Chemistry Exam  >  Chemistry Tests  >  Physical Chemistry  >  Test: Mole Concept, Volumetric & Redox - 1 - Chemistry MCQ

Test: Mole Concept, Volumetric & Redox - 1 - Chemistry MCQ


Test Description

20 Questions MCQ Test Physical Chemistry - Test: Mole Concept, Volumetric & Redox - 1

Test: Mole Concept, Volumetric & Redox - 1 for Chemistry 2024 is part of Physical Chemistry preparation. The Test: Mole Concept, Volumetric & Redox - 1 questions and answers have been prepared according to the Chemistry exam syllabus.The Test: Mole Concept, Volumetric & Redox - 1 MCQs are made for Chemistry 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Mole Concept, Volumetric & Redox - 1 below.
Solutions of Test: Mole Concept, Volumetric & Redox - 1 questions in English are available as part of our Physical Chemistry for Chemistry & Test: Mole Concept, Volumetric & Redox - 1 solutions in Hindi for Physical Chemistry course. Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free. Attempt Test: Mole Concept, Volumetric & Redox - 1 | 20 questions in 90 minutes | Mock test for Chemistry preparation | Free important questions MCQ to study Physical Chemistry for Chemistry Exam | Download free PDF with solutions
Test: Mole Concept, Volumetric & Redox - 1 - Question 1

4.4g of CO2 and 2.24 litre of Hat STP (273.15 K and 1 atm pressure) are mixed in a container. The total of molecules present in the container will be

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 1

Number of moles of CO2 =

∵ 22.4 L of H2 has 1 mole of H2

∴ The number of moles in 2.24L 
The total moles are 0.2.
⇒ Total molecules in 0.2 moles 

Test: Mole Concept, Volumetric & Redox - 1 - Question 2

4I + Hg2+ → HgI42– ; 1 mole of each Hg2+ and I will form:

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 2

Balanced reaction is as follows :
Hg2+(aq) + 4I- ​ (aq) ⇋​   HgI42-​ (aq)​

∵ 4 mole of I- produces = 1 mole of HgI42-​​
∴ 1 mole I- produces = 1/4 moles of HgI42-​​

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Mole Concept, Volumetric & Redox - 1 - Question 3

How many grams of I2 are present in a solution which requires 40 ml of 0.11N Na2S2O3 to react with it via the reaction: 
S2O32– + I2 → S4O62– + 2I?

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 3

Meq. of I= Meq. of Na2S2O3 = 40 × 0.11
∴ (W × 2 × 1000)/254  = 40 × 0.11
W of I= 0.558g

Test: Mole Concept, Volumetric & Redox - 1 - Question 4

Two elements A (atomic mass = 75) and B (atomic mass = 16) combine to yield a compound. The % by weight of A in the compound was found to be 75.08. The formula of compound is:

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 4

Molecular weight of A2B3 = 75 × 2 + 16 × 3 = 198
198g of A2B3 =150g of A
∴ 10g of A2B  = 75.08g of A

So option B is correct. 

Test: Mole Concept, Volumetric & Redox - 1 - Question 5

What weight of HNO3 is needed to convert 5g of Iodine into Iodic acid according to the reaction:
I2 + HNO3→ HIO3 + NO2 +H2O

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 5

The balanced reaction is:
I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O

(127 * 2)g of iodine requires (63 * 10)g of HNO3
Thus, 5g of I2 requires
= [(63 * 10)/(127 * 2)] * 5 = 12.4g 

Test: Mole Concept, Volumetric & Redox - 1 - Question 6

In the reaction, VO + Fe2O3 → FeO + V2O5, the  equilvalent weight of  V2O5 is equal to its

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 6

Test: Mole Concept, Volumetric & Redox - 1 - Question 7

When BrO3 ion reacts with Br ion in acid solution Br2 is liberated. The eq. weight of KBrO3 in this reaction is:

10e + 2Br5+ → Br02

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 7

10e + 2Br5+ → Br02

Test: Mole Concept, Volumetric & Redox - 1 - Question 8

The hydrated salt, Na2SO4.nH2O, undergoes 55.9 % loss in weight on heating and become anhydrous. The value of n will be:

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 8

Molecular Mass of Na2SO4 is = 2 x 23 + 32 + 4 x 16 = 142 gm.

In 100g of the compound, 44.1 g Na2SOcontains 55.9 g of H2O.
(∵ 55.9 % is water)
⇒ 1g Na2SO4 contains = (55.9/44.1)g H2O
∴ 142 g Na2SO4 contains = (55.9/44.1) x 142 = 180 g H2O .

Hence, the number of water molecules = 180/18 = 10
(Since the Molecular mass of H2O is 18).

Thus, the value of n is 10 and the formula is Na2SO4.10H2O.

Test: Mole Concept, Volumetric & Redox - 1 - Question 9

When a metal is burnt, its weight is increased by 24%. The eq. weight of the metal will be:

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 9

Equivalent weight of O^2-(oxide) ion is: Molar mass of oxygen/charge on oxygen
= 16/2 = 8 g eq-1

One equivalent of the metal reacts with one equivalent of oxygen to form metal oxide.

Let M be the equivalent weight of the metal.
Therefore, M grams of the metal will react with 8g of oxygen.
The corresponding metal oxide will have a mass of 8 grams more than the mass of metal M. This 8 gram is 24% of M. 
i.e. M x 24/100 = 8 gram
Thus, M = 100 x 8/24 = 33.33 g

Test: Mole Concept, Volumetric & Redox - 1 - Question 10

One mole of potassium chlorate is thermally decomposed and excess of aluminum is burnt in the gaseous product .How many moles of aluminum oxide are formed?

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 10

2KClO→ 2KCl + 3O2

4Al + 3O→ 2Al2O3

The molar ratio for the first reaction is 2:3 for KClO3:O2

Thus, if 1 mole of  KClO3 is decomposed, 1.5 moles of O2 is produced.

The molar ratio for the second reaction is: 3:2 for O2:Al2O3

3 moles of oxygen produces 2 moles of aluminium oxide. 

Since, we have 1.5 moles of oxygen, number of moles of Al2O3  produced =

 

Test: Mole Concept, Volumetric & Redox - 1 - Question 11

How much Cl2 at STP is liberated when one mole of KMnO4 reacts with HCl ?

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 11

Potassium permanganate reacts with hydrochloric acid according to the following equation:
2KMnO+ 16HCl → 2KCl + 2MnCl+ 8H2O + 5Cl2

⇒ 2 moles of KMnO4 produces 5 moles of Cl2
⇒ 1 mole of KMnO4 will produce 2.5 moles of Cl2

1 mole of CO2 will have 22.4 litre sof volume
∴ Volume of 2.5 moles of Cl= 56 litres

Test: Mole Concept, Volumetric & Redox - 1 - Question 12

Minimum quantity of H2S needed to precipitate 63.5g of Cu2+ is nearly:

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 12

Chemical Reaction: Cu2++ H2S → CuS + 2H+

In this reaction, to precipitate 1 mol of Cu2+ we require 1 mol of H2S.
Moles of Cu2+ = 63.5 g / 63.5 g mol-1 = 1 mol
Hence, to precipitate 1 mol of Cu2+ we require 1 mol of H2S
i.e. Mass of H2S = 1 mol x 34 g mol-1 = 34 g

Test: Mole Concept, Volumetric & Redox - 1 - Question 13

2g of CaCO3 was treated with 0.1M HCl (500 ml). The volume of CO2 evolved at STP after heating the solution is:

Test: Mole Concept, Volumetric & Redox - 1 - Question 14

10g of limestone on heating produces 4.2g of CaO. the percentage purity of CaCO3 in limestone is:

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 14

Test: Mole Concept, Volumetric & Redox - 1 - Question 15

‘x’ g of KClO3 on decomposition gives ‘y’ ml of O2 at STP. The % purity of KClO3 would be

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 15

The balanced eqn. will be:
2KClO3 = 2KCl+3O2

The no. of moles of Oxygen = y/22.4
3 moles of Oxygen is produced from 2 moles of KClO3
⇒ 1 mole will be produced from 2/3 moles of KClO3
∴ y/22.4 will be producing from (2/3) × (y/22.4)
Hence, the mass of KClO3 used = (2/3) × (y/22.4)×M

% Purity = (2 × y × M)/(3 × 22.4 × x)

Test: Mole Concept, Volumetric & Redox - 1 - Question 16

33.6g of an impure sample of sodium bicarbonate when heated strongly gave 4.4g of CO2. The % purity of NaHCO3 would be:

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 16

Balanced Equation:
2NaHCO3 = Na2CO3 + CO2 + H2O

Moles of CO2 = 4.4/44 =0.1
∵ 1 mole of COis produce by 2 Moles of NaHCO3
⇒ Moles of NaHCO3 =0.1×2/1 =0.2
∴ Wt. of NaHCO3 = 0.2 × 84 = 16.8gm
% Purity of NaHCO3 = (16.8/33.6) × 100 = 50%.

Test: Mole Concept, Volumetric & Redox - 1 - Question 17

0.16g of dibasic acid required 25ml of N/10 NaOH for complete neutralization. Molecular wt. of acid is:

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 17

Dibasic acid NaOH; 

N1V= N2V2

M = 2 × E = 2 × 64 = 128

.

 

      

Test: Mole Concept, Volumetric & Redox - 1 - Question 18

The molar coefficients of AsO33– and MnO4- in the reaction are:

AsO33– + MnO4Θ → AsO43– + MnO2

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 18

Test: Mole Concept, Volumetric & Redox - 1 - Question 19

Which of the following is a disproportionation reaction?

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 19

A disproportionation reaction is a reaction in which a single element undergoes reduction and oxidation

i.e in option d), Cu has changed its oxidation state from +1 to 0 (i.e reduction) and +1 to +2 (i.e oxidation)

Test: Mole Concept, Volumetric & Redox - 1 - Question 20

Hydrogen and oxygen combine to form H2O2 and H2O containing 5.93% and 11.2 % hydrogen respectively. The data illustrates:​

Detailed Solution for Test: Mole Concept, Volumetric & Redox - 1 - Question 20

For predicting the law of chemical combination, we have to have find out the ratio of masses in both cases:
Let the total mass in each case be 100 grams.

For H2O2
​​Mass of hydrogen = 5.93 gm
Mass of oxygen = 100-5.93 = 94.07 gm
The ratio of the mass of hydrogen to the mass of oxygen, H:O = 5.93 : 94.07 = 1 : 16

For H2O
​Mass of hydrogen = 11.2 gm
Mass of oxygen = 100-11.2 = 88.8 gm
The ratio of the mass of hydrogen to the mass of oxygen, H:O = 11.2 : 88.8 = 1 : 32

As the ratio of the mass of hydrogen remains the same in both cases and the ratio of masses of oxygen in both cases (16: 32 or 1: 2) is a whole number ratio.

Therefore, the above data illustrate the law of multiple proportions.

83 videos|142 docs|67 tests
Information about Test: Mole Concept, Volumetric & Redox - 1 Page
In this test you can find the Exam questions for Test: Mole Concept, Volumetric & Redox - 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Mole Concept, Volumetric & Redox - 1, EduRev gives you an ample number of Online tests for practice
83 videos|142 docs|67 tests
Download as PDF