Unit 01 Some Basic Concepts of Chemistry
Matter and It’s Nature
Anything which has mass and occupies space is called matter.
Three physical states of matter are solid, liquid and gas.
In solids, molecules are held very close to each other and they have definite shape and definite volume.
In liquids, molecules are close to each other but move around, they have definite volume but not shape.
In gases, molecules are far apart and have neither definite volume nor shape.
At bulk level, matter can be divided in to pure and impure substances.
Elements and compounds are pure substances whereas mixtures fall in the category of impure substances.
Mixtures are further classified as heterogeneous (with non uniform composition) and homogenous mixtures (with uniform composition).
Elements are made up of identical atoms with the same atomic number.
Dalton’s atomic theory
All matter is made of atoms which are indivisible and indestructible particles.
All the atoms of a given element are identical, however, atoms of different elements have different masses and different chemical roperties.
Compounds are formed by the combination of different atoms in the ratio of small whole numbers.
Atoms are neither created nor destroyed in the course of an ordinary chemical reaction.
Atoms and molecules
An atom is the smallest particle of an element which can take part in a chemical reaction. It may or may not be capable of independent existence.
A molecule is the smallest particle of an element or a compound which is capable of independent existence.
Elements such as oxygen, hydrogen, etc. exist in forms of molecules.
When two or more atoms of different elements combine, the molecule of a compound is obtained such as water, ammonia, carbon-dioxide, etc.
The atoms of different elements are present in a compound in a fixed ratio which is characteristic of a particular compound.
The constituents of a compound cannot be separated into simpler substances by physical methods.
Physical quantities and their measurements in chemistry
Mass - quantity of matter present in a substance.
Weight- force exerted by gravity on an object
Volume-the space occupied by a substance, or enclosed within a container
Density - mass per unit volume
Temperature-the degree or intensity of hotness of a substance
Pressure - force exerted per unit area on or against an object
Physical Quantity | S.I. Units |
---|---|
Mass | Kilogram (kg) |
Weight | Newton (N) |
Volume | |
Density | |
Temperature | Kelvin (K) |
Pressure | Pascal |
Precision and accuracy
Precision refers to the closeness of various measurements for the same quantity.
Accuracy is the agreement of a particular value to the true value of the result.
Significant figures
All non zero digits are significant.
The zeros to the right of the decimal point or zeros between two non-zero digits are significant.
Zeros to the left of the first non-zero digit in a number smaller than one are not significant.
In addition or subtraction, the result should be mentioned in the same number of decimal place as that of the term with the least ecimal places.
In multiplication and division, the result should be mentioned in same number of significant figures as the least precise term used in calculation.
Dimensional analysis
While calculating one needs to convent units from one system to other. The method used to accomplish this is called factor label method or unit factor method or dimensional analysis.
Ex: 1 inch
Both these are called unit factors
Laws of Chemical Combinations
Law of Conservation of Mass :
Mass is neither created nor destroyed in chemical reactions. For example
Mass remains conserved, i.e., mass of reactant
Law of Multiple Proportions :
If two elements combine in different ways to form different compounds, the masses of one element that combine with a fixed mass of the other element are in ratio of small whole numbers. For example
Mass ratio of N and |
||
Ratio of masses of 0 that combines with a fixed mass of
Law of Constant Proportions :
Different samples of a pure chemical substance always contain the same proportions of elements by mass. e.g. every sample of water
Law of Reciprocal Proportions :
When definite mass of an element A combines with two different elements
Gay Lussac’s Law :
According to this law, under similar conditions of temperature and pressure, gases react with each other in a simple ratio of their volumes and if the product is also gaseous, it also bears a simple ratio with the volumes of reactants.
Avogadro’s Law :
Equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules.
Atomic mass
Relative atomic mass of an element is defined as the ratio of the mass of one atom of the element to the mass of
Relative atomic mass
Average atomic mass
Where
Similar terms are for molar masses.
Calculation of atomic weight
Atomic wt.
Approx. atomic weight
Valency
Atomic mass
Atomic Mass Unit, u (earlier called atomic mass unit, amu) is equal to
Most of the elements have isotopes, the atomic mass of an element is the weighted average of the masses of its all the naturally occurring isotopes.
If relative abundance is given instead of ratio
Molar mass:
Molar mass of a substance is the mass of one mole of that substance.
Formula mass:
Formula mass of a substance is the sum of the atomic masses of all atoms present in one formula unit of the compound (normally ionic compounds)
Gram-molar volume :
The volume occupied by one gram-molecular mass of any gas at NTP
Vapour density :
V.D.
Solved Examples
Question 1. Convert the
b)
a)
c)
d) 81.0
Show Answer
Answer : (b)
Question 2. What will be the answer of
a) 71.02
b) 71.0
c) 71.2
d) 71.1
Show Answer
Answer : (b)
The answer is to be reported upto same number of decimal places as that of the term with the least number of decimal places. This after rounding off 71.02 becomes 71.0 .
Question 3. A student performs a titration with different burettes and finds titre values of
a) 1
b) 3
c) 2
d) 4
Show Answer
Answer : (b)
Calculate the average value of different titre values. Then use the rule to determine the significant figures for mathematical calculation.
Average value
The number of significant figure is 3
Practice Questions
1. A gaseous mixture contains oxygen and nitrogen in the ratio of
(a)
(b)
(c)
(d)
Show Answer
Answer: (c)2. The total number of electrons in one molecule of carbon dioxide is
(a) 22
(b) 44
(c) 66
(d) 88
Show Answer
Answer: (a)3. The largest number of molecules is in
(a)
(b)
(c)
(d)
Show Answer
Answer: (a)4. When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volumes of hydrogen evolved is
(a)
(b)
(c)
(d)
Show Answer
Answer: (a)5.
(a)
(b)
(c)
(d)
Show Answer
Answer: (a)6. Which has maximum number of molecules?
(a)
(b)
(c)
(d)
Show Answer
Answer: (b)7.
(a)
(b)
(c)
(d)
Show Answer
Answer: (d)8. If
(a)
(b)
(c)
(d)
Show Answer
Answer: (b)9. If oxygen is present in one litre flask at pressure of
(a)
(b)
(c)
(d)
Show Answer
Answer: (a)10. The ratio of masses of oxygen and nitrogen and a particular gaseous mixture is
(a)
(b)
(c)
(d)
Show Answer
Answer: (b)Mole
The quantity of a given substance that contains as many elementary entities, such as atoms, molecules ions or formula units as the number of atoms in exactly
In modern terms, gram-molecule and gram-atom are termed as a mole of molecules and a mole of atoms respectively e.g., 1 gram-molecule of oxygen and I gram-atom of oxygen are expressed as 1 mole of
1 Mole of atoms
The number of moles of a substance can be calculated by various means.
Rules in Brief
(1)
(2)
(3)
(Standard molar volume is the volume occupied by 1 mole of any gas at NTP which is equal to 22.4 litres.)
(4) Number of moles of atoms/molecules/ions/electrons
(5) Number of moles of solute
Percentage Composition
Empirical and Molecular formula:
Steps involved in the calculation of empirical formula:
Convert the mass percentage into grams: considering
Calculate the number of moles of each element.
Calculate the simplest molar ratio: divide the moles obtained by the least value from amongst the values obtained for each element.
Calculate the simplest whole number ratio: multiply all the simplest atomic ratios by a suitable integer.
Write the empirical formula: write the symbols of each element along with their whole number ratios side by side.
Molecular formula = empirical formulaxn
Methods of expressing concentration of a solution
The amount of solute in a solution can be expressed in terms of two systems of units :
(i) Group A units specify the amount of solute in a given volume of solution, such as strength molarity, normality etc.
(ii) Group B units specify the amount of solute for a given mass of solvent or solution, such as mass %, mole fraction, molality etc.
The advantage of Group A units is the ease of solution preparation and that of Group B units is the temperature independence.
In a given solution, let solute be represented by (1) and the solvent by (2)
Expressing concentration of solution in Groups and units :
Following are the terms given in Table 1 and Table 2 widely used to express concentration of solution.
Table 1 : Expressing Concentrations in Group A
Name | Symbol | Definition | Formula | Usual SI unit |
---|---|---|---|---|
Mass percent | Mass of solute present in |
|||
Molarity | Mass of solute present |
Table 2 : Expressing Concentrations in Group B
Name | Symbol | Definition | Formula | Usual SI unit |
---|---|---|---|---|
concentration by mass of solution | Mass of solute present in |
Dimension- | ||
less | ||||
Molality | Mass of solute present in |
mol kg |
||
Fraction of a component out of total mol present in a solution | Dimension- less |
Relationship between molality (M’) and Molarity (M)
Relationship between molality (
Relationship between molarity (
While diluting a solution from one concentration to another, we can use
Molarity of mixture :
Let there be three samples of solution (containing same solvent and solute) with their molarity
Equivalent weight
The number of parts by weight of the substance which combine or displace directly or indirectly 1.008 parts by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine or 108 parts by weight of silver.
Methods to calculate equivalent weight
Hydrogen displacement method
Oxide formation or reduction of the oxide method:
Chloride formation method:
Metal displacement method:
Double decomposition method:
For a reaction
Electrolytic method:
Eq.
On passing the same quantity of electricity through two different electrolytic solutions.
Neutralization method:
Eq.
Eq. wt. of a base
For an organic acid (RCOOH)
Eq. wt. of acid (RCOOH) = Eq. wt. of RCOOAg - 107
Eq.wt. of an acid
Eq. wt. of base
Eq.wt. of salt
Eq.wt. of oxidizing/reducing agent
No. of electrons gained/lost by one molecule
Normality
It is defined as no. of equivalents of a solute present in one litre of solution
Also, no. of equivalents
And no. of Milli equivalents
1. Equivalents and Meq. (milliequivalents) of reactants react in equal amount to give same no. of equivalent or Meq. of products separately.
(i) In a given reaction.
Meq. of
(ii) In a compound
Meq. of
or Eq. of
(iii) In a series of reaction for complete reaction
Meq. of
2. Mole and millimole react according to balanced chemical equation.
3. Molarity
4. On diluting a solution, mole, mM, Equivalents and Meq. of solute do not change.
5. For reporting concentration of
(i) % strength of
(ii) Volume strength of
(iii) Volume strength of
Balancing of a chemical equation
According to the law of conservations of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation.
Step I Write down the correct formulas of reactants and products.
Step II Balance the number of atoms of elements other than
Step III Balance the number of
Step IV Verify the number of atoms for all reactants and products.
Solved Examples
Question 1. How many moles of magnesium phosphate,
(a)
(b)
(c) 0.02
(d)
Show Answer
Strategy : using the mole concept equate the moles of
Solution : 1 mole of
Therefore 0.25 moles of oxygen atoms are present in
Answer : (d)
Question 2. In the reaction
(a)
(b)
(c)
(d)
Show Answer
Strategy : use the mole concept where moles of
One mole of a gas at N.T.P.
Solution : 6 moles of
Convert the moles of
6 moles of
1 mole of
Answer : (c)
Practice Questions
Question 1.
(a)
(b)
(c)
(d)
Show Answer
Answer : (c)Question 2.
(a) 32.7
(b) 48.6
(c) 64.2
(d) 16.3
Show Answer
Answer : (a)Question 3.
(a) 10
(b) 14
(c) 20
(d) 40
Show Answer
Answer : (b)Question 4.
(a) 19.5
(b) 35.5
(c) 39.0
(d) 78.0
Show Answer
Answer : (c)Question 5. The chloride of a metal
(a)
(b)
(c)
(d)
Show Answer
Answer : (c)Question 6. The equivalent weight of a metal is 4.5 and the molecular weight of its chloride is 80 . The atomic weight of the metal is:
(a) 18
(b) 9
(c) 4.5
(d) 36
Show Answer
Answer : (b)Question 7. The sulphate of an element contains
(a) 17.0
(b) 35.0
(c) 51.0
(d) 68.0
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Answer : (b)Question 8. The equivalent weight of an element is 4. Its chloride has a vapour density 59.25. Then the valency of the element is :
(a) 4
(b) 3
(c) 2
(d) 1
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Answer : (b)Question 9. The oxide of an element possesses the formula
(a) 9
(b) 18
(c) 27
(d) none of these
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Answer : (c)Question 10. Approximate atomic weight of an element is 29.89 . If its eq. wt. is 8.9 , the exact atomic wt. is :
(a) 26.89
(b) 8.9
(c) 17.8
(d) 26.7
Show Answer
Answer : (d)Question 11.
(a) 15.9
(b) 47.7
(c) 31.8
(d) 8.0
Show Answer
Answer : (c)Question 12. The equivalent weight of iron in
(a) 18.6
(b) 28
(c) 56
(d) 112.0
Show Answer
Answer : (a)Question 13.
(a)
(b)
(c)
(d)
Show Answer
Answer : (c)Question 14. If
(a) 0.4
(b) 0.2
(c) 0.1
(d) 0.3
Show Answer
Answer : (d)Question 15.
(a) 0.1
(b) 3
(c) 0.3
(d) 0.2
Show Answer
Answer : (a)Question 16.
(a)
(b)
(c)
(d)
Show Answer
Answer : (b)Question 17. The ratio of amounts of
(a)
(b)
(c) zero
(d) infinite
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Answer : (a)Question 18.
(a)
(b)
(c)
(d)
Show Answer
Answer : (d)Question 19. The normality of
(a) 0.1
(b) 0.9
(c) 0.3
(d) 0.6
Show Answer
Answer : (d)Question 20. One litre of
(a) 6.85
(b) 0.685
(c) 0.1043
(d) 6.50
Show Answer
Answer : (b)Question 21. The hydrated salt
(a) 5
(b) 3
(c) 7
(d) 10
Show Answer
Answer : (d)Question 22. The total ionic strength (total molarity of all ions) of a solution which is
(a)
(b)
(c)
(d)
Show Answer
Answer : (b)Question 23. The isotopic abundance of
(a)
(b)
(c)
(d)
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Answer : (a)Question 24.
(a)
(b)
(c)
(d)
Show Answer
Answer : (b)Question 25. What will be the volume of the mixture after the reaction?
(a)
(b)
(c)
(d)
Show Answer
Answer : (b)Question 26. What volume of hydrogen gas at
(a)
(b)
(c)
(d)
Show Answer
Answer : (a)Question 27. The number of water molecules present in a drop of water (volume
(a)
(b)
(c)
(d)
Show Answer
Answer : (a)Question 28. 7.5 grams of a gas occupy 5.6 litres of volume at NTP. The gas is :
(a) NO
(b)
(c)
(d)
Show Answer
Answer : (a)Question 29. For the formation of
(a) 1.12 lit, 1.12 lit
(b) 1.12 lit, 2.24 lit
(c) 3.65 lit, 1.83 lit
(d) 1 lit, 1 lit
Show Answer
Answer : (a)Question 30. How many moles of magnesium phosphate
(a)
(b)
(c) 0.02
(d)
Show Answer
Answer : (d)Question 31. Which among the following is the heaviest?
(a) One mole of oxygen
(b) One molecule of sulphur trioxide
(c)
(d) Ten moles of hydrogen
Show Answer
Answer : (c)Question 32. How many grams of magnesium phosphate,
(a)
(b)
(c)
(d)
Show Answer
Answer : (b)Question 33. The molality of
(a)
(b)
(c)
(d)
Show Answer
Answer : (c)Question 34.
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer
Answer : (c)Question 35. The normality of
(a) 1.79
(b) 3.58
(c) 60.86
(d) 6.086
Show Answer
Answer : (d)Question 36. Vapour density of a volatile substance is 4 . Its molecular weight would be:
(a) 8
(b) 2
(c) 64
(d) 128
Show Answer
Answer : (a)Question 37.
(a) 33.25
(b) 3.325
(c) 12
(d) 20
Show Answer
Answer : (a)Question 38. If 0.50 mole of
(a) 0.70
(b) 0.50
(c) 0.20
(d) 0.10
Show Answer
Answer : (d)Question 39. A molar solution is one that contains one mole of solute in:
(a)
(b) 1.0 L of solvent
(c)
(d) 22.4 L of solution
Show Answer
Answer : (a)Question 40. In which mode of expression, the concentration of a solution remains independent of temperature?
(a) Molarity
(b) Normality
(c) Formality
(d) Molality
Show Answer
Answer : (d)Question 41. How many moles of electron weighs one kilogram?
(a)
(b)
(c)
(d)
Show Answer
Answer : (d)Question 42. Which has maximum number of atoms? (At. wts are given within brackets)
(a)
(b)
(c)
(d)
Show Answer
Answer : (a)Question 43. Dissolving
(a)
(b)
(c)
(d)
Show Answer
Answer : (c)Question 44. The molarity of a solution obtained by mixing
(a)
(b)
(c)
(d)
Show Answer
Answer : (a)Question 45. The highest mass corresponds to which of the following :
(a) 1 molecule of
(b)
(c) an
(d) 1 mole of
Show Answer
Answer : (d)Question 46. The chloride of a metal has the formula
(a)
(b)
(c)
(d)
Show Answer
Answer : (b)Question 47. By heating
(a)
(b)
(c)
(d)
Show Answer
Answer : (a)Question 48. If
(a)
(b)
(c)
(d)
Show Answer
Answer : (a)Question 49. The total number of protons in
(a)
(b)
(c)
(d)
Show Answer
Answer : (c)Question 50. How many millilitres (
(a) 2.5
(b) 5.0
(c) 10.0
(d) 20.0
Show Answer
Answer (b)Significance of Chemical Equations
A chemical equation describes the chemical process both qualitatively and quantitatively. The stoichiometric coefficients in the chemical equation give the quantitative information of the chemical process.
Calculations based on Chemical Equations (Stoichiometry)
Calculation based on chemical equations are known as stoichiometric calculations. It is of four types.
i. Calculations involving mass-mass relationship.
ii. Calculations involving mass-volume relationship.
iii. Calculations involving volume-volume relationship.
iv. Calculations involving mole-mole and mole-mass relationship.
Calculations Involving Mass-Mass Relationship
The following steps are used to solve the problem.
i. Write down balanced molecular equations for the chemical changes.
ii. Write down the number of moles below the formula of each of the reactants and products.
iii. Write down the relative masses of the reactants and products with the help of formula below the respective formula. These shall be the theoretical amounts of reactants and products.
iv. By the application of unitary method, the unknown factor or factors are determined.
Calculations Involving Mass-Volume Relationship
These calculations are based on the facts that I mole or I g molecule of the substance occupies
Volume of a gas at any temperature and pressure can be converted into mass or vice-versa with the help of the equation:
Where wis the mass of the gas,
Calculations Involving Volume-Volume Relationship
These calculations are based on two laws:
i. Avogadro’s law and
ii. Gay-Lussac’s law
For example,
Under similar conditions of temperature and pressure, gases react in simple ratio of their volumes.
Under similar conditions of temperature and pressure, equal moles of gases occupy equal volumes.
Example:
Mole ratio | 2 | 1 | 2 |
Mass ratio | |||
Volume ratio | |||
Volumeat STP | |||
Molecular ratio |
1. Calculations Based on Mass-Mass Relationship
Calculate the weight of iron which will be converted into its oxide
Solution. The balanced chemical equation is :
Now
2. Based on Mass-Volume Relationship
A sample of lime stone contains
Show Answer
Solution : Mass of pure
3. Based on Volume-Volume Relationship
One litre mixture of
Show Answer
Solution : Suppose Volume of
1 litre
a litre
Volume of
Total volume of the gases
Another method to solve objective questions of stoichiometry is POAC method.
4. Principle of Atom Conservation (POAC):
If atoms are conserved, moles of atoms shall also be conserved. This is known as principle of atom conservation. This principle is in fact the basis of the mole concept.
Let us see how this principle works. Consider the unbalanced chemical equation :
Apply the principle of atom conservation (POAC) for Katoms.
Moles of K atoms in reactant = Moles of K atoms in products
Or
Moles of
Now, since 1 molecule of
Thus, moles of
And moles of
Moles of
The above equation gives the weight relationship between
Again applying principle of atoms conservation of 0 atoms,
Moles of
But since 1 mole of
Moles of
Therefore
Or
The above equation thus gives weight-volume relationship of reactant and product.
Question. A sample of
i. Wt. of oxygen produced
ii. Wt. of
iii. Wt. of
Show Answer
Solution :
i. Mole of oxygen
Wt. of oxygen
ii.
Apply POAC for 0 atoms,
Moles of
( 1 mole of
Wt. of
iii. Again applying POAC for K atoms
( 1 mole of
Limiting Reactant- The reactant which is completely consumed in a chemical reaction.
The reactant producing the least number of moles of the product is the limiting reactant.
Hence
Or The reactant with the least number of equivalents (or milli equivalants) is the limiting reactant.
Question.
Show Answer
Solution : Reaction involved in the process is :
Molarity of
Number of moles of
Moles of
1 mole
0.125 moles
Required moles of
Thus
Mass of
Neutralization
The term neutralization is used for a reaction between an acid and a base or alkali.
Acid + base
For example,
Neutralization is a quantitative reaction.
A general definition is based on Bronsted & Lowry acid - base theory.
HA represents an acid and
When the acid has been neutralized there are no molecules of
In all cases : equivalents of an acid = equivalents of a base
When neutralization point is reached,
End point is the point of completion of the reaction indicated by suitable indicator. Hence it has additional drop of titrating reagent but, we use
Acid Base Titrations
Acid or acid mixture can be titrated against a suitable base or vice-versa using a suitable indicator. These are summarized in the following Table.
Acid-Base Titrations
Combination | Suitable Indicator | Colour Change | ||
---|---|---|---|---|
In acid | In Alkali | |||
1 | Strong acid / strong base | Methyl orange | Red | Yellow |
Bromothymol blue | Yellow | Blue | ||
2 | Strong acid / weak base | Methyl orange | Red | Yellow |
3 | Weak acid / strong base | Phenolphthalein | Colourless | Pink |
4 | Weak acid / weak base | Not suitable for a titration |
Titration of mixture of bases with two indicators
Every indicator has a working range
Indicator | pH range | Behaving as |
---|---|---|
Phenolphthalein | Weak organic acid | |
Methyl orange | Weak oraganic base |
Thus methyl orange with lower pH range can indicate complete neutralization of all types of bases.
Extent of reaction of different bases with acid
Phenolphthalein | Methyl Orange | |
---|---|---|
Stage is indicated |
||
No reaction is indicated |
Suppose volume of given standard acid solution (say
for complete reaction of
for complete reaction of
for complete reaction of
There may be different combination of mixture of bases. We may opt two methods
Method I: We carry two titrations separately with two different indicators
Method II: We carry single titration but adding second indicator after first end point is reached
Results with Two Indicators
Method I | Method II | ||||
---|---|---|---|---|---|
Mixture | Volume of HCI used with indicator | Volume of HCI used with indicator | |||
Phenolphthalein | Methyl orange | Phenolphthalein | Methyl orange added after firs end point is reached | ||
1 | |||||
2 | y |
||||
3 |
Solved Examples
Question 1. A solution containing
(a)
(b)
(c)
(d)
Show Answer
Strategy: write a chemical equation of the above process. Calculate the number of moles of
Solution :
Number of moles of
Number of moles of
The complexis
Question 2. For a reaction
(a) 5 moles
(b) 8 moles
(c) 16 moles
(d) 4 moles
Show Answer
Answer : 1 mole of
Now 2 moles of
Practice Question
Question 1. In the reaction,
(a) 1.0 mole of
(b) all the oxygen is consumed
(c) 1.5 mole of
(d) all the ammoniais consumed
Show Answer
Answer: (b)Question 2.
(a)
(b)
(c)
(d)
Show Answer
Answer: (b)Question 3. A solution containing
(a) Both
(b) Only
(c) Only
(d)
Show Answer
Answer: (a)Question 4. A solution containing
(a) Both
(b) Only
(c) Only
(d)
Show Answer
Answer: (d)Question 5. A solution containing
(a) Both
(b) Only
(c) Only
(d)
Show Answer
Answer: (d)Question 6. A mass of
(a) 2
(b) 5
(c) 7
(d) 10
Show Answer
Answer: (a)Question 7.
(a)
(b)
(c)
(d)
Show Answer
Answer: (c)Question 8. A mixture of
(a)
(b)
(c)
(d)
Show Answer
Answer: (a)Question 9. The mass of
(a)
(b)
(c)
(d)
Show Answer
Answer: (b)Question 10. Twenty milliliter of a solution is
(a)
(b)
(c)
(d)
Show Answer
Answer: (b)Question 11. An aqueous solution of
(a)
(b)
(c)
(d)
Show Answer
Answer: (a)Question 12. Ten milliliter of
(a)
(b)
(c)
(d)
Show Answer
Answer: (c)Question 13. An aqueous solution of
(a)
(b)
(c)
(d)
Show Answer
Answer: (a)Question 14. In the standardization of
(a) (molar mass)/2
(b) (molar mass)/6
(c) (molar mass)/3
(d) same as molar mass
Show Answer
Answer: (b)Question 15. An aqueous solution of
(a)
(b)
(c)
(d)
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Answer: (a)Question 16. No. of oxalic acid molecules in
(a)
(b)
(c)
(d)
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Answer: (a)Question 17.
(a)
(b)
(c)
(d)
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Answer: (a)Question 18. To neutralize
(a)
(b)
(c)
(d)
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Answer: (d)Question 19.
(a) 32
(b) 64
(c) 128
(d) 256
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Answer: (c)Question 20. The volume of water to be added to
(a)
(b)
(c)
(d)
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Answer: (a)Question 21. The mass of
(a)
(b)
(c)
(d)
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Answer: (c)Question 22. If LPG cylinder contains a mixture of butane and isobutene, then the amount of oxygen that would be required for combustion of
(a)
(b)
(c)
(d)
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Answer: (d)Question 23.
(a)
(b)
(c)
(d)
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Answer: (c)Question 24. If potassium chlorate is
(a)
(b)
(c)
(d)
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Answer: (a)Question 25. The mass of oxygen that would be required to produce enough
(a)
(b)
(c)
(d)
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Answer: (b)Question 26.
(a)
(b)
(c)
(d)
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Answer: (a)Question 27.
(a)
(b)
(c)
(d)
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Answer: (d)Question 28. Minimum volume of
(a)
(b)
(c)
(d)
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Answer: (d)Question 29. In Haber process, gaseous nitrogen and hydrogen react to form ammonia whose volume as compared to that of reactants (N.T.P.) would be :
(a) one fourth
(b) one half
(c) the same
(d) three fourth
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Answer: (b)Question 30.
(a) 40
(b) 50
(c) 60
(d) 70
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Answer: (b)Practice Questions
Subjective Type:
Question 1. How many
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Answer:Question 2.
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Answer:
Question 3. A solution containing
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Answer:
Question 4. A solution contains
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Answer:
Question 5. What weight of
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Answer:Question 6. Calculate the normality of the resulting solution made by adding
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Answer:Question 7. What weight of
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Answer:Question 8. How much
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Answer:Question 9. What is the normality and nature of a mixture obtained by mixing
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Answer: 0.1NQuestion 10.
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Answer:Question 11. A sample of mixture of
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Answer:Question 12.
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Answer:Question 13. One gram of a sample of lime stone was treated with
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Answer:Question 14. If a
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Answer:Question 15.
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Answer:Question 16.
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Answer: 2Question 17.
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Answer:Question 18. Find out the molarity of a solution obtained by mixing
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Answer: 0.133 MQuestion 19.
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Answer:Question 20. A mixture of
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Answer:Question 21. A sample of
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Answer:Question 22. A gas mixture of
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Answer:Question 23.
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Answer:Question 24. The reaction
i. The weight of copper deposited
ii. Molarity of copper sulphate in the original solution
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Answer: (1)Question 25.