Some Basic Concept Of Chemistry Chemistry Class 11 ncert questions Answers


As a learner, we always feel the lack of a good questions bank which contains not only millions of questions but also it was well organised with exact solutions. NCERT Solutions for Class 11 Chemistry extra and important Questions Answers Some Basic Concept Of Chemistry. Our NCERT Solutions is a powerfull application for cbse learning students and other learners who always search variety of questions with answers from ncert solutions or from questions bank sample papers and board exams questions. Students need stored questions answers as questions banks which can be used anytime anywhere as we can't learn so many ncert solutions.


Some Basic Concept Of Chemistry Chemistry Class 11 ncert questions Answers


All chapters of NCERT Book as ncert solutions have exercise questions, textual questions and so many addtional questions like short answered questions, long answered questions and very long questions, here we included all types of questions answers format that need for a students and other stock holders like teachers and tutors. NCERT Solutions for Class 11 Chemistry extra and important Questions Answers Some Basic Concept Of Chemistry. Our NCERT Solutions is a powerfull application for cbse learning students and other learners who always search variety of questions with answers from ncert solutions or from questions bank sample papers and board exams questions.

NCERT Solutions for Class 11 Chemistry Questions with Solutions

Class 11 Chemistry Questions With Answers

Class 11 Chemistry Questions And Answers:

Answer:

It indicates that the element exist in the form of isotopes and its atomic mass is average atomic mass.

Answer:

1 mole of CuS04 contains 1 mole (1 g atom) of Cu

Molar mass of CuS04= 63.5 + 32 + 4 x 16 = 159.5 g mol-1

Thus, Cu that can be obtained from 159.5 g of CuS04 = 63.5 g

Therefore, Cu can be obtain from 159.5 g of CuSO4 = 63.5 x 100 /159.5 

                                                                                    = 39.81g Answer

Answer:

Mass of a molecule is that of a single molecule also known as its actual mass. But molecular mass is the mass of Avagadro’s number(6.022 x 1023) of molecules.

Answer:

In the text part we use the word mole while as a unit, we call it mol.

Answer:

No, one gram mole of a gas occupies 22.4 L only under N.T.P or S.T.P conditions. i.e. at 273K temperature and under 760 mm pressure. If these conditions are not used, then the volume is not 22.4 L.

Answer:

Empirical formula gives the simplest ration of the atoms of different elements in the molecules of a substance but the molecular formula gives their actual ratio.

Answer:

This indicates that the compounds contains in it oxygen also and its percentage is (100 – 92) = 8.

Answer:

A chemical equation has to be balanced in order to satisfy the law of conservation of mass. According to the law, there is no change in mass when the reactants change into products therefore, the chemical equations has to be balanced

Answer:

Methane (CH4) is regarded as the limiting reagent because air or oxygen is always present in excess. The amounts of C02 and H20 formed in the reaction depend upon the amount of methane only. Therefore, it is regarded as limiting reagent.

Answer:

by definition, molarity (M) of a solution is the number of moles of the solute dissolved per litre of the solution and molality(m) is the number of moles of the solute dissolved per kilogram of the solvent. Therefore, in molarity, Volume of the solution is considered while in molality, mass of the solvent is taken into consideration.

Answer:

Avagadro's number of molecules particles are present in one mole of a gas i.e. molecular mass of a gas expressed in grams. Any change in temperature and pressure has no influence on the number of particles.

Answer:

No, there is a difference there are 0.5 mole of NaOH and 0.5 M(Molarity) means there are 0.5 moles present in Volume of solution (in Litre).

Answer:

No, it is not possible. The fractional atomic mass of an element is it’s an average mass and not the actual mass. In the case, the element chlorine exists as two isotopes with atomic mass 35 u. and 37 u respectively in the ratio of 3: 1. The average come out to be fractional i.e. 35.5 u.

Answer:

No, atomic masses of the elements are not actual masses. There are only relative because the actual masses are very small. For detail consult text part.

 

Answer:

The coefficient of the reactant and product species involved in a chemical equation represent by the balanced form, are known as stiochiometric coefficient. For example.

N(g) + 3H ⇒ 2 NH3 (g)

Answer:

Limiting reactant i.e. the reactant present in fixed amount is so named because it limits the participation of other reactant ever if present in excess in a particular reaction.

Answer:

No, The actual masses are not fractional. It is the average of the atomic masses of the isotopes of an element which may be fractional. For example, the two isotopes of chlorine have atomic masses 35u and 37 u respectively the average atomic mass of chlorine 35.5 u.

 

Answer:

Air which contains certain suspended particles such as dust particles is heterogeneous mixture and not a homogeneous mixture.

 

Answer:

Since water (H2O) is a compound It is chemically and physically different from hydrogen and oxygen, its properties are quite different from those of hydrogen and oxygen.

 

Answer:

According to the Dalton atomic theory. An atom is the ultimate particles of matter which cannot be further divided into anything simpler than itself.

 

Answer:

Molarity of a solution normally decreases with rise in temperature.

Answer:

It means that 0.1 gram equivalent (4 g) of NaOH is dissolved per litre of the solution.

 

Answer:

Yes, because HCL is a monobasic acid.  Its molecular mass and equivalent mass are both same. (36.5).

 

Answer:

Methane (CH4) is regarded as the limiting reactant in the combustion reaction because the other reactant i.e. oxygen is always present in excess. The extent of the combustion reaction will depend upon the amount of methane only. 

 

Answer:

This illustrate the law of constant composition because sulphur and oxygen are combined in fixed ratio by mass n sulphur dioxide (S02) which is formed in this case.

 

Answer:

(i) H = 1, 0 = 16 

Molar mass = 1 x 2 + 16 x 1 = 18 u or g/mol

(ii) C = 12  , 0 = 16 

Molar mass = 12 x 1 + 2 x 16 = 42 u or g/mol

(iii) C = 12, H = 1 

Moalr mass = 12 x 1 + 4 x 1 = 16 u or g/mol  

Answer:

Na = 23u , S = 32u , 0 = 16u

Na2SO4 = 2 x 23 + 32 + 16 x 4 = 142 u 

% of Na in Na2SO4 =  \frac{46}{142}\times 100   = 32.99 % 

% of S in Na2SO4 =  \frac{32}{142} \times 100 = 22.54%

% of 0 in Na2SO4 =  \frac{64}{100} \times 100 = 45.07 %

 

Answer:

Iron = 69.9% , dioxide = 30.1%

Atomic Mass of Fe = 55.85

Atomic mass of O = 16

Mole of Oxygen =   \fn_cm \frac{% of di oxide }{atomic mass of oxygen} = \frac{30.1 }{16} = 1.88

Mole of fe  = \fn_cm \frac{% \: of \, Fe}{Atomic\, mass \, of \: oxygen} \, = \, \frac{69.9}{55.85} \, = \, 1.25

Molecular ratio in ratio simplest form = 2 : 3

 

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Class 11 Chemistry Some Basic Concept Of Chemistry 2022-2023 Latest Syllabus

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