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Equivalent Weight - MCQ(Solved), Chemistry (IIT-JEE and AIPMT) PDF Download

Q: 4. How do you calculate equivalent weight?

Ans. The formula to calculate equivalent weight is:Equivalent weight = Atomic weight / ValenceFor example, the equivalent weight of sulfuric acid (H2SO4) is calculated as follows:Equivalent weight = Molecular weight / Acidic hydrogen ionsEquivalent weight of H2SO4 = 98 / 2 = 49So, the equivalent weight of sulfuric acid is 49 g/equivalent.

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1 
 
 
 
EQUIVELENT WEIGHT 
Q.1 How many litres of a 0.5 N solution of an oxidising agent are reduced by 2 litres of a 2.0 N solution 
of a reducing agent? 
 (A) 8  (B) 4   (C) 6   (D) 7  
Q.2 How many grams of I
2
 are present in a solution which requires 40 mL of 0.11 N Na
2
S
2
O
3
 to react 
with it? (I = 127)        S
2
O
3
2 –
 + I
2
 ?? ? S
4
O
6
2 –
 + 2I¯ 
 (A) 12.7 g  (B) 0.558 g   (C) 25.4 g  (D) 11.4 g 
Q.3 What mass of MnO
2
 is reduced by 35 mL of 0.16 N oxalic acid in acid solution ?  
(Mn = 55)            MnO
2
 + H
+
 + H
2
C
2
O
4
 ?CO
2
 + H
2
O + Mn
2+
  
 (A) 8.7 g  (B) 0.24 g   (C) 0.84 g  (D) 43.5 g 
Q.4 Volume of 0.1 M K
2
Cr
2
O
7
 required to oxidise 35 mL of 0.5 M FeSO
4
 solution is - 
 (A) 29 mL  (B) 87 mL     (C) 175 mL  (D) 145 mL 
Q.5 0.116 g of C
4
H
4
O
4 
(A) is neutralised by 0.074 g of Ca(OH)
2 
. Hence, protonic hydrogen (H
+
) in (A) 
will be -  
 (A) 1     (B) 2     (C) 3     (D) 4  
Q.6 A 20.0 mL solution of Na
2
SO
3 
required 30 mL of 0.01 M K
2
Cr
2
O
7 
solution for the oxidation to 
Na
2
SO
4 
. Hence, molarity of Na
2
SO
3 
solution is -  
 (A) 0.015 M    (B) 0.045 M    (C) 0.030 M   (D) 0.0225 M   
Q.7 100 g of a mixture of NaOH and Na
2
SO
4 
is neutralized by 100 mL of 0.5 M H
2
SO
4
. What amount 
of Na
2
SO
4 
is present in the mixture ?  
 (A) 82 g   (B) 96 g    (C) 88 g   (D) 92 g   
Q.8 The number of moles of Cr
2
O
7
2 – 
need to oxidize 0.034 mole of N
2
H
5
+ 
by the reaction,  
   N
2
H
5
+ 
+ Cr
2
O
7
2 – 
— ? N
2 
+ Cr
3+ 
+ H
2
O is -  
 (A) 0.136   (B) 0.272    (C) 0.816  (D) 0.0227  
Q.9 If 200 mL of 0.1 M KHC
2
O
4 
, 300 mL of 0.2 M H
2
C
2
O
4 
and 500 mL of 0.3 M NaCl is  
neutralized by 140 mL of Ca(OH)
2 
. What is its molarity?  
 (A) 1     (B) 0.5    (C) 1.5    (D) 2   
Q.10 The equivalent mass of sodium thiosulphate (Na
2
S
2
O
3
.5H
2
O) in the reaction, (I = 127) 
   2Na
2
S
2
O
3
 + I
2
 ? 2NaI + Na
2
S
4
O
6 
is : 
 (A) 248  (B) 124   (C) 596   (D) 62 
Q.11 One mole of KMnO
4
 on treatment with H
2
SO
4
 produces......equivalent of oxygen – 
  (A) 2   (B) 4   (C) 2.4   (D) 5 
Q.12 The equivalent weight of divalent metal is W. The molecular weight of its chloride is – 
  (A) W + 35.5  (B) W + 71  (C) 2W + 71  (D) 2W + 35.5 
Q.13 The specific heat of an element of atomic weight 32 is likely to be - 
 (A) 0.25 cal/g (B) 0.24 cal/g  (C) 0.20 cal/g  (D) 0.15 cal/g 
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1 
 
 
 
EQUIVELENT WEIGHT 
Q.1 How many litres of a 0.5 N solution of an oxidising agent are reduced by 2 litres of a 2.0 N solution 
of a reducing agent? 
 (A) 8  (B) 4   (C) 6   (D) 7  
Q.2 How many grams of I
2
 are present in a solution which requires 40 mL of 0.11 N Na
2
S
2
O
3
 to react 
with it? (I = 127)        S
2
O
3
2 –
 + I
2
 ?? ? S
4
O
6
2 –
 + 2I¯ 
 (A) 12.7 g  (B) 0.558 g   (C) 25.4 g  (D) 11.4 g 
Q.3 What mass of MnO
2
 is reduced by 35 mL of 0.16 N oxalic acid in acid solution ?  
(Mn = 55)            MnO
2
 + H
+
 + H
2
C
2
O
4
 ?CO
2
 + H
2
O + Mn
2+
  
 (A) 8.7 g  (B) 0.24 g   (C) 0.84 g  (D) 43.5 g 
Q.4 Volume of 0.1 M K
2
Cr
2
O
7
 required to oxidise 35 mL of 0.5 M FeSO
4
 solution is - 
 (A) 29 mL  (B) 87 mL     (C) 175 mL  (D) 145 mL 
Q.5 0.116 g of C
4
H
4
O
4 
(A) is neutralised by 0.074 g of Ca(OH)
2 
. Hence, protonic hydrogen (H
+
) in (A) 
will be -  
 (A) 1     (B) 2     (C) 3     (D) 4  
Q.6 A 20.0 mL solution of Na
2
SO
3 
required 30 mL of 0.01 M K
2
Cr
2
O
7 
solution for the oxidation to 
Na
2
SO
4 
. Hence, molarity of Na
2
SO
3 
solution is -  
 (A) 0.015 M    (B) 0.045 M    (C) 0.030 M   (D) 0.0225 M   
Q.7 100 g of a mixture of NaOH and Na
2
SO
4 
is neutralized by 100 mL of 0.5 M H
2
SO
4
. What amount 
of Na
2
SO
4 
is present in the mixture ?  
 (A) 82 g   (B) 96 g    (C) 88 g   (D) 92 g   
Q.8 The number of moles of Cr
2
O
7
2 – 
need to oxidize 0.034 mole of N
2
H
5
+ 
by the reaction,  
   N
2
H
5
+ 
+ Cr
2
O
7
2 – 
— ? N
2 
+ Cr
3+ 
+ H
2
O is -  
 (A) 0.136   (B) 0.272    (C) 0.816  (D) 0.0227  
Q.9 If 200 mL of 0.1 M KHC
2
O
4 
, 300 mL of 0.2 M H
2
C
2
O
4 
and 500 mL of 0.3 M NaCl is  
neutralized by 140 mL of Ca(OH)
2 
. What is its molarity?  
 (A) 1     (B) 0.5    (C) 1.5    (D) 2   
Q.10 The equivalent mass of sodium thiosulphate (Na
2
S
2
O
3
.5H
2
O) in the reaction, (I = 127) 
   2Na
2
S
2
O
3
 + I
2
 ? 2NaI + Na
2
S
4
O
6 
is : 
 (A) 248  (B) 124   (C) 596   (D) 62 
Q.11 One mole of KMnO
4
 on treatment with H
2
SO
4
 produces......equivalent of oxygen – 
  (A) 2   (B) 4   (C) 2.4   (D) 5 
Q.12 The equivalent weight of divalent metal is W. The molecular weight of its chloride is – 
  (A) W + 35.5  (B) W + 71  (C) 2W + 71  (D) 2W + 35.5 
Q.13 The specific heat of an element of atomic weight 32 is likely to be - 
 (A) 0.25 cal/g (B) 0.24 cal/g  (C) 0.20 cal/g  (D) 0.15 cal/g 
 
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2 
 
Q.14 In a compound A
x 
B
y
  
 (A) mole of A = mole of B = mole of A
x
 B
y
 
 (B) equiv of A = equiv of B = equiv of A
x 
B
y
 
 (C) y × moles of A = y × mole of B = (x + y) mole of A
x 
B
y 
 
(D) y × mole of A = y × mole of B 
Q.15 Equivalent mass of a bivalent metal is 32.7. Molecular mass of its chloride is - 
 (A) 68.2  (B) 103.7   (C) 136.4  (D) 166.3 
Q.16 What volume of 0.05 M  in acid medium in needed for complete oxidation of 200 mL of 0.6 M 
FeC
2
O
4
 solution ?  
 (A) 0.6 L  (B) 1.2 L   (C) 2.4 L  (D) 3.6 L 
Q.17 Milli equivalents of a solute in a solution can be given by -  
 (A) Meq. = M × V
in mL 
(B) Meq. = N × V
in mL 
(C) Meq. =  × 1000
 
(D) (B)  and (C) both
 
Q.18 How many milli litre of 0.5 N SnCl
2
 solution will reduce 600 mL of 0.1 N HgCl
2
 to Hg
2
Cl
2
 - 
 (A) 120 mL  (B) 60 mL  (C) 30 mL  (D) 240 mL 
Q.19 KMnO
4
 reacts with oxalic acid according to the reaction : 
 2KMnO
4
 + + 16H
+
 ? 2Mn
2+
 + 10CO
2
 + 8H
2
O 
 Then, 20 mL of 0.1 M KMnO
4
 is equivalent to: 
 (A) 30 mL of 0.5 M C
2
H
2
O
4
 (oxalic acid)  (B) 50 mL of 0.1 M C
2
H
2
O
4
 (oxalic acid) 
 (C) 20 mL of 0.5 M C
2
H
2
O
4
 (oxalic acid)  (D) 10 mL of 0.1 M C
2
H
2
O
4
 (oxalic acid) 
Q.20 The number of moles of Cr
2
O
7
2 –
 required to oxidize one mole of FeC
2
O
4
 in acidic medium - 
 (A) 0.5   (B) 0.6   (C) 0.2   (D) 0.3 
Q.21 The number of moles of MnO
4
–
 ions needed to react with one mole of sulphite ions in acidic 
medium are - 
 (A) 0.8    (B) 0.4   (C) 0.6   (D) 0.5 
Q.22 When the ion Cr
2
O
7
2 –
 acts as an oxidant in acidic aqueous solution the ion Cr
3+
 is formed. How 
many moles of Sn
2+
 would be oxidised to Sn
4+
 by one mole of Cr
2
O
7
2 –
 ions - 
 (A) 2/3   (B) 3/2   (C) 2   (D) 3 
Q.23 What mass of Na
2
S
2
O
3
.5H
2
O is needed to make 500 cm
3
 of 0.200 N solution for the  
reactions  2S
2
O
3
2 –
 + I
2
 ? S
4
O
6
2 –
 + 2I
–
 
 (A)  4.8 g  (B) 14.8 g  (C) 24.8 g  (D) 16.8 g 
Q.24 The mass of  that is completely oxidised by 31.6 g of  (Molar mass  = 158 g mol
-1
) in acidic 
medium is - 
 (A) 12 g  (B) 15 g   (C) 17 g  (D) 1 g 
Q.25 The mass of sodium bromate (molar mass 151 g mol
-1
) required to prepare 50 mL of 0.6N solution 
based on the reaction   is  
 (A) 0.642 g  (B) 0.755 g   (C) 0.814 g  (D) 1.51 g 
Q.26 If in a reaction  is reduced to NO, the mass of HNO
3
 absorbing one mole of electrons would be - 
 (A) 12.6 g  (B) 21.0 g   (C) 31.5 g  (D) 63.0 g 
Q.27 If in a given reaction  is reduced to , the mass of  absorbing one mole of electrons would be - 
 (A) 31.0 g  (B) 12.4 g   (C) 6.29 g  (D) 7.75 g 
Q.28 For the reaction  what is the equivalent weight of Br
2
 
 (A) M/8   (B) M/5   (C) 3M/5  (D) 5M/8. 
Q.29 Equivalent weight of iron in  Fe2O3  would be -  (Fe = 56) 
 (A) 28  (B) 56   (C) 18.6  (D) 112. 
Q.30 For the reaction,       if one mole of  oxidises 1.67 moles of to , then 
the value of x in the reaction is - 
O H 3 Br e 6 H 6 BrO
2 3
? ? ? ?
? ? ? ?
O H BrO Br OH Br
2 3 2
? ? ? ?
? ? ?
? ? ? ?
? ? ?
2
3 4
x
Mn MO MnO M
Page 3


 
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rd 
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1 
 
 
 
EQUIVELENT WEIGHT 
Q.1 How many litres of a 0.5 N solution of an oxidising agent are reduced by 2 litres of a 2.0 N solution 
of a reducing agent? 
 (A) 8  (B) 4   (C) 6   (D) 7  
Q.2 How many grams of I
2
 are present in a solution which requires 40 mL of 0.11 N Na
2
S
2
O
3
 to react 
with it? (I = 127)        S
2
O
3
2 –
 + I
2
 ?? ? S
4
O
6
2 –
 + 2I¯ 
 (A) 12.7 g  (B) 0.558 g   (C) 25.4 g  (D) 11.4 g 
Q.3 What mass of MnO
2
 is reduced by 35 mL of 0.16 N oxalic acid in acid solution ?  
(Mn = 55)            MnO
2
 + H
+
 + H
2
C
2
O
4
 ?CO
2
 + H
2
O + Mn
2+
  
 (A) 8.7 g  (B) 0.24 g   (C) 0.84 g  (D) 43.5 g 
Q.4 Volume of 0.1 M K
2
Cr
2
O
7
 required to oxidise 35 mL of 0.5 M FeSO
4
 solution is - 
 (A) 29 mL  (B) 87 mL     (C) 175 mL  (D) 145 mL 
Q.5 0.116 g of C
4
H
4
O
4 
(A) is neutralised by 0.074 g of Ca(OH)
2 
. Hence, protonic hydrogen (H
+
) in (A) 
will be -  
 (A) 1     (B) 2     (C) 3     (D) 4  
Q.6 A 20.0 mL solution of Na
2
SO
3 
required 30 mL of 0.01 M K
2
Cr
2
O
7 
solution for the oxidation to 
Na
2
SO
4 
. Hence, molarity of Na
2
SO
3 
solution is -  
 (A) 0.015 M    (B) 0.045 M    (C) 0.030 M   (D) 0.0225 M   
Q.7 100 g of a mixture of NaOH and Na
2
SO
4 
is neutralized by 100 mL of 0.5 M H
2
SO
4
. What amount 
of Na
2
SO
4 
is present in the mixture ?  
 (A) 82 g   (B) 96 g    (C) 88 g   (D) 92 g   
Q.8 The number of moles of Cr
2
O
7
2 – 
need to oxidize 0.034 mole of N
2
H
5
+ 
by the reaction,  
   N
2
H
5
+ 
+ Cr
2
O
7
2 – 
— ? N
2 
+ Cr
3+ 
+ H
2
O is -  
 (A) 0.136   (B) 0.272    (C) 0.816  (D) 0.0227  
Q.9 If 200 mL of 0.1 M KHC
2
O
4 
, 300 mL of 0.2 M H
2
C
2
O
4 
and 500 mL of 0.3 M NaCl is  
neutralized by 140 mL of Ca(OH)
2 
. What is its molarity?  
 (A) 1     (B) 0.5    (C) 1.5    (D) 2   
Q.10 The equivalent mass of sodium thiosulphate (Na
2
S
2
O
3
.5H
2
O) in the reaction, (I = 127) 
   2Na
2
S
2
O
3
 + I
2
 ? 2NaI + Na
2
S
4
O
6 
is : 
 (A) 248  (B) 124   (C) 596   (D) 62 
Q.11 One mole of KMnO
4
 on treatment with H
2
SO
4
 produces......equivalent of oxygen – 
  (A) 2   (B) 4   (C) 2.4   (D) 5 
Q.12 The equivalent weight of divalent metal is W. The molecular weight of its chloride is – 
  (A) W + 35.5  (B) W + 71  (C) 2W + 71  (D) 2W + 35.5 
Q.13 The specific heat of an element of atomic weight 32 is likely to be - 
 (A) 0.25 cal/g (B) 0.24 cal/g  (C) 0.20 cal/g  (D) 0.15 cal/g 
 
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rd 
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E-mail: query.expositors@gmail.com; Web: www.expositorsacademy.in 
2 
 
Q.14 In a compound A
x 
B
y
  
 (A) mole of A = mole of B = mole of A
x
 B
y
 
 (B) equiv of A = equiv of B = equiv of A
x 
B
y
 
 (C) y × moles of A = y × mole of B = (x + y) mole of A
x 
B
y 
 
(D) y × mole of A = y × mole of B 
Q.15 Equivalent mass of a bivalent metal is 32.7. Molecular mass of its chloride is - 
 (A) 68.2  (B) 103.7   (C) 136.4  (D) 166.3 
Q.16 What volume of 0.05 M  in acid medium in needed for complete oxidation of 200 mL of 0.6 M 
FeC
2
O
4
 solution ?  
 (A) 0.6 L  (B) 1.2 L   (C) 2.4 L  (D) 3.6 L 
Q.17 Milli equivalents of a solute in a solution can be given by -  
 (A) Meq. = M × V
in mL 
(B) Meq. = N × V
in mL 
(C) Meq. =  × 1000
 
(D) (B)  and (C) both
 
Q.18 How many milli litre of 0.5 N SnCl
2
 solution will reduce 600 mL of 0.1 N HgCl
2
 to Hg
2
Cl
2
 - 
 (A) 120 mL  (B) 60 mL  (C) 30 mL  (D) 240 mL 
Q.19 KMnO
4
 reacts with oxalic acid according to the reaction : 
 2KMnO
4
 + + 16H
+
 ? 2Mn
2+
 + 10CO
2
 + 8H
2
O 
 Then, 20 mL of 0.1 M KMnO
4
 is equivalent to: 
 (A) 30 mL of 0.5 M C
2
H
2
O
4
 (oxalic acid)  (B) 50 mL of 0.1 M C
2
H
2
O
4
 (oxalic acid) 
 (C) 20 mL of 0.5 M C
2
H
2
O
4
 (oxalic acid)  (D) 10 mL of 0.1 M C
2
H
2
O
4
 (oxalic acid) 
Q.20 The number of moles of Cr
2
O
7
2 –
 required to oxidize one mole of FeC
2
O
4
 in acidic medium - 
 (A) 0.5   (B) 0.6   (C) 0.2   (D) 0.3 
Q.21 The number of moles of MnO
4
–
 ions needed to react with one mole of sulphite ions in acidic 
medium are - 
 (A) 0.8    (B) 0.4   (C) 0.6   (D) 0.5 
Q.22 When the ion Cr
2
O
7
2 –
 acts as an oxidant in acidic aqueous solution the ion Cr
3+
 is formed. How 
many moles of Sn
2+
 would be oxidised to Sn
4+
 by one mole of Cr
2
O
7
2 –
 ions - 
 (A) 2/3   (B) 3/2   (C) 2   (D) 3 
Q.23 What mass of Na
2
S
2
O
3
.5H
2
O is needed to make 500 cm
3
 of 0.200 N solution for the  
reactions  2S
2
O
3
2 –
 + I
2
 ? S
4
O
6
2 –
 + 2I
–
 
 (A)  4.8 g  (B) 14.8 g  (C) 24.8 g  (D) 16.8 g 
Q.24 The mass of  that is completely oxidised by 31.6 g of  (Molar mass  = 158 g mol
-1
) in acidic 
medium is - 
 (A) 12 g  (B) 15 g   (C) 17 g  (D) 1 g 
Q.25 The mass of sodium bromate (molar mass 151 g mol
-1
) required to prepare 50 mL of 0.6N solution 
based on the reaction   is  
 (A) 0.642 g  (B) 0.755 g   (C) 0.814 g  (D) 1.51 g 
Q.26 If in a reaction  is reduced to NO, the mass of HNO
3
 absorbing one mole of electrons would be - 
 (A) 12.6 g  (B) 21.0 g   (C) 31.5 g  (D) 63.0 g 
Q.27 If in a given reaction  is reduced to , the mass of  absorbing one mole of electrons would be - 
 (A) 31.0 g  (B) 12.4 g   (C) 6.29 g  (D) 7.75 g 
Q.28 For the reaction  what is the equivalent weight of Br
2
 
 (A) M/8   (B) M/5   (C) 3M/5  (D) 5M/8. 
Q.29 Equivalent weight of iron in  Fe2O3  would be -  (Fe = 56) 
 (A) 28  (B) 56   (C) 18.6  (D) 112. 
Q.30 For the reaction,       if one mole of  oxidises 1.67 moles of to , then 
the value of x in the reaction is - 
O H 3 Br e 6 H 6 BrO
2 3
? ? ? ?
? ? ? ?
O H BrO Br OH Br
2 3 2
? ? ? ?
? ? ?
? ? ? ?
? ? ?
2
3 4
x
Mn MO MnO M
 
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3 
 
 (A) 5  (B) 3   (C) 2   (D) 1 
Q.31 Amount of oxalic acid required to prepare 250 mL of N/10 solution (Mol. mass of oxalic acid = 126) 
is:        
 (A) 1.5759 g (B) 3.15 g   (C) 15.75 g  (D) 63.0 g 
Q.32 The equivalent mass of phosphoric acid (H
3
PO
4
) is 49. It behaves as ... acid. 
 (A) monobasic  (B) dibasic   (C) tribasic  (D) tetrabasic 
Q.33 What is the normality of 0.3 M H
3
PO
4
 when it undergoes the reaction as 
 H
3
PO
4
 + 2OH¯ ?  + 2H
2
O  
 (A) 0.3 N  (B) 0.15 N   (C) 0.60 N  (D) 0.90 N 
Q.34 The value of n in the equation, 
 Cr2O7
2-
 + 14H
+
 + ne¯ ? 2Cr
3+
 + 7H
2
O is: 
 (A) 2  (B) 3   (C) 4   (D) 6 
Q.35 Mixture of 1mol MgCl
2
 and 2 mol H
2
SO
4
 can be neutralised by: 
 (A) 1 mol KOH (B) 2 mol Ca(OH)
2
 (C) 2 mol NaOH  (D) 2 mol KOH 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Qus. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. A B B A B B B D B A D C C B C B D A B A
Qus. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
Ans. B D C C B B D C C C A B C D B

FAQs on Equivalent Weight - MCQ(Solved), Chemistry (IIT-JEE and AIPMT)

1. What is equivalent weight in chemistry?

Ans. Equivalent weight in chemistry is defined as the mass of a substance that can combine with or displace one mole of hydrogen ions (H+) or any other equivalent of a given chemical species. It is also known as gram equivalent weight.

2. How is equivalent weight related to valency?

Ans. Equivalent weight is related to valency as it is calculated by dividing the atomic weight of an element by its valency. This means that equivalent weight is directly proportional to the valency of an element.

3. What is the difference between normality and molarity?

Ans. Normality and molarity are both units of concentration in chemistry. However, normality is defined as the number of equivalents of solute per liter of solution, while molarity is defined as the number of moles of solute per liter of solution. Normality takes into account the valence of the solute, while molarity does not.

4. How do you calculate equivalent weight?

Ans. The formula to calculate equivalent weight is: Equivalent weight = Atomic weight / Valence For example, the equivalent weight of sulfuric acid (H2SO4) is calculated as follows: Equivalent weight = Molecular weight / Acidic hydrogen ions Equivalent weight of H2SO4 = 98 / 2 = 49 So, the equivalent weight of sulfuric acid is 49 g/equivalent.

5. What is the significance of equivalent weight in redox reactions?

Ans. Equivalent weight plays a significant role in redox reactions as it helps to determine the amount of one substance that can be oxidized or reduced by another substance. In a redox reaction, the reducing agent donates electrons and undergoes oxidation, while the oxidizing agent accepts electrons and undergoes reduction. The equivalent weight of each substance involved in the reaction is used to balance the equation and determine the stoichiometry of the reaction.

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