REDOX AND EQUIVALENT CONCEPT QUIZ-15

REDOX AND EQUIVALENT CONCEPT QUIZ-15

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. Turn bull’s blue is :

•  Fe₃ [Fe(CN)₆]₂
•  K₄ Fe(CN)₆
•  K₃ Fe(CN)₆
•  Na₄ Fe(CN)₆

Solution
It is the formula of turns bull’s blue.

Q2.In alkaline condition KMnO₄ reacts as follows :2KMnO₄+2KOH ⟶2K₂ MnO₄+H₂ O+O. The eq. wt. of KMnO₄ is :

•  52.7
•  158
•  31.6
•  79

Solution
le+Mn⁽⁷⁺⁾ ⟶Mn⁽⁶⁺⁾ ∴E=M/1

Q3.  In an oxidation process for a cell M₁ ⟶M₁⁽ⁿ⁺⁾+ne, the other metal (M₂) being univalent showing reduction takes up the …..electrons to complete redox reaction.

•   (n-1)
•  1
•  n
•  2

Solution
Electrons released at anode = Electrons used at cathode.

Q4. Sulphur has the highest oxidation state in :

•  SO₂
•  SO₃
•  H₂ SO₃
•  H₂ S

Solution
S has +6 ox. no. in SO₃

Q5.How many gram of I₂are present in a solution which requires 40 mL, of 0.11 N Na₂ S₂ O₃ to react with it, S₂ O₃^(2-)+I₂ ⟶S₄ O₆^(2-)+2I^-?

•  12.7 g
•  0.558 g
•  25.4 g
•  11.4 g

Solution
 Meq.of I_2=Meq.of Na₂ S₂ O₃=40 × 0.11 ∴w/(254/2) × 1000=40 × 0.11 w_{(I2 )}=0.558 g

Q6. The most stable oxidation state of chromium is :

•  + 5
•  +3
• +2
•  +4

Solution
It is a fact.

Q7.If equal volumes of 1M KMnO₄ and 1 M K₂ Cr₂ O₇solutions are allowed to oxidise F⁽²⁺⁾ to F⁽³⁺⁾ in acidic medium volume of oxidant required for one mole of F⁽²⁺⁾will be :

•  V_{(KMnO4 )}>V_{(K2 Cr2 O7 )}
•  V_{(KMnO4 )}(K₂ Cr₂ O₇ )
•  (KMnO₄ )=V_{(K2 Cr2 O7)}
•  Nothing can be predicted

Solution

Q8.KMnO₄ in acid medium is always reduced to :

•  Mn⁽⁴⁺⁾
•  TMn⁽²⁺⁾
•  Mn⁽⁶⁺⁾
•  Mn

Solution
Mn⁽⁷⁺⁾ 5e ⟶Mn⁽²⁺⁾

Q9.Oxidation state of carbon in graphite is:

•  Zero
•  +1
•  +4
•  +2

Solution
Graphic is uncombined state of carbon.

Q10. Total number of AlF₃ molecules in a sample of AlF₃ containing 3.01 × 10²³ ions of F^-is :

•  9 ×10²⁴
•  3×10²⁴
•  7×10²³
• 10²³

Solution
∵3 ions of F^- from 1 molecule of AIF₃ ∴3 × 10²³ ions of F^- from 10²³ molecules of AIF₃