⁶

Q.26. The number of molecules present in a drop of water. If its volume is 0.05 ml are
(a) 1.66 × 10²¹
(b) 1.60 × 10²²
(c) 1.66 × 10²³ 
(d) 1.60 × 10²⁴
Ans. (a)
Solution.
H₂O density = gm/ml =1 ml=1 gm
⇒ mole H₂O = 0.05/18 × N_A
= 0.05/18 × 6.023 × 10²³
= 1.66 × 10²¹

Q.27. Which of the following will not have a mass of 10 g? 
(a) 0.1 mol CaCO₃
(b) 1.51 × 10²³ Ca²⁺ ions
(c) 0.16 mol of Cl^-ions
(d) 7.525 × 10²² Br atom.
Ans. (c)
Solution.
(a) CaCO₃ = 0.1 mol = 10g
(b) N_A Ca²⁺- 408  1.51 × 10²³ - 1.51 × 10²³/6 × 10²³ = 1/4 × 40 = 10 g
(c) 1 mole Cl =35.5  0.16 = ? 0.16 × 35.5 = 5.68 gm × 2 ≅ 10
(d) 7.525 × Br atom
1 mole – 6.023× 10²³
?   ---   7.525 × 10²² 
⟹ 7.525 × 10²²/6.2023× 10²³ = 1.2 × 10^-1

Q.28. x L of N₂ at STP contains 3 × 10²² molecules. The number of molecules in x/2 L of ozone at STP will be 
(a) 3 × 10²² 
(b) 1.5 × 10²² 
(c) 1.5 × 10²¹ 
(d) 1.5 × 10¹¹
Ans. (b)
Solution.
xLN₂ at stp → 3 x 10²² molecules

∴ volume occupied by gas at S.T.P doesn’t depend on nature of gas.

Q.29. 10 ml of a gaseous hydrocarbon combustion gives 40 ml of CO₂ and 50 ml of H₂O vapour under the same condition. The hydrocarbon is 
(a) C₄H₆ 
(b) C₆H₁₀ 
(c) C₄H₈ 
(d) C₄H₁₀
Ans. (d)
Solution.
_XH_y + 0₂[x + y/4] C0₂ → + H₂O
Volume of CO₂ formed 
10x = 40ml CO₂  
X = 4 
X = c = 4 
C₄H₁₀ 
volume of H₂O
10 y/2 = 50ml H₂O
y = 100/4 = 10
y = H = 1
OR
P.O.A.C method  
P.O.A.C for carbon C atom
10x = 40
X = 4
C₄H₁₀
D.O.A.C for H atom
10y/2 = 50
y = 100/10
y = 10

Q.30. 15 ml hydrocarbon requires 45 ml of O₂ for complete combustion and 30 ml of CO₂ is formed. The formula of the hydrocarbon is 
(a) C₃H₆ 
(b) C₂H₆
(c) C₄H₁₀
(d) C₂H₄
Ans. (d)
Solution.
C_xH_y + 0₂ → C0₂ + H₂O
x + Y/4    x     y
(1) volume of CO₂ formed
15x = 30 ml of CO₂
X = 2
(2) The volume of O₂ required 15x + Y/4 = 45 ml of O₂
⟹ 30+15Y/4 = 45
⟹ 120+15y/4 = 45
⟹ 120+15y = 180
⟹ 15y = 180-120
15 y = 60
Y = 60/15 = 4
C₂H₄ 
x = 2   y = 4

Q.31. Complete combustion of a sample of a hydrocarbon gives 0.66 2 grams of CO₂ and 0.362 grams of H₂O the formula of a compound is 
(a) C₃H₈ 
(b) CH₄ 
(c) C₂H₆ 
(d) C₂H₄
Ans. (a)
Solution.
hydrocarbon+ O₂ → CO₂ + H₂O



Q.32. The simplest formula of a compound containing 50% of element x (atomic weight = 10) and 50% of element y (atomic weight = 20) is 
(a) xy
(b) x₂y
(c) xy₂ 
(d) x₂y₃
Ans. (b)
Solution.
%x = 50/10
x = 5/2.5
x = 2
y% = 50/20
y = 2.5/2,5
y = 1
x₂y

Q.33. The empirical formula of an organic compound is CH. 6.023 × 10²² molecules of same organic compound weigh 7.8 g. The molecular formula is
(a) C₂H₂ 
(b) C₆H₆ 
(c) C₂H₄ 
(d) None
Ans. (b)
Solution.
empirical formula weight = 12+1 = 13g
6 × 10²²− 7.8g
6 × 10²³   -  ?
= 7.8 × 6 × 10²³/6 × 10²²
= 78 gm
n = 78/13 = 6
C_nH_n = C₆H₆

Q.34. An organic compound contains C = 21.56%, H = 4.56% and Br = 73.36%. Its molecular weight is 109. Its molecular formula is 
(a) C₂H₅Br
(b) C₃H₇Br
(c) C₄H₈Br
(d) C₆H₆Br
Ans. (a)
Solution.
C% = 21.56/12 = 1.7/0.917 = 1.853 ≅ 2
H% = 4.56/1 = 4.56/0917 = 4.97 ≅ 5
Br = 73.36/80 = 0.917/0.917 = 1
n = log/log = n
e.f = m.f
= C₂H₅Br

Q.35. 0.078 gram of hydrocarbon occupies 22.4 ml volume at STP. The empirical formula of hydrocarbon is CH. The molecular formula of hydrocarbon is 
(a) C₅H₅ 
(b) C₆H₆ 
(c) C₂H₂ 
(d) C₈H₈
Ans. (b)
Solution.
22.4 ml--- 0.078 gm hydrocarbon

= 78 g
n = 78/13 = 6      CH = 13
12+1
C₆H₆

Q.36. Empirical formula of a compound is CH₂. The mass of one litre of this organic gas is exactly equal to that of one litre of nitrogen. Therefore the molecular formula of the organic gas is 
(a) C₃H₈ 
(b) C₂H₆ 
(c) C₂H₄ 
(d) C₃H₆
Ans. (c)
Solution.
e.f = CH₂    12 + 2 = 14
n= 28/14 = 2
C₂H₄

Q.37. When 1.2 g of carbon is completely burnt in 6 litres of oxygen at STP, the remaining volume of oxygen is 
(a) 3.76 lit
(b) 2.6 lit
(c) 5.8 lit
(d) 37.6 lit
Ans. (a)
Solution.
C + 1/2 O₂ →  (Incomplete combustion)
1C → (1 mole oxygen)
0.1C → 0.1 × 1 = 0.1 mole of oxygen
⟹ 22.4 litre → 1 mole O₂
1 C --------- 22.4 litre 
0.1 ---------- 22.4 litre 
6 − 2.24 = 3.76 liter 

Q.38. 0.5 mole of H₂SO₄ is mixed with 0.2 mole of Ca (OH)₂. The maximum number of moles of CaSO₄ formed is 
(a) 0.5 
(b) 0.2 
(c) 0.4 
(d) 0.25
Ans. (b)
Solution.
H₂SO₄ + Ca(OH)₂ →  CaSO₄ + 2H₂O
d = 0    0.5    0.2    0
t = ?   0.5 - x   0.2 - x   x
x = 0.2
LR
L.R = 0.5/1
L.R = 0.2/1 = 0.2 = x
nx = aa
1 × = 0.2 × 100/100
x = 0.2

Q.39. 70 gram of a sample of magnesite on treatment with excess of HCl gave 11.2 litre of CO₂ at STP. The percentage purify of the sample is
(a) 80
(b) 70
(c) 60
(d) 50
Ans. (d)
Solution.
Magnetic MgCO₃+ Fe]=140gm
Fe + MgCO₃ → MgCl₂ + H₂CO₃
    ↓
H₂O = CO₂
1 mole----------1CO₂
1 CO₂------------1 mol MgCO₃
0.5--------------0.5 mol
70/140 × 100
% purity = 50%

Q.40. When 100 gram of ethylene polymerises to polythene according to the equation nCH₂ = CH₂ → (CH₂ - CH₂)_n - the weight of polythene produced will be 
(a) n/2 gram
(b) 100 gram
(c) 100/n gram
(d) 100n gram
Ans. (b)
Solution.
nCH₂ = CH₂ → (-CH₂ - CH₂ -)_n
100g    100g
n28gm -------- nx28g

Q.41. Air contains 20% by volume of oxygen. The volume of air required for the complete combustion of one litre of methane under the same conditions is 
(a) 2 litre
(b) 4 litre
(c) 10 litre
(d) 0.4 litre
Ans. (c)
Solution.
CH₄ +20₂ → CO₂ +2H₂O
1 litre  2 litre
2 liter O₂ → 1 litre CH₄
20 litre ---- 20/2 = 10 litre CH₄

Q.42. The hydrated Na₂SO₄nH₂O undergoes 56% loss in weight on heating and become anhydrous. The value of n will be 
(a) 5
(b) 3
(c) 7
(d) 10
Ans. (d)
Solution.
Na₂SO₄H₂O → Na₂SO₄+ xH₂O
142 + 18x = 56/100%
Lose in weigh due to H₂O = 18x/142 + 18x = 56/100
⟹ 1800x = 142 × 56 + 18x × 56
⟹ 1800x = 7952 + 1000x
⟹ 792x = 795x
x = 10.004
x = 10

Q.43. 1.25 g of a solid dibasic acid is completely neutralized by 25 mL of 0.25 molar Ba (OH)₂ solution . The molecular mass of the acid is 
(a) 100
(b) 150
(c) 120
(d) 200
Ans. (d)
Solution.
Neutralisation rxn
Solid dibasic acid =
N₁V₁ = N₂V₂
1000 × w/e = N₁V₁

mw = 160 × 1.25
mw = 200

Q.44. 0.126 g of acid requires 20 ml of 0.1 N NaOH for complete neutralization. The equivalent mass of the acid is 
(a) 45
(b) 53
(c) 40
(d) 63
Ans. (d)
Solution.
w/e × 1000 = 20 × 0.1
⟹ 0.126/E × 1000 = 2
E = 0.126 × 1000/2
E = 63

Q.45. The mole fraction of the solute in one mole aqueous solution is 
(a) 0.009
(b) 0.018
(c) 0.027
(d) 0.036
Ans. (b)
Solution.

 


X_A = 1/56.5 = 0.18

Q.46. How many grams of phosphoric acid would be needed to neutralize 100 g of magnesium hydroxide? 
(a) 66.7 g
(b) 252 g
(c) 112 g
(d) 168 g
Ans. (c)
Solution.
⇒ w/E₁ × 1000 = w₂/E₂ × 1000

⇒ 3x/98 = 200/58
⇒ x = 200 × 98 / 58 × 3 = 19600/174
= 112g

Q.47. Normality of solution of FeSO₄. 7H₂O containing 5.56 g / 200 mL which converts to ferric from in a reaction is  (Fe = 56, s = 32, O = 16, H = 1) 
(a) 1
(b) 0.1
(c) 0.01
(d) 10
Ans. (b)
Solution.
Molarity of FeSO₄7H₂O
156 + 126 = 278g mw
Mole = 5.56/278g mol
200-----5.56/278
1000------------?  
5.56 × 1000/278 × 200 × 10^-2
= 20/200 = 0.1
Fe²⁺ → Fe³⁺  VF----1
N = M_XV_F
N = 0.1 × 1
N = 0.1N

Q.48. 100 g of a sample of HCl solution of relative density 1.17 contains 31.2 g of HCl. What volume of this HCl solution will be required to neutralize exactly 5 litres of N/20 KOH solution?
(a) 25 ml
(b) 29.2 ml
(c) 34.2 ml
(d) 250 ml
Ans. (a)
Solution.
D = m/v
1.74 × g/m =100/V
V = 100/1.17
V = 85.4m volume of solution
⟹ 85.4 - 31.2/36.5 = 0.84
1000ml -31.2 × 1000/36.5 × 85.4
M = 10m
N= m × V_F
N=10N----1
Molarity of HCl solution = 10m
Normality = molarity × V_F
= 10 ×1
= 10N
N₁V₁ = N₂V₂
⟹ 10 × V_litre  = 5/10
V_litre = 5/100 = 0.025 litre
= 25 ml

Q.49. 300 ml of 1 M HCI and 100 ml of 1 M NaOH are mixed. The chloride ion concentration in the resulting solution is 
(a) 1 M
(b) 0.5 M
(c) 0.75 M
(d) 0.25 M
Ans. (c)
Solution.
HCL ⇋ H⁺ + Cl^-
0.3       0.3       0.3
NaOH ⇋ Na⁺ + OH
0.1       0.1   0.1
Cl^- = 0.3/0.4 = 0.75M
Cl^- = c0ncentration of Cl^-1/total volume

Q.50. 200 ml of 1 M H₂SO₄, 300 ml 3 M HCI and 100 ml of 2 M HCI is mixed and made up to 1 litre. The proton concentration in the resulting solution is 
(a) 1.25 M
(b) 1.5 M
(c) 2.5 M
(d) 0.75 M
Ans. (b)
Solution.
All are acids so it shows additive properties
H⁺ = (N₁V₁+N₂V₂+N₃V₃)/(V₁+V₂+V₃)
400+900+200/1000 = 1500/1000 = 1.5

Q.51. The volume of 0.25 M NaOH to be added to 250 mL of 0.15 M NaOH so that the resultant solution is 0.2 M would be
(a) 250 mL
(b) 350 mL
(c) 450 mL
(d) 550 mL
Ans: a
Solution: Let V be the volume of 0.25 M NaOH solution. Total amount of NaOH after mixing the two solutions is n = V (0.25 mol L^–1) + (0.25 L) (0.15 mol L^–1)
Total volume of the solution = V + 0.25 L
Molarity of the resultant solution

Equating this to 0.2 M, we get

Solving for V, we get V = 0.25 L = 250 mL

Q.52. The mass of H₂O₂ that is completely oxidized by 30.2 g of KMnO₄ (molar mass = 158 g mol^–1) in acidic medium is
(a) 12 g
(b) 15 g
(c) 17 g
(d) 1 g
Ans: c
Solution: The reaction is 2KMnO₄ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O + 2K⁺
2 x 151 g of KMnO₄ reacts completely with 5 x 34 g of H₂O₂
30.2 g of KMnO₄ reacts completely with