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Mole & Redox Reaction - Class 12 MCQ


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30 Questions MCQ Test - Mole & Redox Reaction

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Mole & Redox Reaction - Question 1

What is the mass of Fe304 if it reacts completely with 25 ml of 0.3m K2Cr2O7?

Detailed Solution for Mole & Redox Reaction - Question 1

neqFe3O4 = neq K2Cr2O7
wM × nf = 0.3 * 25 * 6
w = (0.3×25×6×232)/1
w = 10.44 g.

Mole & Redox Reaction - Question 2

One mole of a mixture of CO and CO2 requires exactly 20 g of NaOH in solution for complete conversion of all the CO2 into Na2CO3 .How much NaOH would it require for conversion into Na2CO3, if the mixture (one mole) is completely oxidised to CO2?    

Detailed Solution for Mole & Redox Reaction - Question 2

2 NAOH + CO₂ — > Na₂CO₃ + H₂O

Mole ratio :

1 : 2

We need 2 moles of NaOH to convert one mole of CO₂ to Na₂CO₃.

Moles of NaOH :

20 / 40 = 0.5 moles

Moles of CO₂ :

0.5/ 2 = 0.25

Moles of CO are:

1 - 0.25 = 0.75 moles.

Oxidation of CO :

CO + 1/2 O₂ —> CO₂

Mole ratio is 1 : 1

Moles of CO₂ formed are 0.75 moles.

Thus is half the moles of NaOH required.

NaOH required is thus :

2 × 0.75 =1.5 moles

1 mole — > 40 g

1.5 moles —>?

1.5 × 40 = 60g

60 g is required

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Mole & Redox Reaction - Question 3

1.60 g of a metal were dissolved in HNO3 to prepare its nitate. The nitrate on strong heating gives 2 g of its oxide. The equivalent weight of metal is:    

Detailed Solution for Mole & Redox Reaction - Question 3

Eq. of metal = Eq. of oxide

Mole & Redox Reaction - Question 4

The maximum amount of BaSO4 precipitated on mixing 20 mL of 0.5 M BaCI2 with 20 mL of 1 M H2SO   

Detailed Solution for Mole & Redox Reaction - Question 4


mm taken    =10    20        0    0
mm formed    0    10        10    20
Milli mole of BaSO4=10
or
Mole of BaSO4 = 10-2

Mole & Redox Reaction - Question 5

What is the [OH-) in the final solution prepared by mixing 20.0 mL of 0.050 M HCI with 30.0 mL of 0.10 M Ba(OH)2?    

Detailed Solution for Mole & Redox Reaction - Question 5



=6        =1
5        0        1    1
∴ 

Mole & Redox Reaction - Question 6

Oxidation number of carbon in C3O2, Mg2C3 are respectively:    

Detailed Solution for Mole & Redox Reaction - Question 6

Carbon has negative oxidation no. in Mg3C2 and positive oxidation number in C3O2; O is more electronegative than C. Mg is more electropositive than C.

Mole & Redox Reaction - Question 7

Which is not a redox reaction?

Detailed Solution for Mole & Redox Reaction - Question 7

No change in oxidation number of any species.

Mole & Redox Reaction - Question 8

Which is not correct in case of Mohr’s salt?    

Detailed Solution for Mole & Redox Reaction - Question 8

Ox. no. of Fe in Mohr’s salt,
[FeSO4.(NH4)2SO4.6H2O] is +2 

Mole & Redox Reaction - Question 9

In the reaction;   the element oxidised is/are

Detailed Solution for Mole & Redox Reaction - Question 9

Mole & Redox Reaction - Question 10

HCO-3 contains carbon in the oxidation state:    

Detailed Solution for Mole & Redox Reaction - Question 10

1+ a +3 x (-2) = -1
∴ a = +4    

Mole & Redox Reaction - Question 11

Oxidation number of Fe in Fe0.94O is:    

Detailed Solution for Mole & Redox Reaction - Question 11

0.94 x a +1 x (-2) = 0
∴  a = 200/94

Mole & Redox Reaction - Question 12

The brown ring complex [Fe(H2O)5NO+]SO4 has oxidation number of Fe:    

Detailed Solution for Mole & Redox Reaction - Question 12

No in iron complex has +1 ox. no. in MnSO4

Mole & Redox Reaction - Question 13

The pair of compounds which cannot exist in solution is:    

Detailed Solution for Mole & Redox Reaction - Question 13

NaHCO3 being an acid salt will react with NaOH as, one mole of substance.

Mole & Redox Reaction - Question 14

0.5 mole of H2SO4 is mixed with 0.2 mole of Ca(OH)2 The maximum number of mole of CaSO4 formed is:    

Detailed Solution for Mole & Redox Reaction - Question 14

Eq. of H2SO4 = 0.5x2=1.0;
Eq. of Ca(OH)2 = 0.2x2 = 0.4;
Equal Eq. reacts and thus, Eq. of CaSO4formed = 0.4
∴ Mole of CaSO4 formed = 0.4/2 = 0.2

Mole & Redox Reaction - Question 15

Volume of 0.1 M NaOH needed  for the neutralisation of 20 mL of 0.05 M oxalic acid is:    

Detailed Solution for Mole & Redox Reaction - Question 15

Meq. of NaOH = Meq. oxalic acid;
0.1x1xV = 20x0.05x2;

∴ V = 20mL

Mole & Redox Reaction - Question 16

How many garm of KMnO4 should be taken to make up 250 mL of a solution of such strength that 1 mL is equivalent to 5.0 mg of Fe in FeSO4?    

Detailed Solution for Mole & Redox Reaction - Question 16

Meq. of KMnO4 in 1 mL = Meq. of Fe = 
∴ Meq. of KMnO4 in 250 mL = 
Thus, 

Mole & Redox Reaction - Question 17

25 mL of 0.50 M H2O2 solution is added to 50 mL of 0.20 M KMnO4 in acidic solution. Which of the following statements is true?    

Detailed Solution for Mole & Redox Reaction - Question 17

Meq. of H2O2 = 25x0.5x2 = 25;
Meq. of KMnO4 = 50x0.2x5 = 50;
25 Meq. or 5 milli mole of KMnO4 are left

Mole & Redox Reaction - Question 18

What volume of 0.40 M Na2S2O3 would be required to react with the I2 liberated by adding excess of KI to 50 mL of 0.20 M CuSO4 solution?    

Detailed Solution for Mole & Redox Reaction - Question 18

Meq. of Na2S2O3 = Meq .of CuSO4
V x0.4x1=50x0.2x1
V = 25 mL

Mole & Redox Reaction - Question 19

The number of mole of potassium salt, i.e, KHC2O4.H2C2O4.2H2O oxidised by one mole of permanganate ion is:    

Detailed Solution for Mole & Redox Reaction - Question 19

Mole & Redox Reaction - Question 20

5 g of a sample of bleaching  powder is treated with excess acetic acid and KI solution. The liberated I2 required 50 mL of N/10 hypo. The percentage of available chlorine in the sample is:    

Detailed Solution for Mole & Redox Reaction - Question 20

Meq. of bleaching powder = Meq. of CI2 = Meq. of hypo

Mole & Redox Reaction - Question 21

4.4 g of CO2  and  2.24 litre of H2 at STP are mixed in a container. The total number of molecules present in the container will be:    

Detailed Solution for Mole & Redox Reaction - Question 21

Mole & Redox Reaction - Question 22

20 g of an acid furnishes 0.5 moles of H3O+ ions in its aqueous solution. The value of 1 g eq. of the acid  will be.    

Detailed Solution for Mole & Redox Reaction - Question 22

0.5 mole of H2O+=20 g; Also H3O+  is monovalent, thus
Mol.wt.=Eq.wt.
∴  1 mole of H3O+ = 40 g = GEW of acid

Mole & Redox Reaction - Question 23

The chloride of a metal contains 71 % chlorine by weight and the vapour density of it is 50. The atomic weight of the metal will be:    

Detailed Solution for Mole & Redox Reaction - Question 23

Mol.wt. of metal chloride = 50x2=100
Let metal chloride be MCIn then
Eq. of metal = Eq. of chloride, or 
Now    a+35.5n =100
or    n.E+35.5n=100;
n = 2            
Therefor,    a=2xE = 2x(29/2) = 29

Mole & Redox Reaction - Question 24

The simplest formula of a compound containing 50% of element X (at. wt. 10) and 50% of element  Y (at. wt. 20) is:    

Detailed Solution for Mole & Redox Reaction - Question 24

g atom of X = 50/10 = 5;
g atom of Y = 50/20 = 2.5;
Ratio of g atom of X and Y = 2:1.

Mole & Redox Reaction - Question 25

One g  of a mixture of Na2CO3 and NaHCO3 consumes y equivalent of HCI for complete neutralisation. One g of the mixture is strongly heated, then cooled and the residue treated with HCI how many equivalent of HCI would be required for com-plete neutralisation?    

Detailed Solution for Mole & Redox Reaction - Question 25

Mole & Redox Reaction - Question 26

The pair of species having same percentage of carbon is:    

Detailed Solution for Mole & Redox Reaction - Question 26

Both have same empirical formula CH2O

Mole & Redox Reaction - Question 27

0.5 g of fuming H2SO4 (oleum) is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is:    

Detailed Solution for Mole & Redox Reaction - Question 27

Meq. of H2SO4+Meq. of SO3=Meq. of NaOH

∴  a = 0.103
∴  % of SO3

Mole & Redox Reaction - Question 28

Equivalent weight of bivalent metal is 32.7. Molecular weight of its chloride is:    

Detailed Solution for Mole & Redox Reaction - Question 28

mol.wt. of MCI2=2x32.7+71=136.4

Mole & Redox Reaction - Question 29

If 20% nitrogen is present in a compound, it’s minimum molecular weight will be.    

Detailed Solution for Mole & Redox Reaction - Question 29

20 g N, then mol.wt. =100
14g N, then mol.wt.= 

Mole & Redox Reaction - Question 30

0.52 g of a dibasic acid required 100 mL of 0.1 N NaOH for complete neutralisation. The equivalent weight of acid is:    

Detailed Solution for Mole & Redox Reaction - Question 30

Meq. of acid = Meq. of NaOH

∴  E = 52

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