redox-and-equivalent-concept-quiz-3

Redox And Equivalent Concept Quiz-3

Dear Readers,

The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry..

Q1. The least count of burette used normally in laboratory is :

•  0.1 mL
•  0.01 mL
•  0.2 mL
•  0.02 mL

Solution
(a) Usually burettes have least count of 0.1 mL.  

Q2.The oxidation number of Mn in MnO₂ is :

•  +4
•  +6
•  +2
•  -4

Solution

(a) a+2 × (-2)=0 ∴a= +4

Q3. Which substance serves as a reducing agent in the following reaction, 

 14H⁺ +Cr₂ O₇^2-+3Ni ⟶2Cr³⁺+7H₂ O+3Ni²⁺ ?  

•  H₂ O
•  Ni
•  H⁺
•  Cr₂O₇^2-

Solution
(b) Ni ⟶Ni²⁺+2e; Ni is oxidized and thus, reductant.

Q4. Titrations in which liberated I₂ is estimated to carry out the volumetric estimations are known as ….titrations.

•  Iodometric
•  Iodimetric
•  Acidimetric
•  Alkalimetric

Solution
(a) It is definition of iodimetric titrations.

Q5.In presence of dil. H₂ SO₄. The equivalent weight of KMnO₄ is :

•  1/5 of its molecular weight
•  1/6 of its molecular weight
•  1/10 of its molecular weight
•  1/2 of its molecular weight

Solution
  (a) Mn⁷⁺+5e ⟶Mn²⁺ ∴ E=M/5  

Q6. How many milligram of iron (Fe²⁺) are equal to 1 mL of 0.1055 N K₂ Cr₂ O₇ equivalent?

•  5.9 mg
•  0.59 mg
•  59 mg
•  59×10^-3 mg

Solution
(a) Meq.of Fe=Meq.of K₂ Cr₂ O₇ 

 w/(56/1) × 1000=1 × 0.1055 

 ∴ w=5.9 × 10^-3 g=5.9 mg  

Q7.What volume of 0.1 M KMnO₄ is needed to oxidise 100 mg of FeC₂ O₄ in acidic solution?

•  4.1 mL
•  8.2 mL
•  10.2 mL
•  4.6 mL

Solution

(a) Meq.of KMnO_4=Meq.of FeC₂ O₄ 

 Fe²⁺ C₂²⁺ O₄ ⟶Fe³⁺+2C⁴⁺ O₂+3e 

 0.1 × 5 × V=(100 × 10^-3/(144/3)× 1000 

 ∴ V=4.1 mL

Q8.The number of equivalent per mole of H₂ S used in its oxidation to SO₂ is :

•  3
•  6
•  4
•  2

Solution
(b) S^2-⟶S⁴⁺ +6e ∴Eq.=mole × 6  

Q9.In the reaction, 2KMnO₄+16HCl ⟶2KCl+2MnCl₂+8H₂ O+5Cl₂, the reduction product is :

•  Cl₂
•  MnCl₂
•  KCl
•  H₂ O

Solution

(b) Mn⁷⁺+5e ⟶Mn²⁺

Q10. What weight of HNO₃ is needed to convert 5 g of iodine into iodic acid according to the reaction, 

I₂+HNO₃ ⟶HIO₃+NO₂+H₂ O ?

•  12.4 g
•  24.8 g
•  0.248 g
•  49.6 g

Solution

(a) Meq.of HNO₃=Meq.of I₂ 

 w/(63/1) × 1000=5/(254/10) × 1000 

 ∴w=12.4 g