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Test: Some Basic Concepts Of Chemistry 1 -From Past 28 Years Questions


18 Questions MCQ Test Chemistry 28 Years Past year papers for NEET/AIPMT Class 11 | Test: Some Basic Concepts Of Chemistry 1 -From Past 28 Years Questions


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This mock test of Test: Some Basic Concepts Of Chemistry 1 -From Past 28 Years Questions for Class 11 helps you for every Class 11 entrance exam. This contains 18 Multiple Choice Questions for Class 11 Test: Some Basic Concepts Of Chemistry 1 -From Past 28 Years Questions (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Some Basic Concepts Of Chemistry 1 -From Past 28 Years Questions quiz give you a good mix of easy questions and tough questions. Class 11 students definitely take this Test: Some Basic Concepts Of Chemistry 1 -From Past 28 Years Questions exercise for a better result in the exam. You can find other Test: Some Basic Concepts Of Chemistry 1 -From Past 28 Years Questions extra questions, long questions & short questions for Class 11 on EduRev as well by searching above.
QUESTION: 1

In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end? [2003]

Solution:

It is given that only 50% of the expected product is formed hence  only 10 litre of NH3 is formed N2 used = 5 litres, left = 30 – 5 = 25 litres H2 used = 15 litres,  left = 30 – 15 = 15 litres

QUESTION: 2

The maximum number of molecules is present in

Solution:

No. of molecules in different cases
(a)   22.4 litre at STP contains = 6.023×1023molecule of H2

∴ 15 litre at STP contains

(b)  22.4 litre at STP contains                            
= 6.023×1023 molecule of N2

5 litre at STP contians

(c)  2 gm of  H2= 6.023×1023 molecules of  H2  0.5 gm of H2

(d) Similarly 10 g of O2 gasmolecules

Thus (a) will have maximum number of molecules

QUESTION: 3

The mass of carbon anode consumed (givin g only carbondioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is  (Atomic mass: Al = 27) [2005]

Solution:

2Al2O3 + 3C → Al3CO2

Gram equivalent  of Al2O3 gm equivalent of C
Now equivalent weight of Al 

No. of gram equivalent of Al  = 

   = 30 × 103
Hence, No. of gram equivalent of C = 30 × 103

Again, No. of gram equivalent of C

⇒  mass = 90 × 103g = 90 kg

QUESTION: 4

The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is:   [2005]
Solution:

 

Change in oxidation number of Mn in basic medium is 1. Hence mole of KI is equal to mole of KMnO4.

QUESTION: 5

Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g mL– 1. Volume of acid required to make one litre of 0.1MH2SO4 solution is [2007]
Solution:

Molarity of H2SO4 solution

Suppose V ml of this H2SO4 is used to prepare 1 lit. of 0.1M H2SO4
∴ V x 18.02 = 1000 x 0.1

QUESTION: 6

An element , X has the following isotopic composition :
200X:90 %
199X : 8.0%
202X ; 2.0 %
The weighted average atomic mass of the naturally occuring element X is closest to
Solution:

Average isotopic mass of

199.96 amu

QUESTION: 7

The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is [2007]
Solution:

The balance chemical equation is :

From the equation it is clear that Moles of MnO4– require to oxidise 5 moles of SO3– – are 2
Moles of MnO4– require to oxidise 1 mole of SO3– – are 2/5.

QUESTION: 8

Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]
Solution:

Density

1 gram cm–3 

Volume

∴ Volume occupied by 1 gram water = 1 cm3 or Volume occupied by

molecules of water = 1 cm3

[∴ 1g water = moles of water]

Thus volume occupied by 1 molecules of water

i.e. the correct answer is option (c).

QUESTION: 9

Number of moles of MnO4- required to oxidize one mole of ferrous oxalate completely in acidic medium will be : [2008]
Solution:

QUESTION: 10

What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1L of propane gas (C3H8) measured under the same conditions? [2008]
Solution:

Writing the equation of  combustion of propane (C3H8), we get

From the above equation we find that we need 5 L of oxygen at NTP to completely burn 1 L of propane at N.T.P.
If we change the conditions for both the gases from N.T.P. to same conditions of temperature and pressure. The same results are obtained. i.e. 5 L is the correct answer.

QUESTION: 11

An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be: [2008]
Solution:

Thus empirical formula is CH3O.

QUESTION: 12

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl ? [2008]
Solution:

Writing the equation for the reaction, we get

From this equation we find 223 g of PbO reacts with 73 g of HCl to form 278 g of PbCl2.
If we carry out the reaction between 3.2 g HCl and 6.5 g PbO.
Amount of PbO that reacts with 3.2 g HCl

Since amount of PbO present is only 6.5 g so PbO is the limiting reagent.
Amount of PbCl2 formed by 6.5 g of PbO

Number of moles of PbCl2 formed

QUESTION: 13

10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be: 
Solution:

In this reaction oxygen is the limiting agent.
Hence amount of H2O produced depends on the amount of O2 taken 0.5 mole of O2 gives H2O = 1 mol
∴ 2 mole of O2 gives H2O = 4 mol

QUESTION: 14

What is the [OH–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2? [2009]
Solution:

No. of milli equivalent of HCl = 20 × 0.05          
= 1.0
No. of milli equivalent of Br (OH)2   = 30 × 0.1 × 2 = 6.0
After neutralization, no. of milli equivalents in 50 ml. of solution = (6 – 1) = 5
Total volume of the solution = 20 + 30 = 50 ml
∴ No. of milli equivalent of OH– is 5 in 50 ml

QUESTION: 15

The number of atoms in 0.1 mol of a triatomic gas is :[2010](NA = 6.02 ×1023 mol–1)
Solution:

The number of atoms in 0.1 mole of a triatomic gas = 0.1 × 3 × 6.023 × 1023.
= 1.806 × 1023

QUESTION: 16

Which has the maximum number of molecules among the following ? [2011 M]
Solution:

QUESTION: 17

6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is: [NEET 2013]
Solution:

QUESTION: 18

In an experiment it showed that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of AgNO3, which of the following will be the formula of the chloride (X stands for the symbol of the element other than chlorine): [NEET Kar. 2013]
Solution:

Millimoles of solution of chloride = 0.05 × 10 = 0.5
Millimoles of AgNO3 solution = 10 × 0.1 = 1
So, the millimoles of AgNO3 are double than the chloride solution

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