Stoichiometry Notes - einstein classes [PDF]

mass unit (u or amu). One amu = atom. /g. 10. 02 .6. 1. 23. × . Molar mass of a substance is the mass of 6.02 × 1023 p

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CS – 1

S TO I C H I O M E T RY Syllabus :

w E w I w NS .e T in E st IN ei C nc L A la S ss S E es S .c om

The concept of atoms and molecules; Dalton’s atomic theory; Mole concept; Chemical formulae; Balanced chemical equations; Calculations (based on mole concept) involving common oxidationreduction, neutralization, and displacement reactions; Concentration in terms of mole fraction, molarity, molality and normality.

CONCEPTS

C1

In this chapter we will discuss the calculations based on chemical equations. It has been classified into two parts : 1.

Mole Concept

2.

Equivalent Concept

C2A MOLE CONCEPT :

In mole concept we deal with different types of relations like weight-weight, weight-volume, or volume-volume relationship between reactants or products of the reaction. Some basic formulas and definitions used in mole concept are as follows :

Mole is the unit representing 6.022 × 1023 numbers (atoms, molecules or ions). A mole also represents gram molecular mass or gram atomic mass. Masses of atoms, molecules and nuclear particles are expressed in a specially defined unit, called atomic mass unit (u or amu). One amu =

1

6.02  10 23

g / atom .

Molar mass of a substance is the mass of 6.02 × 1023 particles present in it.

An empirical formula represents the simplest whole number ratio of various atoms present in a compound, whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

Number of moles of a substance(n) 

weight of substance atomic or molecular weight

Also, Number of moles of a substance(n) 

Given number of molecules Avogadro number

In gas phase reaction number of moles of a gas (n) =

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CS – 2 At STP/NTP one mole of any gas contains 22.4 L i.e. at 273 K and 1 atm pressure. Mole concept is based on balanced chemical reaction. If amounts of both reactants are given then find the limiting reagent. It is that reagent which is consumed completely in a irreversible chemical reaction. Class Discussion Problems : 1.

Hydrogen reacts with nitrogen to produce ammonia according to this equation : 3H2(g) + N2(g)  2NH3(g) Determine how much ammonia would be produced when 200.0 g of hydrogen react.

2.

If 20.0 g of CaCO3 is treated with 20.0 g of HCl, how many grams of CO2 can be generated according to the following equation : CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g) Given : Molar mass of (CaCO3 = 100 g mol–1, HCl = 36.5 g mol–1, CO2 = 44.0 g mol–1) Zinc and hydrochloric acid react according to the reaction :

w E w I w NS .e T in E st IN ei C nc L A la S ss S es ES .c om

3.

Zn(s) + 2HCl (aq)  ZnCl2(aq) + H2(g)

If 0.30 mol Zn are added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced ?

4.

A compound contains 4.07% hydrogen, 22.47% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas ? [At. wt. : H = 1, C = 12, Cl = 35.5]

5.

Find the number of atom in 1g of 12C. Home Work Problems :

In a reaction A + B2  AB2. Identify the limiting reagent, if any, in the following reaction mixtures.

6.

7.

(i)

300 atoms of A + 200 molecules of B

(ii)

2 mol A + 3 mol B

(iii)

100 atoms of A + 100 molecules of B

(iv)

5 mol A + 2.5 mol B

(v)

2.5 mol A + 5 mol B

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation : N2(g) + 3H2(g)  2NH3(g). [At. wt. : N = 14, H = 1]

8.

(i)

Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen.

(ii)

Will any of the two reactants remain unreacted ?

(iii)

If yes, which one and what would be the mass ?

Which one of the following will have largest number of atoms ? (i)

1 g Au(s)

(ii)

1 g Na(s)

(iii)

1 g Li(s)

(iv)

1 g of Cl2(g)

[At. Wt. Au = 197, Na = 23, Li = 7, Cl = 35.5]

[Answers : (1) 1132.2 g (2) 8.80 g of CO2 (3) 0.26 mol (4) CH2Cl, C2H4Cl2 (5) 5.016 × 1022 (6) (i) B is the limiting reagent (ii) A is the limiting reagent (iii) No limiting reagent (iv) B is the limiting reagent (v) A is limiting reagent (7) (i) 2428.57 g NH3 (ii) H2 will remain unreacted (iii) 571.4 g (8) 1 g of Li has the largest number of atoms]

C3

DIFFERENT WAYS OF EXPRESSING THE CONCENTRATION TERMS : Important Definitions :

mass percent 

mass of solute  100 mass of solution

Molarity(M) 

No. of moles of solute , unit of molarity are mol/lit., M or molar.. Vol. of solution in L

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CS – 3 Normality ( N ) 

No. of gramequiva lents of solute , unit of normality are g-eq./lit., N or normal. Vol. of solution in L

No. of gram equivalents of solute(neq ) 

Equivalent weight 

Molality (m ) 

Weight of solute Equivalent weight

Molecular weight (or ) Atomic weight (or ) Ionic weight n factor

No. of moles of solute , unit of molality are mol/kg, m or molal. wt . of solvent nA . n A  nB

w E w I w NS .e T in E st IN ei C nc L A la S ss S E es S .c om

Mole fraction( x A ) 

ppm 

mass of solute  10 6 mass of solution

The relation between different concentration terms :

1.

neq = nmol × n-factor

2.

neq = Normality × Volume (L)

3.

Number of moles(nmol) = Molarity × Volume (L)

4.

Normality = Molarity × n-factor

5.

M

10xd M

6.

m

7.

m

x B  1000 (1  x B )M A

M  1000 1000d  MM 

(d  density of solution in g/ml, M  molar mass of solute, xB and xA are mole fraction of solute and solvent respectively, MA  molar mass of solvent) Calculation of ‘n’ Factor for Different Compounds : 1.

2.

Acids : n = basicity H3PO4 n = 3

H3PO3 n = 2

H3PO2 n = 1

H3BO3 n = 1

Bases : n = acidity of base

e.g. Ammonia and all amines are monoacidic bases, NaOH(n = 1), Na2CO3(aq) n = 2, NaHCO3(n = 1)

3.

Salt : (Which does not undergo redox reactions)

n factor = Total cationic or anionic charge, e.g. Na3PO4 n = 3, Ba3(PO4)2 n = 6

4.

Oxidizing Agents or Reducing Agents : ‘n’ factor = change in oxidation number Or number of electron lost or gained from one mole of the compound.

Class Discussion Problems :

1.

If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution ?

2.

A sample of drinking water was found to be severely contaminated with chloroform, CHCl3 supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

3.

(i)

Express this in per cent by mass.

(ii)

Determine the molality of chloroform in the water sample.

Commercially available concentrated hydrochloric acid contains 38% HCl by mass. (a)

What is the molarity of this solution ? The density is 1.19 g mL–1.

(b)

What volume of concentrated HCl is required to make 1.00 L of 0.10 M HCl ?

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CS – 4 4.

500 mL of 0.250 M Na2SO4 solution is added to an aqueous solution of 15.00 g BaCl2 resulting in the formation of the white precipitate of insoluble BaSO4. How many moles and how many grams of BaSO4 are formed ? (At. wt. Ba = 137, S = 32, Na = 23, O = 16, Cl = 35.5)

5.

Calculate the volume of 1.00 mol L–1 aqueous sodium hydroxide that is neutralised by 200 mL of 2.00 mol L–1 aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralisation reaction is : NaOH(aq) + HCl (aq)  NaCl (aq) + H2O(l)

6.

What volume of 0.30 M Na2SO4 solution is required to prepare 2.0 L of a solution 0.40 M in Na+ ? Home Work Problems :

7.

What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2 L ?

8.

Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction

w E w I w NS .e T in E st IN ei C nc L A la S ss S es ES .c om

CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O (l)

What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl ?

The density of 3 M solution of NaCl is 1.25 g mL–1. Calculate the molality of the solution.

9.

[Answers : (1) 25.22 mL (2) (i) 1.5 × 10–3% (ii) 1.255 × 10–4 m (3) (a) 12.38 M (b) 807.7 mL (4) 0.072 mol BaCl2, 16.80 g BaSO4 (5) 400 mL, 23.40 g (6) 1.3 L (7) 0.0292 mol L–1 (8) 0.94 g CaCO3 (9) 2.79 m]

C4

EQUIVALENT CONCEPT

It is based on law of equivalence which is explained as follows :

Law of chemical equivalents : In a chemical reaction the equivalents of all the species (reactants or products) are equal to each other provided none of these compounds is in excess. N1V1 = N2V2 (when normalities and volumes are given).

If the number of equivalence of both the reactants are different then reactant with the lesser number of equivalence will be the limiting reagent. Application of equivalent concept : It is used in acid base titration, back titration and double titration, similarly in redox titration. Equivalent concepts can be used on all reaction whether they are balanced or not balanced but mole concept is used in solving the problems when the reactions are balanced. Basic principles of tirations :

In volumetric analysis, a given amount (weight or volume) of an unknown substance is allowed to react with a known volume of a standard solution slowly. A chemical reaction takes place between the solute of an unknown substance and the solute of the standard solution. The completion of the reaction is indicated by the end point of the reaction, which is observed by the colour change either due to the indicator or due to the solute itself. Whether the reactions during the analysis are either between an acid and or base or between oxidising agent and reducing agent., the law of equivalence is used at end point. Following are the different important points regarding this process : (i)

In case of acid base titration at the equivalence point (neq)acid = (neq)base

(ii)

In case of redox titration

(neq)oxidant = (neq)reductant

(iii)

If a given volume of solution is diluted then number of moles or number of equivalence of solute remains same but molarity or normality of the solution decreases.

(iv)

If a mixture contains more than one acids and is allowed to react completely with the base then at the equivalence point, (neq) acid1 + (neq) acid2 + ... = (neq) base

(v)

Similarly if a mixture contains more than one oxidising agents then at equivalence point, (neq) O.A1 + (neq) O.A2 +... = (neq) reducing agent.

(vi)

If it is a difficult to solve the problem through equivalence concept then use the mole concept.

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CS – 5 Back titration : This is a method in which a substance is taken in excess and some part of it has to react with another substance and the remaining part has to be titrated against standard reagent. Double titration : This is a titration of specific compound using different indicators. Let us consider a solid mixture of NaOH, Na2CO3 and inert impurities. When the solution containing NaOH and Na2CO3 is titrated using phenolphthalein indicator following reaction takes place at the phenolphthalein end point – NaOH + HCl  NaCl + H2O Na2CO3 + HCl  NaHCO3 + H2O 1 eq. of Na 2 CO 3  eq. of HCl 2 (n  2)

w E w I w NS .e T in E st IN ei C nc L A la S ss S E es S .c om

Here, eq. of NaOH 

When methyl orange is used, Na2CO3 is converted into NaCl + CO2 + H2O Hence, eq. of NaOH + eq. of Na 2 CO 3 = eq. of HCl (n 2)

TITRATION OF MIXTURE OF BASES WITH TWO INDICATORS Every indicator has a working range Indicator pH range Behaving as Phenolphthalein 8 — 10 weak organic acid Methyl orange 3 — 4.4 weak organic base Thus methyl orange with lower pH range can indicate complete neutralisation of all types of bases. Extent of reaction of different bases with acid (HCl) using these two indicators summarised below Phenolphthalein Methyl Orange NaOH 100% reaction is indicated 100 % reaction is indicated NaOH + HCl  NaCl + H2O NaOH + HCl  NaCl + H2O Na2CO3 50% reaction upto NaHCO3 100% reaction is indicated stage is indicated Na2CO3 + 2HCl  2NaCl + H2O Na2CO3 + HCl  NaHCO3 + NaCl + CO2 NaHCO3 No reaction is indicated NaHCO3 + HCl  NaCl + H2O + CO2 100% reaction is indicated

Redox titration :

Class Discussion Problems :

1.

What is the normality of 0.300 M H3PO3 when it undergoes the following reaction ? H3PO3 + 2OH–  HPO32– + 2H2O

2.

What volumes of 12.0 N and 3.00 N HCl must be mixed to give 1.00 L of 6.00 N HCl ?

3.

Determine the normality of an H3PO4 solution, 40.0 mL of which neutralized 120 mL of 0.531 N NaOH.

4.

A 48.4 mL sample of HCl solution requires 1.240 g of pure CaCO3 for complete neutralization. Calculate the normality of the acid. (Molecular wt. of CaCO3 = 100 g/mol)

5.

In standardizing HCl, 22.5 mL was required to neutralize 25.0 mL of 0.100 N Na2CO3 solution. What is the normality of the HCl solution ? How much water must be added to 200 mL of it to make it 0.100 N ? Neglect volume changes.

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CS – 6 6.

Given the unbalanced equation KMnO4 + KI + H2SO4  K2SO4 + MnSO4 + I2 + H2O (a) How many g KMnO4 are needed to make 500 mL of 0.250 N solution ? (b) How many g KI are needed to make 25.0 mL of 0.360 N solution ? (At. wt. K = 39, Mn = 55, O = 16, I = 127)

7.

Determine the mass of KMnO4 required to make 80.0 mL of N/8 KMnO4 when the latter acts as an oxidizing agent in acid solution and Mn2+ is a product of the reaction. Home Work Problems :

8.

Calculate the normality of each of the following solutions : (a) 7.88 g of HNO3 per L solution (b) 26.5 g of Na2CO3 per L solution (if acidified to form CO2). [At. wt. C = 12, N = 14, O = 16, Na = 23] (a) What volume of 5.00 N H2SO4 is required to neutralize a solution containing 2.50 g NaOH ? (b) How many g pure H2SO4 are required ? [Molecular wt. of NaOH = 40, H2SO4 = 98]

10.

Exactly 50.0 mL of a solution of Na2CO3 was titrated with 65.8 mL of 3.00 N HCl.

w E w I w NS .e T in E st IN ei C nc L A la S ss S es ES .c om

9.

If the density of the Na2CO3 solution is 1.25 g/mL, what percent Na2CO3 by weight does it contain ? (At. wt. Na = 23, C = 12, O = 16)

11.

Calculate the mass of oxalic acid, H2C2O4, which can be oxidized to CO2 by 100.0 mL of an MnO4– solution 10.0 mL of which is capable of oxidizing 50.0 mL of 1.00 N I– to I2. (At. wt. H = 1, C = 12, O = 16)

12.

Find the equivalent weight of KMnO4 in the reaction Mn2+ + MnO4– + H 2O  MnO 2 + H + (unbalanced). How many g MnSO4 is oxidized by 1.25 g KMnO4 ? (molar mass MnSO4 = 151 g) [Answers : (1) 0.600 N (2) 0.333 L of conc. solution, 0.667 L of dil. solution (3) 1.59 N (4) 0.512 N (5) 0.111 N, 22 mL (6) (a) 3.95 g (b) 1.49 g KI (7) 0.316 g (8) 0.1251 N, 0.500 N (9) (a) 12.5 mL (b) 3.07 g (10) 16.8% Na2CO3 (11) 22.5 g (12) 1.79 g MnSO4]

C5

VOLUME STRENGTH OF H2O2

x volume of H2O2 means x litre of O2 is liberated by 1 volume of H2O2 on decomposition

2H 2 O 2   2H 2 O  O 2 68 gm

22.4 lit at STP

Volume strength of H2O2 solution = N × 5.6......

(where N is the normality of the H2O2 solution

Class Discussion Problems :

1.

Calculate the strength of ‘20 V’ of H2O2 in terms of normality

2.

In a 50 ml solution of H2O2 an excess of KI and dilute H2SO4 were added. The I2 so liberated required 20 ml of 0.1 N Na2S2O3 for complete reaction. Calculate the strength of H2O2 in grams per litre. [Answers : (1) 3.58 N (2) 0.68 g/litre]

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CS – 7 Electron exchanged Eq. wt. or change in O.N.

Changed to

Reactions

1.

MnO4— (O.A.)

Mn2+ in acidic medium

MnO4— + 8H+ + 5e—  Mn2+ + 4H2O

5

E

2.

MnO4— (O.A.)

MnO2 in basic medium

MnO4— + 3e— + 2H2O  MnO2 + 4OH—

3

E

3.

MnO4— (O.A.)

MnO42— in neutral medium

MnO4— + e— + 2H2O  MnO22—

1

E

4.

Cr2O72—(O.A.)

Cr3+ in acidic medium

Cr2O72— + 14H+ + 6e—  2Cr3+ + 7H2O

6

E

5.

MnO2(O.A.)

Mn2+ in acidic medium

MnO2 + 4H+ + 2e—  Mn2+ + 2H2O

2

E

6.

Cl—

Cl2 + 2e—  2Cl—

2

E

Cu+

Cu2+ + e–  Cu+

1

E

8.

Cl2(O.A.) (in bleaching powder) CuSO4 (O.A.) (in iodometric titration) S2O32— (R.A.)

S4O62—

2S2O32—  S4O62— + 2e—

9.

H2O2(O.A.)

H2 O

H2O2 + 2H+ + 2e—  2H2O

2

10.

H2O2(R.A.)

O2

2

11.

Fe2+ (R.A.)

Fe3+

H2O2  O2 + 2H+ + 2e— (O.N. of oxygen in H2O2 is (–1) per atom) Fe2+  Fe3+ + e—

w E w I w NS .e T in E st IN ei C nc L A la S ss S E es S .c om

Species

7.

1.

2 (for two molecules) E 

3.

E

1

Reaction

Relation between O.A. and R.A.

I2

I2 + 2Na2S2O3  2NaI + Na2S4O62— I2 + 2S2O32—  2I— + S4O62—

I2  2I—  2Na2S2O3

4.

5.

CaOCl2

MnO2

IO3—

2CuSO4  I2  2Na2S2O3

CaOCl2 + H2O  Ca(OH)2 + Cl2 Cl2 + 2KI  2KCl + I2 Cl2 + 2I—  2Cl— + I2

CaOCl2  Cl2  I2  2I—  2Na2S2O3

 MnO2 + 4HCl (conc.)  MnCl2 +  Cl2 + 2H2O Cl2 + 2KI  2KCl + I2 or MnO2 + 4H+ + 2Cl—  Mn2+ + 2H2O + Cl2 Cl2 + 2I—  I2 + 2Cl—

IO3— + 5I— + 6H+  3I2 + 3H2O

Eq. wt. of CuSO4 =

M 1

M 1

2CuSO4 + 4KI  Cu2I2 + 2K2SO4 + I2 or 2Cu2+ + 4I—  Cu2I2 + I2 white ppt.

M 1

M 2

Eq. wt. of CaOCl2 =

MnO2  Cl2  I2  2I—  2Na2S2O3 Eq. wt. of MnO2 =

M 2

IO3—  3I2  6I  6Na2S2O3 Eq. wt. IO3— =

Einstein Classes,

M 1

M 2 M E 2

Estimation of

CuSO4

M 1

E

Eq. wt. (Na2S2O3) = E 

2.

M 5 M 3 M 1 M 6 M 2 M 2

M 6

Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18

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CS – 8 6.

H2 O 2

H2O2 + 2I— + 2H+  I2 + 2H2O

H2O2  I2  2I—  2Na2S2O3 Eq. wt. H2O2 =

7.

Cl2

Cl2 + 2I—  2Cl— + I2

M 2

Cl2  I2  2I—  2Na2S2O3

M 2

Eq. wt. of Cl2 = 8.

O3

O3 + 6I— + 6H+  3I2 + 3H2O + O2

O3  3I2

M 2

w E w I w NS .e T in E st IN ei C nc L A la S ss S es ES .c om

Eq. wt. of O3 = 9.

ClO—

ClO— + 2I— + 2H+  H2O + Cl— + I2

ClO—  I2  2I  2Na2S2O3 Eq. wt. of ClO— =

10.

Cr2O72—

Cr2O72— + 14H+ + 6I—  3I2 + 2Cr3+ + 7H2O

Cr2O72—  3I2  6I—

Eq. wt. of Cr2O72— =

11.

Fe2+

Fe2+  Fe3+ + e— MnO + 8H+ + 5e—  Mn2+ + 4H2O

M 2

M 6

5Fe2+ = MnO4—

— 4

Eq. Wt. Fe2+ =

M 1

Eq. Wt. MnO4— =

12.

Fe2+

Cr2O

13.

C2O42—

2— 7

Fe2+  Fe3+ + e— + 14H+ + 6e—  2Cr3+ + 7H2O

C2O42—  2CO2 + 2e— MnO4— + 8H+ + 5e  Mn2+ + 4H2O

M 5

6Fe2+  Cr2O72—

Eq. Wt. Cr2O72— =

M 6

5C2O42—  2MnO4—

Eq. wt. C2O42— =

M 2

Eq. wt. MnO4— = M/5

14.

H2 O 2

H2O2  2H+ + O2 + 2e— 5e– + MnO4—  Mn2+

5H2O2  2MnO4—

Eq. wt. H2O2 =

M 2

M 5 M Eq. wt. As2O3 = 4

Eq. wt. MnO4— =

15.

As2O3

Einstein Classes,

As2O3 + 5H2O  2AsO43— + 10H+ + 4e—

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CS – 9

EXERCISE TYPE – 1 (SINGLE CORRECT CHOICE TYPE) 1.

(a)

50

(b)

30

(c)

20

(d)

None

If 0.5 mol of BaCl2 is mixed with 0.20 mol of Na3PO4, the maximum amount of Ba 3(PO4)2 that can be formed is

11.

The normality of an H3PO4 solution 40.0 mL of which neutralized 120 mL of 0.531 N NaOH is (a)

1.00 N

(b)

1.59 N

(c)

3.00 N

(d)

5.02 N

To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorus acid (H3PO3), the volume of 0.1 M aqueous KOH solution required is (a)

10 mL

(b)

20 mL

(c)

40 mL

(d)

60 mL

w E w I w NS .e T in E st IN ei C nc L A la S ss S E es S .c om

2.

1 g of the carbonate of a metal was dissolved in 25 ml of N-HCl. The resulting liquid required 5 ml of N-NaOH for neutralization. The eq. wt. of the metal carbonate is

10.

3.

4.

5.

6.

7.

8.

(a)

0.70 mol

(b)

0.50 mol

(c)

0.20 mol

(d)

0.10 mol

12.

The equivalent mass of MnSO4 is half of its molar mass when it is converted to (a)

Mn2O3

(c)



(b)

MnO2

23

(b)

12

(c)

24

(d)

13

(a)

11.2 litres

(b)

22.4 litres

(c)

44.8 litres

(d)

56.0 litres

(a)

2.5

(b)

1

(c)

0.5

(d)

0.25

(d)

MnO

2– 4

When one gram mol of KMnO4 reacts with HCl, the volume of chlorine liberated at NTP will be

34 g of hydrogen peroxide is present in 1120 ml of solution. This solution is called

The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous oxalate in acidic solution is

A solution contains Na2CO3 and NaHCO3. 10 ml of the solution required 2.5 ml of 0.1 M H2SO4 for neutralisation using phenolphthalein as indicator. Methyl orange in then added when a further 2.5 ml of 0.2 M H 2SO 4 was required. The amount of Na2CO3 and NaHCO3 in 1 litre of the solution is

(a)

3/5

(b)

2/5

(a)

5.3 and 4.2 g

(b)

3.3 g and 6.2 g

(c)

4/5

(d)

1

(c)

4.2 g and 5.3 g

(d)

6.2 g and 3.3 g

(a)

10 vol solution (b)

20 vol solution

(c)

30 vol solution (d)

32 vol solution

14.

(a)

0.135 M

(b)

0.141 M

0.5 g of fuming H2S2O7(Oleum) is diluted with water. This solution is completely neutralized by 26.7 ml of 0.4 N NaOH. The percentage of free SO3 in sample is

(c)

0.210 M

(d)

0.246 M

(a)

30.6%

(b)

40.6%

(c)

20.6%

(d)

50%

[Na+] in a solution prepared by mixing 30.00 mL of 0.12 M NaCl with 70 mL of 0.15 M Na2SO4 is

10 mL of H2O2 solution

15.

(volume strength = x)

N required 10 mL of MnO4– solution in acidic 0.56

16.

medium. Hence x is

9.

(a)

Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation 2H2O 2(aq.)  2H2O(l) + O2(g) under conditions where one mole of gas occupies 24 dm3, 100 cm3 of X M solution of H2O2 produces 3 dm3 of O2 X is thus

MnO4

13.

5.3 g of M2CO3 is dissolved in 150 mL of 1 N HCl. Unused acid required 100 mL of 0.5 N NaOH. Hence equivalent weight of M is

(a)

0.56

(b)

5.6

(c)

0.1

(d)

10.0

Molality of 18 M H2SO4 (d = 1.8 gmL–1) is (a)

36 mol kg–1

(b)

200 mol kg–1

(c)

500 mol kg–1

(d)

18 mol kg–1

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17.

0.7 g of a sample of Na2CO3.xH2O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralization. The value of x is (a)

7

(b)

3

(c)

2

(d)

5

100 mL of 1 M KMnO4 oxidised 100 mL of H2O2 in acidic medium (when MnO4– is reduced to Mn2+); volume of same KMnO4 required to oxidise 100 mL of H2O2 in basic medium (when MnO4–. is reduced to MnO2) will be

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CS – 10 (a)

(c)

(b)

300 mL 3

(d)

500 mL 3

The no. of of Fe2+ ions

(S)

3.01 × 1022

(T)

6.02 × 1019

formed when excess of iron is treated with 5 ml of 0.04 N HCl

100 mL

For a given mixture of NaHCO 3 and Na 2CO 3, volume of given HCl required is x mL with phenolphthalein indicator and further y mL required with methyl orange indicator. Hence volume of HCl for complete reaction of NaHCO3 is (a)

2x

(b)

(c)

y

(d)

TYPE-4 (MULTIPLE CORRECT CHOICE TYPE) 1.

Assuming complete dissociation of H 2 SO 4 as (H2SO4  2H+ + SO42–), the number of sulphate ions present in 50 ml of 0.1 N H2SO4 solution are (a)

x 2

15.05 × 1020

(b)

1.505 × 1021

w E w I w NS .e T in E st IN ei C nc L A la S ss S es ES .c om

18.

100 mL 3

(D)

19.

2.

(y – x)

1 mol of ferric oxalate is oxidised by x mol of MnO4– and also 1 mol of ferrous oxalate is oxidised by y mol of MnO4– in acidic medium. The ratio

20.

(c)

(a)

2:1

(b)

1:2

(c)

3:1

(d)

1:3

x is y

3.

3 mol of a mixture of FeSO4 and Fe2(SO4)3 required 100 mL of 2 M KMnO4 solution is acidic medium. Hence mol fraction of FeSO4 in the mixture is (a)

1 3

(b)

2 3

(c)

2 5

(d)

3 5

4.

TYPE-2 (COMPREHENSION TYPE)

5.

1. 2. 3.

TYPE-3 (MATRIX-MATCH TYPE)

(A)

Column-A

Column-B

Total no. of electrons in

(P)

3.01 × 1021

(Q)

6.02 × 1023

No. of sulphate ions

In the following redox reaction 2MnO4– + 10Cl– + 16H+  2Mn2+ + 5Cl2 + 8H2O. Pick up the correct statements. (a)

MnO4– is reduced

(b)

Cl– is oxidising agent

(c)

MnO4– is an oxidising agent

(d)

Cl– is reduced

Which of the following is disproportionation ? (a)

2Cu+  Cu + Cu2+

(b)

3Cl2 + 6OH–  ClO3– + 5Cl– + 3H2O

(c)

2H2S + 8O2  2H2O + 3S

(d)

Na +

1 Cl  NaCl 2 2

Oxidation number of C is zero in (a)

CHCl3

(b)

CH2Cl2

(c)

C6H12O 6

(d)

CO

Among the species given below which can act as oxidising as well as reducing agent ? (a)

SO2

(b)

SO3

(c)

H2 O 2

(d)

H 2S

(A)

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B)

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C)

Statement-1 is True, Statement-2 is False

(D)

Statement-1 is False, Statement-2 is True

H2SO4 solution are

500cm3 of 0.2 molar NaCl (R)

3.01 × 10–3 mol

Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

present in 50mL of 0.1M

(C)

(d)

TYPE-5 (Assertion-Reason Type)

1.6 g methane are

(B)

5 × 1023

6.02 × 1022

is added to 100cm3 of

0.5 molar AgNO3. Thus no. of ions of AgCl formed are

1.

STATEMENT-1 : 22.4 L of ethane at N.T.P. contains one mole of hydrogen molecules. STATEMENT-2 : One mole of hydrogen

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CS – 11 molecules at N.T.P. occupies 22.4 L of volume. 2.

STATEMENT-1 : The masses of oxygen which combine with fixed mass of nitrogen in N2O, NO, N2O3 bears a simple multiple ratio. STATEMENT-2 : The combination according to law of multiple proportions.

3.

6.

A 10.0 mL portion of (NH4)2SO4 solution was treated with excess NaOH. The NH 3 gas evolved was absorbed in 50.00 mL of 0.1000 N HCl. To neutralize the remaining HCl, 21.50 mL of 0.0980 N NaOH was required. What is the molar concentration of the (NH4) 2SO 4 ? How many g (NH4)2 SO4 are in 1 L solution ?

7.

A 1.2 g of a mixture containing H2C2O4.2H2O and KHC2O4.H 2O and impurities of a neutral salt, consumed 18.9 ml of 0.5 N NaOH for neutralization. On titrating with KMnO4 solution 0.4 g of the same substance needed 21.55 ml of 0.25 N KMnO4. Calculate the percentage composition of the substance.

8.

A mixture of FeO and Fe 2 O 3 is reacted with acidified KMnO4 solution having a concentration of 0.2278 M, 100 ml of which was used. The solution then was treated with Zn dust which converted the Fe3+ of the solution to Fe2+ . The Fe2+ required 1000 ml of 0.13 M K2Cr2O7 solution. Find the mass % of FeO and Fe2O3.

9.

5 ml of 8 N nitric acid, 4.8 ml of 5 N HCl acid and a certain volume of 17 M sulphuric acid are mixed together and made up to 2 litre. Thirty ml of this acid mixture exactly neutralise 42.9 ml of sodium carbonate solution containing one gram of Na2CO3. 10H2O in 100 ml of water. Calculate the mass in gram of the sulphate ions in solution.

10.

A mixture of calcium carbonate and sodium chloride weighing 3.20 g was added to 100 ml of 1.02 N HCl. After the reaction had ceased the liquid was filtered, the residue washed and the filtrate was made up to 200 ml. 20 ml of this dilute solution required 25 ml of N/5 NaOH for neutralization. Calculate the % of CaCO3 in the mixture. (Ca = 40).

STATEMENT-1 : 8.075 × 10–2 kg of Glauber’s salt is dissolved in water to obtain 1 dm3 of solution . The molarity of the resultant solution is 0.0025 M. STATEMENT-2 : The volume in mL of 0.5 M H2SO4 needed to dissolve 0.5 g of copper (II) carbonate is 24.3 mL (At. Wt of Cu = 63.5) STATEMENT-1 : Basicity of an acid can change in different reactions.

w E w I w NS .e T in E st IN ei C nc L A la S ss S E es S .c om

4.

and 10.0 g O2 were put into a reaction vessel. (a) How many grams of acetic acid can be produced by this reaction from these amounts of reactants ? (b) How many grams of the excess reactant remaining after the reaction is complete ?

STATEMENT-2 : As the equivalent weight always remains same.

5.

STATEMENT-1 : If equal volume of C M KMnO4 and C M K2Cr2O7 solutions are allowed to oxidised Fe2+ to Fe3+ in acidic medium, then Fe2+ oxidised by KMnO 4 is more then K 2 Cr 2 O 7 of same concentrations.

STATEMENT-2 : Number of moles of electrons gained by 1 mole of K2Cr2O7 is more than the 1 mole of KMnO4 TYPE – 6 (SUBJECTIVE TYPE PROBLEMS)

1.

If 40.00 mL of 1.600 M HCl and 60.00 mL of 2.000 M NaOH are mixed, what are the molar concentrations of Na +, Cl – , and OH – in the resulting solution ? Assume a total volume of 100.00 mL.

2.

The acidic substance in vinegar is acetic acid CH3COOH. When 6.00 g of a certain vinegar was titrated with 0.100 M NaOH, 40.11 mL of base had to be added to reach the equivalence point. What percent by mass of this sample of vinegar is acetic acid ?

3.

One gram of a mixture of CaCO3 and MgCO3 gives 240 ml of CO2 at N.T.P. Calculate the percentage composition of mixture (Ca = 40, Mg = 24, C = 12, O = 16).

4.

A solution contains 4 g of Na2CO3 and NaCl in 250 ml. 25 ml of this solution required 50 ml of N/10 HCl for complete neutralization. Calculate % composition of mixture.

5.

In an Industrial process for producing acetic acid, oxygen gas is bubbled into acetaldehyde CH3CHO, containing manganese (II) acetate (catalyst) under pressure at 600C. 2CH3CHO(l) + O2(g)  2CH3COOH(l) In a laboratory test of this reaction, 20.0 g CH3CHO

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ANSWERS TYPE – 1 a d b d a a d d c b

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

c a a a c c b d a a

8.

(a) 27.24 gm (b) 0.145 M, 19.1 g/L KHC2O4.H2O = 80.9 %, H2C2O4.2H2O = 14.7 % % FeO = 13.34 %, % Fe2O3 = 86.66 %

9. 10.

6.528 gm 81.25%

w E w I w NS .e T in E st IN ei C nc L A la S ss S es ES .c om

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

5. 6. 7.

CS – 12 2.73 gm

TYPE – 2

1. 2. 3.

TYPE – 3

A-R ; B-P ; C-S ; D-T

TYPE – 4

1. 2. 3. 4. 5.

a, b a, c a, b b, c a, c

TYPE – 5

1. 2. 3. 4. 5.

D A B C D

TYPE – 6

1. 2. 3. 4.

0.56 M OH–, 0.640 Cl–, 1.2MNa+ 4.01 % CaCO3 = 62.5 %, MgCO3 = 37.5% Na2CO3 = 66.25%, NaCl = 33.75%

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