CS– 1
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
STOICHIOMETRY
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
C2 MOLE CONCEPT :
In mole concept we deal with different types of relations like weight-weight, weight-volume, orvolume-volume relationship between reactants or products of the reaction.
Mole concept is based on balanced chemical chemical reaction. Some basic definitions used in moleconcept are as follows :
Limiting Reagent : A reagent which is consumed completely during the chemical reaction.
weightmolecularoratomic
substanceofweightn)substance(aofmolesofNumber
Also, number Avogadro
molecules ofnumber Givenn)substance(aofmolesofNumber
In gas phase reaction number of moles of a gas (n) = RT
PV,
At STP/NTP one mole of any gas contains 22.4 L i.e. at 273 K and 1 atm pressure.
In aq. solution n = MV [M - molarity, V - volume of solution]
Practice Problems :
1. Chlorine can be produced by reacting H2SO
4 acid with a mixture of MnO
2 and NaCl. The reactions
follows the equation : 2NaCl + MnO2 + 3H
2SO
4 2NaHSO
4 + MnSO
4 + Cl
2 + H
2O what volume of
chlorine at STP can be produced from 100 g of NaCl ? (At. wt. Na = 23, Cl = 35.5)
(a) 19.15 lt (b) 30 lt (c) 29 lt (d) 5 lt
2. A solution contains 5 g of KOH was poured into a solution containing 6.8 g of AlCl3, find the mass of
precipitate formed [At. wt. : H-1, Al-27, Cl-35.5, K-39]
(a) 2.3 g (b) 23 g (c) 32 g (d) 0.32 g
[Answers : (1) a (2) a]
C3 DIFFERENT WAYS OF EXPRESSING THE CONCENTRATION TERMS :
Important Definitions :
100solutionofmass
soluteofmasspercentmass
L in solution of Vol.
solute of moles of No. )Molarity(M , unit of molarity are mol/lit., M or molar..
L in solution of Vol.
solute of lentsgramequiva of No. )N(Normality , unit of normality are g-eq./lit., N or normal.
solventof.wt
soluteofmolesof.No)m(Molality , unit of molality are mol/kg, m or molal.
BA
AA
nn
n)(fractionMole
x .
610solutionofmass
soluteofmassppm
CS – 2
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
weight Equivalent
solute ofWeight )solute(n of sequivalent gram of .No eq
factorn
weightIonic)or(weightAtomic)or(weightMolecularweightEquivalent
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
xd10M
6.
MMd1000
1000Mm
7.AB
B
M)x1(
1000xm
(d density of solution in g/ml, M molar mass of solute, xB and x
A are mole fraction of solute and
solvent respectively, MA molar mass of solvent)
Calculation of ‘n’ Factor for Different Compounds :
1. Acids : n = basicity
H3PO
4n = 3 H
3PO
3n = 2
H3PO
2n = 1 H
3BO
3n = 1
2. Bases : n = acidity of base
e.g. Ammonia and all amines are monoacidic bases,
NaOH(n = 1), Na2CO
3(aq) n = 2, NaHCO
3(n = 1)
3. Salt : (Which does not undergo redox reactions)
n factor = Total cationic or anionic charge, e.g. Na3PO
4 n = 3, Ba
3(PO
4)
2 n = 6
4. Oxidizing Agents or Reducing Agents : ‘n’ factor = change in oxidation number Ornumber of electron lost or gained from one mole of the compound.
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 orproducts) are equal to each other provided none of these compounds is in excess.
N1V
1 = N
2V
2 (when normalities and volumes are given).
If the number of equivalence of both the reactants are different then reactant with the lesser number ofequivalence 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 ornot balanced but mole concept is used in solving the problems when the reactions are balanced.
Basic principles of tirations :
In voltmetric analysis, a given amount (weight or volume) of an unknown substance is allowed to react witha known volume of a standard solution slowly. A chemical reaction takes place between the solute of anunknown substance and the solute of the standard solution. The completion of the reaction is indicated bythe end point of the reaction, which is observed by the colour change either due to the indicator or due to thesolute itself. Whether the reactions during the analysis are either between an acid and or base or betweenO.A. and R.A., 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
CS– 3
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
(iii) If a given volume of solution is diluted then number of moles or number of equivalence ofsolute 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 thenat the equivalence point, (n
eq) acid
1 + (neq) acid
2 + ... = (n
eq) base
(v) Similarly if a mixture contains more than one oxidising agents then at equivalence point,
(neq
) O.A1 + (n
eq) O.A
2 +... = (n
eq) reducing agent.
(vi) If it is a difficute to solve the problem through equivalence concept then use the mole concept.
Back titration :
This is a method in which a substance is taken in excess and some part of its has to react with anothersubstance 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,Na
2CO
3 and inert impurities.
When the solution containing NaOH and Na2CO
3 is titrated using phenolphalein indicator following reac-
tion takes place at the phenolphthalein end point –
NaOH + HCl NaCl + H2O
Na2CO
3 + HCl NaHCO
3 + H
2O
Here, eq. of HClof.eqCONaof.eq2
1NaOH
)2n(32
When methyl orange is used, Na2CO
3 is converted into NaCl + CO
2 + H
2O
Hence, eq. of NaOH + eq. of )2n(
32CONa
= eq. of HCl
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 typesof bases. Extent of reaction of different bases with acid (HCl) using these two indicatorssummarised below
Phenolphthalein Methyl Orange
NaOH 100% reaction is indicated 100 % reaction is indicated
NaOH + HCl NaCl + H2O NaOH + HCl NaCl + H
2O
Na2CO
350% reaction upto NaHCO
3100% reaction is indicated
stage is indicated Na2CO
3 + 2HCl 2NaCl + H
2O
Na2CO
3 + HCl NaHCO
3 + NaCl + CO
2
NaHCO3
No reaction is indicated NaHCO3 + HCl NaCl + H
2O +
CO2
100% reaction is indicated
CS – 4
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Species Changed to Reactions Electron exchangedor change in O.N.
1.
Eq. wt.
MnO4— (O.A.) Mn2+ in acidic
mediumMnO
4— + 8H+ + 5e— Mn2+ +
4H2O
5
5
ME
2. MnO4— (O.A.) MnO
2 in basic
mediumMnO
4— + 3e— + 2H
2O MnO
2 +
4OH—
33
ME
3. MnO4— (O.A.) MnO
42— in
neutral mediumMnO
4— + e— + 2H
2O MnO
22— 1
1
ME
4. Cr2O
72—(O.A.) Cr3+ in acidic
mediumCr
2O
72— + 14H+ + 6e— 2Cr3+ +
7H2O
66
ME
5. MnO2(O.A.) Mn2+ in acidic
mediumMnO
2 + 4H+ + 2e— Mn2+ +
2H2O
22
ME
6. Cl2(O.A.)
(in bleachingpowder)
Cl— Cl2 + 2e— 2Cl— 2
2
ME
7. CuSO4 (O.A.)
(in iodometrictitration)
Cu+ Cu2+ + e– Cu+ 11
ME
8. S2O
32— (R.A.) S
4O
62— 2S
2O
32— S
4O
62— + 2e— 2 (for two molecules) M
5
ME
9. H2O
2(O.A.) H
2O H
2O
2 + 2H+ + 2e— 2H
2O 2
2
ME
10. H2O
2(R.A.) O
2H
2O
2 O
2 + 2H+ + 2e—
(O.N. of oxygen in H2O
2 is (–1)
per atom)
22
ME
11. Fe2+ (R.A.) Fe3+ Fe2+ Fe3+ + e— 11
ME
Estimation of Reaction Relation between O.A. and R.A.
1. I2
I2 + 2Na
2S
2O
3 2NaI + Na
2S
4O
62—
I2 + 2S
2O
32— 2I— + S
4O
62—
I2 2I— 2Na
2S
2O
3
Eq. wt. (Na2S
2O
3) =
1
ME
2. CuSO4
2CuSO4 + 4KI Cu
2I
2 + 2K
2SO
4 + I
2
or 2Cu2+ + 4I— Cu2I
2 + I
2
white ppt.
2CuSO4 I
2 2Na
2S
2O
3
Eq. wt. of CuSO4 =
1
M
3. CaOCl2
CaOCl2 + H
2O Ca(OH)
2 + Cl
2
Cl2 + 2KI 2KCl + I
2
Cl2 + 2I— 2Cl— + I
2
CaOCl2 Cl
2 I
2 2I— 2Na
2S
2O
3
Eq. wt. of CaOCl2 =
2
M
4. MnO2 MnO
2 + 4HCl (conc.) MnCl
2 +
Cl2 + 2H
2O
Cl2 + 2KI 2KCl + I
2
or MnO2 + 4H+ + 2Cl— Mn2+ + 2H
2O
+ Cl2
Cl2 + 2I— I
2 + 2Cl—
MnO2 Cl
2 I
2 2I— 2Na
2S
2O
3
Eq. wt. of MnO2 =
2
M
5. IO3
— IO3— + 5I— + 6H+ 3I
2 + 3H
2O IO
3— 3I
2 6I 6Na
2S
2O
3
Eq. wt. IO3— =
6
M
CS– 5
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Practice Problems :
1. [Na+] in a solution prepared by mixing 30.00 mL of 0.12 M NaCl with 70 mL of 0.15 M Na2SO
4 is
(a) 0.135 M (b) 0.141 M (c) 0.210 M (d) 0.246 M
2. The equivalent mass of MnSO4 is half of its molar mass when it is converted to
(a) Mn2O
3(b) MnO
2(c) MnO
4– (d) MnO
42–
3. The anion nitrate can be converted into ammonium ion. The equivalent mass of NO3– ion in this
reaction would be
(a) 6.20 g (b) 7.75 g (c) 10.5 g (d) 21.0 g
4. When BrO3– ion reacts with Br– ion in acid solution Br
2 is liberated. The equivalent weight of KBrO
3
in this reaction is
(a) M/8 (b) M/3 (c) M/5 (d) M/6
5. The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous
oxalate in acidic solution is
(a) 3/5 (b) 2/5 (c) 4/5 (d) 1
6. 5 ml of N-HCl, 20 ml of N/2-H2SO
4 and 30 ml of N/3 – HNO
3 are mixed together and the volume
made to 1 litre.
(i) The normality of the resulting solution is
(a) N/5 (b) N/10 (c) N/20 (d) N/40
(d)
(ii) The wt. of pure NaOH required to neutralize the above solution is
(a) 10 g (b) 2 g (c) 1 g (d) 2.5 g
7. 0.7 g of a sample of Na2CO
3.xH
2O 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
6. H2O
2H
2O
2 + 2I— + 2H+ I
2 + 2H
2O H
2O
2 I
2 2I— 2Na
2S
2O
3
Eq. wt. H2O
2 =
2
M
7. Cl2
Cl2 + 2I— 2Cl— + I
2Cl
2 I
2 2I— 2Na
2S
2O
3
Eq. wt. of Cl2 =
2
M
8. O3
O3 + 6I— + 6H+ 3I
2 + 3H
2O + O
2O
3 3I
2
Eq. wt. of O3 =
2
M
9. ClO— ClO— + 2I— + 2H+ H2O + Cl— + I
2ClO— I
2 2I 2Na
2S
2O
3
Eq. wt. of ClO— = 2
M
10. Cr2O
72— Cr
2O
72— + 14H+ + 6I— 3I
2 + 2Cr3+ +
7H2O
Cr2O
72— 3I
2 6I—
Eq. wt. of Cr2O
72— =
6
M
CS – 6
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
8. 100 mL of 1 M KMnO4 oxidised 100 mL of H
2O
2 in acidic medium (when MnO
4– is reduced to Mn2+);
volume of same KMnO4 required to oxidise 100 mL of H
2O
2 in basic medium (when MnO
4–. is
reduced to MnO2) will be
(a) mL3
100(b) mL
3
500(c) mL
3
300(d) 100 mL
9. 100 mL of a mixture of NaOH and Na2SO
4 is neutralised by 100 mL of 0.5 M H
2SO
4. Hence amount
of NaOH in 100 mL mixture is
(a) 0.2 g (b) 0.4 g (c) 0.6 g (d) 1.0 g
10. 3 mol of a mixture of FeSO4 and Fe
2(SO
4)
3 required 100 mL of 2 M KMnO
4 solution is acidic
medium. Hence mol fraction of FeSO4 in the mixture is
(a)3
1(b)
3
2(c)
5
2(d)
5
3
11. 5.3 g of M2CO
3 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) 23 (b) 12 (c) 24 (d) 13
[Answers : (1) d (2) b (3) b (4) c (5) a (6) (i) d (ii) c (7) c (8) b (9) b (10) a (11) a]
C5 VOLUME STRENGTH OF H2O
2
x volume of H2O
2 means x litre of O
2 is liberated by 1 volume of H
2O
2 on decomposition
STPatlit4.2222
gm6822 OOH2OH2
Volume strength of H2O
2 solution = N × 5.6...... (where N is the normality of the H
2O
2
solution
Practice Problems :
1. (a) Calculate the strength of ‘20 V’ of H2O
2 in terms of :
(i) normality (ii) grams per litre (iii) molarity and (iv) percentage
(b) Calculate the volume strength of 2.0 N H2O
2 solution.
2. In a 50 ml solution of H2O
2 an excess of KI and dilute H
2SO
4 were added. The I
2 so liberated required
20 ml of 0.1 N Na2S
2O
3 for complete reaction. Calculate the strength of H
2O
2 in grams per litre.
[Answers : (a) (i) 3.58 N (ii) 60.86 g/lit. (iii) 1.79 M (iv) 6.086% (W/V) (b) 11.2 V (2) 0.68 g/litre]
CS– 7
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
SINGLE CORRECT CHOICE TYPE
1. 1 g of the carbonate of a metal was dissolved in25 ml of N-HCl. The resulting liquid required 5 mlof N-NaOH for neutralization. The eq. wt. of themetal carbonate is
(a) 50 (b) 30
(c) 20 (d) None
2. If 0.5 mol of BaCl2 is mixed with 0.20 mol of Na
3PO
4,
the maximum amount of Ba3(PO
4)
2 that can be
formed is
(a) 0.70 mol (b) 0.50 mol
(c) 0.20 mol (d) 0.10 mol
3. When one gram mol of KMnO4 reacts with HCl,
the volume of chlorine liberated at NTP will be
(a) 11.2 litres (b) 22.4 litres
(c) 44.8 litres (d) 56.0 litres
4. 34 g of hydrogen peroxide is present in 1120 ml ofsolution. This solution is called
(a) 10 vol solution (b) 20 vol solution
(c) 30 vol solution (d) 32 vol solution
5. To prepare a solution that is 0.50 M KCl startingwith 100 mL of 0.40 M KCl
(a) add 0.75 g KCl
(b) add 20 mL of water
(c) add 0.10 mol KCl
(d) evaporate 10 mL water
6. In hot alkaline solution, Br2 disproportionates to
Br– and BrO3–
3Br2 + 6OH– 5Br– + BrO
3– + 3H
2O
hence equivalent weight of Br2 is (molecular
weight = M)
(a)6
M(b)
5
M
(c)5
M3(d)
5
M5
7. When 80 mL of 0.20 M HCl is mixed with 120 mLof 0.15 M KOH, the resultant solution is the sameas a solution of
(a) 0.16 M KCl and 0.02 M HCl
(b) 0.08 M KCl
(c) 0.08 M KCl and 0.01 M KOH
(d) 0.08 M KCl and 0.01 M HCl
8. Molality of 18 M H2SO
4 (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
9. 1 g equiv. of a substance is the weight of that amountof a substance which is equivalent to
(a) 0.25 mol of O2
(b) 0.50 mol of O2
(c) 1 mol of O2
(d) 8 mol of O2
10. The molality of a H2SO
4 solution is 9. The weight of
the solute in 1 kg H2SO
4 solution is
(a) 900.0 g (b) 468.65 g
(c) 882.0 g (d) 9.0 g
11. The density of 1 M solution of NaCl is 1.0585 g/mL.The molality of the solution is
(a) 1.0585 (b) 1.00
(c) 0.10 (d) 0.0585
12. Which is false about H3PO
2
(a) it is tribasic acid
(b) one mole is neutralised by 0.5 molCa(OH)
2
(c) NaH2PO
2 is normal salt
(d) it disproportionates to H3PO
3 and PH
3 on
heating.
13. When KMnO4 acts as an oxidising agent and
ultimately forms [MnO4], MnO
2, Mn
2O
3, Mn2+ then
the number of electrons transferred in each caserespectively is
(a) 3, 5, 7, 1 (b) 1, 5, 3, 7
(c) 4, 3, 1, 5 (d) 1, 3, 4, 5
14. With increase of temperature, which of thesechanges ?
(a) mole fraction
(b) Fraction of solute present in water
(c) molality
(d) weight fraction of solute
15. In a compound C, H and N atoms are present in9:1:3. 5 by weight of compound is 108. Molecularformula of compound is
(a) C2H
6N
2(b) C
3H
4N
(c) C9H
12N
3(d) C
6H
8N
2
CS – 8
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
ANSWERS (SINGLE CORRECTCHOICE TYPE)
11. b
12. a
13. d
14. b
15. d
16. d
17. a
18. b
19. b
20. c
1. a
2. d
3. d
4. a
5. a
6. c
7. c
8. c
9. a
10. b
21. c
22. b
23. c
24. b
25. c
16. 8 g of sulphur are burnt to form SO2 which is
oxidised by Cl2 water. The solution is treated with
BaCl2 solution. The amount of BaSO
4 precipitated
is
(a) 1 mol (b) 0.5 mol
(c) 0.24 mol (d) 0.25 mol
17. One mole of a mixture of CO and CO2 requires
exactly 20 gram of NaOH in solution for completeconversion of all the CO
2 into Na
2CO
3. How many
grams of NaOH would it require for conversion intoNa
2CO
3 if the mixture (one mole) is completely
oxidised to CO2
(a) 60 grams (b) 80 grams
(c) 40 grams (d) 20 grams
18. One gram of a mixture of Na2CO
3 and NaHCO
3
consumes y gram equivalents of HCl for completeneutralisation. One gram of the mixture is stronglyheated, the cooled and the residue treated with HCl.How many gram equivalents of HCl would berequired for complete neutralization ?
(a) 2 y gram equivalent
(b) y gram equivalents
(c) 3y/4 gram equivalents
(d) 3y/2 gram equivalents
19. If equal volumes of 1 M KMnO4 and 1 M K
2Cr
2O
7
solutions are allowed to oxidise Fe (II) to Fe(III),then Fe (II) oxidised will be
(a) more by KMnO4
(b) more by K2Cr
2O
7
(c) equal in both cases
(d) none of these
20. 0.5 g of fuming H2S
2O
7(Oleum) is diluted with
water. This solution is completely neutralized by26.7 ml of 0.4 N NaOH. The percentage of free SO
3
in sample is
(a) 30.6% (b) 40.6%
(c) 20.6% (d) 50%
21. One mole of N2H
4 loses 10 mol of electrons to form
a new compound Y. Assuming that all the nitrogenappears in the new compound. What is theoxidation state of nitrogen in Y.
(a) – 1 (b) – 3
(c) + 3 (d) + 5
22. 25 ml of a solution of barium hydroxide ontitration with a 0.1 molar solution of hydrochloricacid gave a titre value of 35 ml. The molarity ofbarium hydroxide solution was
(a) 0.35 (b) 0.07
(c) 0.14 (d) 0.28
23. To neutralise completely 20 mL of 0.1 M aqueoussolution of phosphorus acid (H
3PO
3), the volume
of 0.1 M aqueous KOH solution required is
(a) 10 mL (b) 20 mL
(c) 40 mL (d) 60 mL
24. Excess of KI reacts with CuSO4 solution and then
Na2S
2O
3 solution is added to it. Which of the
statements is incorrect for this reaction ?
(a) Cu2I
2 formed
(b) CuI2 is formed
(c) Na2S
2O
3 is oxidised
(d) evolved I2 is reduced
25. Two solutions of a substance (non electrolyte) aremixed in the following manner 480 ml of 1.5 M firstsolution and 520 mL of 1.2 M second solution. Whatis the molarity of the final mixture ?
(a) 1.20 M (b) 1.50 M
(c) 1.344 M (d) 2.70 M
CS– 9
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPEComprehension-1A 2.0 g sample of a mixture containing sodiumcarbonate, sodium bicarbonate and sodiumsulphate is gently heated till the evolution of CO
2
ceases. The volume of CO2 at 750 mm Hg pressure
and at 298 K is measured to be 123.9 ml. A 1.5 g ofthe same sample requires 150 ml of M/10 HCl forcomplete neutralisation.
1. The percentage composition of sodium carbonatein the mixture is(a) 42% (b) 18%(c) 53% (d) none
2. The percentage composition of sodiumbicarbonate in the mixture is(a) 42% (b) 18%(c) 53% (d) none
3. The amount of HCl used for completeneutralization of 2g of sample is(a) 0.73 (b) 0.42(c) 0.53 (d) 0.18Comprehension-2Reducing sugars are sometimes characterized by anumber R
Cu, which is defined as the number of mg
of copper reduced by 1 g of the sugar, in which thehalf-reaction for the copper isCu2+ + OH— Cu
2O + H
2O (unbalanced)
It is sometimes more convenient to determine thereducing power of a carbohydrate by an indirectmethod. In this method 43.2 mg of thecarbohydrate was oxidized by an excess ofK
3Fe(CN)
6. The Fe(CN)
64— formed in this reaction
required 5.29 cm3 of 0.0345 N Ce (SO4)
2 for
reoxidation to Fe(CN)6
3— [the normality of thecerium(IV) sulfate solution is given with respect tothe reduction of Ce4+ to Ce3+].
4. The number of milimole of Ce(SO4)
2 used for
re-oxidation to Fe(CN)63– is
(a) 0.183 (b) 0.0915(c) 1.83 (d) 9.14
5. 43.2 mg of sugar was reduced by(a) 11.01 mg Cu2+ (b) 11.6 mg Cu2+
(c) 1.101 mg Cu2+ (d) 1.16 mg Cu2+
6. The RCu
value for the sample is(a) 269 (b) 2.69(c) 26.9 (d) 0.269Comprehension-3An acid solution of a KReO
4 sample containing
26.83 mg of combined rhenium was reduced bypassage through a column of granulated zinc. Theeffluent, including was washing from the column,was then titrated with 0.1000 N KMnO
4; 11.45 mL
of the standard permanganate was required for thereoxidation of all the rhenium to the perrhenate ion,
ReO4
—. Assuming that rhenium was the onlyelement reduced.
7. Number of equivalence of KMnO4 used
(a) 1.145 (b) 1.145 × 10–3
(c) 11.45 × 10–3 (d) 3.25 × 10–3
8. What is the oxidation state to which rhenium wasreduced by the zinc column ?(a) 0 (b) –1(c) –2 (d) –3
MATRIX-MATCH TYPEMatching-1Column - A Column - B
(A) 0.1M of MnO4– (P) oxidised 0.25M
in acidic medium C2O
42–
(B) 0.6 mol of KMnO4
(Q) oxidised 0.5Min acidic medium Fe2+
(C) Molarity of pure water (R) oxidised(density of water = 1g/ml) 0.166M
FeC2O
4
(D) 0.083 molar Cr2O
72– (S) oxidises 1 mol
in acidic medium of Fe(C2O
4)
2
(T) 5.55 × 10Matching-2Column-A Column-B
(A) Total no. of electrons in (P) 3.01 × 1021
1.6 g methane are(B) No. of sulphate ions (Q) 6.02 × 1023
present in 50mL of 0.1MH
2SO
4 solution are
(C) 500cm3 of 0.2 molar NaCl (R) 6.02 × 1022
is added to 100cm3 of0.5 molar AgNO
3. Thus no.
of ions of AgCl formed are(D) The no. of of Fe2+ ions (S) 3.01 × 1022
formed when excess ofiron is treated with 5 mlof 0.04 N HCl
(T) 6.02 × 1020
MULTIPLE CORRECT CHOICE TYPE1. Assuming complete dissociation of H
2SO
4 as
(H2SO
4 2H+ + SO
42–), the number of sulphate ions
present in 50 ml of 0.1 M H2SO
4 solution are
(a) 5 × 10–3 mol (b) 3.01 × 1021
(c) 5 × 1023 (d) 3.01 × 10–3 mol2. In the following redox reaction 2MnO
4– + 10Cl– +
16H+ 2Mn2+ 5Cl2 + 8H
2O. Pick up the correct
statements.(a) MnO
4– is reduced
(b) Cl– is oxidising agent(c) MnO
4– is an oxidising agent
(d) Cl– is reduced
CS – 10
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
3. Which of the following represents redox reaction(s) ?(a) Cu + Cu2+ 2Cu+
(b) MnO4– + Mn2+ + OH– MnO
2 + H
2O
(c) Cr2O
72– + 2OH– 2CrO
42– + H
2O
(d) 2CrO42– + 2H+ Cr
2O
72– + H
2O
4. Which of the following are redox reactions ?(a) Zn + 2HCl ZnCl
2 + H
2
(b) Al(OH)3 + 3HCl AlCl
3 + 3H
2O
(c) Disproportionation of Cu+ ions in a givensolution
(d) Ag+(aq.) + I–(aq.) Ag I (s)5. Which of the following is disproportionation ?
(a) 2Cu+ Cu + Cu2+
(b) 3Cl2 + 6OH– ClO
3– + 5Cl– + 3H
2O
(c) 2H2S + 8O
2 2H
2O + 3S
(d) Na + 2
1Cl
2 NaCl
6. Oxidation number of C is zero in(a) CHCl
3(b) CH
2Cl
2
(c) C6H
12O
6(d) CO
7. Among the species given below which can act asoxidising as well as reducing agent ?(a) SO
2(b) SO
3
(c) H2O
2(d) H
2S
8. Which of the following changes involve oxidation ?(a) change of Zn to ZnSO
4 by reaction with
H2SO
4
(b) change of Cl2 to chloride ion
(c) change of H2S to S
(d) change of sodium sulphite to sodiumsulphate.
Assertion-Reason TypeEach question contains STATEMENT-1 (Assertion)and STATEMENT-2 (Reason). Each question has4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.
(A) Statement-1 is True, Statement-2 is True;Statement-2 is a correct explanationfor Statement-1
(B) Statement-1 is True, Statement-2 is True;Statement-2 is NOT a correctexplanation for Statement-1
(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True
1. STATEMENT-1 : 22.4 L of ethane at N.T.P.contains one mole of hydrogen molecules.STATEMENT-2 : One mole of hydrogenmolecules at N.T.P. occupies 22.4 L of volume.
2. STATEMENT-1 : The masses of oxygen whichcombine with fixed mass of nitrogen in N
2O, NO,
N2O
3 bears a simple multiple ratio.
STATEMENT-2 : The combination according tolaw of multiple proportions.
3. STATEMENT-1 : 8.075 × 10–2 kg of Glauber’s saltis dissolved in water to obtain 1 dm3 of solution ofdensity 1077.2 kg m–3. The molarity of theresultant solution is 0.25 M .STATEMENT-2 : The volume in mL of 0.5 MH
2SO
4 needed to dissolve 0.5 g of copper (II)
carbonate is 24.3 mL4. STATEMENT-1 : Basicity of an acid can change
in different reactions.STATEMENT-2 : As the equivalent weight alwaysremains same.
5. STATEMENT-1 : If equal volume of C M KMnO4
and C M K2Cr
2O
7 solutions are allowed to oxidised
Fe2+ to Fe3+ in acidic medium, then Fe2+ oxidisedby KMnO
4 is more then K
2Cr
2O
7 of same
concentrations.STATEMENT-2 : Number of moles of electronsgained by 1 mole of K
2Cr
2O
7 is more than the 1
mole of KMnO4
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1. c 2. a 3. a 4. b 5. b 6. a
7. b 8. b
MATRIX-MATCH TYPE
1. [A-P, Q, R ; B-S ; C-T ; D-P] 2. [A-Q ; B-P ; C-S ; D-R]
MULTIPLE CORRECT CHOICE TYPE
1. a, b 2. a, c 3. a, b 4. a, c 5. a, b 6. b, c
7. a, c 8. a, c, d
ASSERTION-REASON TYPE
1. D 2. A 3. B 4. C 5. D
CS– 11
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
INITIAL STEP EXERCISE
(SUBJECTIVE)
1. 2.68 × 10–3 mol of a solution containing an ion An+
require 1.61 × 10–3 mol of MnO4– for the oxidation
of An+ to AO3– in acid medium. What is the value of
n ?2. How much 1.00 M HCl should be mixed with what
volume of 0.250 M HCl in order to prepare 1.00 Lof 0.500 M HCl ?
3. Calculate the final concentration of HNO3 if
0.20 mol HNO3 is added to a beaker containing
2.0 L of 1.1 M HNO3 and enough pure water is
added to give a final volume of 3.0 L.4. If 40.00 mL of 1.600 M HCl and 60.00 mL of
2.000 M NaOH are mixed, what are the molarconcentrations of Na+, Cl–, and OH– in theresulting solution ? Assume a total volume of100.00 mL.
5. Calculate the molarity of the original H3PO
4
solution if 20.0 mL of H3PO
4 solution is required to
completely neutralize 40.0 mL of 0.0500 M Ba(OH)2
solution.6. What volume of 96.0% H
2SO
4 solution (density 1.83
g/mL) is required to prepare 2.00 L of 3.00 M H2SO
4
solution ?7. How many mL of 0.5000 M KMnO
4 solution will
react completely with 20.00 g of K2C
2O
4.H
2O
according to the following equation ?16H+ + 2MnO
4– + 5C
2O
42– 10CO
2 + 2Mn2+ + 8H
2O
8. When 50.00 mL of a nitric acid solution was titratedwith 0.334 M NaOH, it requires 42.80 mL of thebase to achieve the equivalence point. What is themolarity of the nitric acid solution ? What mass ofHNO
3 was dissolved in 90.00 mL of solution ?
9. The acidic substance in vinegar is acetic acidCH
3COOH. When 6.00 g of a certain vinegar was
titrated with 0.100 M NaOH, 40.11 mL of base hadto be added to reach the equivalence point. Whatpercent by mass of this sample of vinegar is aceticacid ?
10. Calculate the percent of BaO in 29.0 g of a mixtureof BaO and CaO which just reacts with 100.8 mLof 6.00 M HCl. BaO + 2HCl BaCl
2 + H
2O ;
CaO + 2HCl CaCl2 + H
2O
11. Calculate the normality of each of the followingsolutions : (a) 7.88 g of HNO
3 per L solution (b)
26.5 g of Na2CO
3 per L solution (if acidified to form
CO2).
12. What volumes of 12.0 N and 3.00 N HCl must bemixed to give 1.00 L of 6.00 N HCl ?
13. One gram of a mixture of CaCO3 and MgCO
3 gives
240 ml of CO2 at N.T.P. Calculate the percentage
composition of mixture (Ca = 40, Mg = 24, C = 12,O = 16).
14. (a) What volume of 5.00 N H2SO
4 is required
to neutralize a solution containing2.50 g NaOH ?
(b) How many g pure H2SO
4 are required ?
15. A 0.250 g sample of a solid acid was dissolved inwater and exactly neutralized by 40.0 mL of 0.125N base. What is the equivalent weight of the acid ?
16. Exactly 50.0 mL of Na2CO
3 solution is equivalent
to 56.3 mL of 0.102 N HCl in an acid-baseneutralization. How many g CaCO
3 would be
precipitated if an excess of CaCl2 solution were
added to 100 mL of this Na2CO
3 solution ?
17. How many cm3 of concentrated sulfuric acid, ofdensity 1.84 g/cm3 and containing 98.0% H
2SO
4 by
weight, should be taken to make 1.00 L of normalsolution. Assume complete ionisation.
18. A 40.8 mL sample of an acid is equivalent to50.0 mL of Na
2CO
3 solution, 25.0 mL of which is
equivalent to 23.8 mL of a 0.102 N HCl. What is thenormality of the first acid ?
19. Given the unbalanced equationKMnO
4 + KI + (H)
2SO
4 (K)
2SO
4 + MnSO
4 + I
2 +
H2O
(a) How many g KMnO4 are needed to make
500 mL 0.250 N solution ?(b) How many g KI are needed to make
25.0 mL 0.360 N solution ?20. A solution contains 4 g of Na
2CO
3 and NaCl in
250 ml. 25 ml of this solution required 50 ml ofN/10 HCl for complete neutralization. Calculate %composition of mixture.
21. The density of a 2.0 M solution of acetic acid(MW = 60) in water is 1.02g/mL. Calculate the molefraction of acetic acid.
22. The density of a 2.03 M solution of acetic acid inwater is 1.017 g/mL. Calculate the molality of thesolution.
23. Calculate the (a) molar concentration and (b)molality of a sulfuric acid solution of density1.198 g/cm3, containing 27.0% H
2SO
4 by weight.
24. A solution contains 57.5 mL ethyl alcohol (C2H
5OH)
and 600 mL benzene (C6H
6). How many g alcohol
are in 1000 g benzene ? What is the molality of thesolution ? Density of C
2H
5OH is 0.800 g/mL; of
C6H
6, 0.900 g/mL.
25. One gram of an alloy of aluminium andmagnesium when treated with excess of dil. HClfroms magnesium chloride, aluminium chloride andhydrogen. The evolved hydrogen collected overmercury at 00 C has a volume of 1.20 litres at0.92 atm. pressure. Calculate the composition of thealloy. [H = 1, Mg = 24, Al = 27].
CS – 12
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
26. A gas mixture of 3.0 litres of propane and butaneon complete combustion at 250C produced 10 litresof CO
2. Find out the composition of the gas
mixture.27. In an Industrial process for producing acetic acid,
oxygen gas is bubbled into acetaldehyde CH3CHO,
containing manganese (II) acetate (catalyst) underpressure at 600C.2CH
3CHO(l) + O
2(g) 2CH
3COOH(l)
In a laboratory test of this reaction, 20.0 g CH3CHO
and 10.0 g O2 were put into a reaction vessel. (a)
How many grams of acetic acid can be producedby this reaction from these amounts of reactants ?(b) How many grams of the excess reactantremaining after the reaction is complete ?
28. The hourly energy requirements of an astronautcan be satisfied by the energy released when34 grams of sucrose are “burned” in his body. Howmany grams of oxygen would he needed to becarried in spaces capsule to meet his requirementfor one day ?
29. 1.84 g of a mixture of CaCO3 and MgCO
3 are heated
strongly till no further loss of weight takes place.The residue weighs 0.96 g. Find the percentagecomposition of the mixture.(Mg = 24, Ca = 40, C = 12, O = 16)
30. 1.0 g of a mixture of Potassium chloride andPotassium iodide dissolved in water andprecipitated with Silver nitrate, gave 1.618 g ofsilver halides. Calculate the percentage of each inthe mixture (K = 39, Cl = 35.5, Ag = 108, I = 127).
31. 20% surface sites have adsorbed N2. On heating N
2
gas evolved from sites and were collected at0.001 atm and 298 K in a container of volume is2.46 cm3. Density of surface sites is6.023 × 1014/cm2 and surface area is 1000 cm2, findout the number of surface sites occupied permolecule of N
2.
32. A plant virus is found to consist of uniformcylindrical particles of 150 Å in diameter and5000 Å long. The specific volume of the virus is0.75 cm3/g. If the virus is considered to be a singleparticle, find its molar mass.
33. (a) Calculate the amount of calcium oxiderequired when it reacts with 852 g ofP
4O
10.
243104 )PO(Ca2OPCaO6
[At. wt. : Ca = 40, P = 31, O = 16](b) How many milliliters of 0.5 M H
2SO
4 are
needed to dissolve 0.5 g of copper (II)carbonate ?
224423 COOHCuSOSOHCuCO
34. 1.5 gm of an impure sample of (NH4)
2SO
4 was boiled
with excess of Caustic soda solution in a Kjeldahl’sflask and the ammonia evolved was passed into 200ml of semi-normal H
2SO
4 solution. Thepartially
neutralized acid was made to 500 ml with distilledwater. 25ml of this diluted acid required 40.8 ml ofdecinormal Caustic soda for completeneutralization. Calculate the percentage purity ofAmmonium sulphate.
35. A sample of a metal (M) carbonate was neutralizedby 10 ml of 0.1 N-hydrochloric acid and theresulting chloride gave 0.0517 g of phosphate[M
3(PO
4)
2]. Calculate the eq. wt. of M. (the formula
of Phosphoric acid is H3PO
4). Determine the atomic
weight of M ?
FINAL STEP EXERCISE
(SUBJECTIVE)
1. One litre of a mixture of O2 and O
3 at STP was
allowed to react with excess of acidified solution ofKI. The iodine liberated require 40mL of M/10sodium thiosulfate solution for titration. What isthe mass percentage of ozone in mixture.
222
KI3 OOIO
2. What volume of 3.00 M HNO3 can react completely
with 15.0 g of a brass (90.0% Cu, 10.0% Zn)according to the following equations ?Cu + 4H+(aq) + 2NO
3–(aq) 2NO
2(g) + Cu2+ + 2H
2O
4Zn + 10H+(aq) + NO3–(aq) NH
2+ + 4Zn2+ + 3H
2O
What volume of NO2 gas at 250C and 1.00 atm
pressure would be produced ?
3. A mixture of FeO and Fe3O
4 when heated in air to a
constant weight gains 5 % in its weight. Find thecomposition of initial mixture. (Fe = 56, O = 16)
4. A 10.0 g sample of “gas liquor” is boiled with anexcess of NaOH, and the resulting ammonia ispassed into 60 cm3 if 0.90 N H
2SO
4. Exactly 10.0
cm3 of 0.40 N NaOH is required to neutralize theexcess sulfuric acid (not neutralized by the NH
3).
Determine the percent ammonia in the “gas liquor”examined.
5. A 10.0 mL portion of (NH4)
2SO
4 solution was treated
with excess NaOH. The NH3 gas evolved was
absorbed in 50.00 mL of 0.1000 N HCl. Toneutralize the remaining HCl, 21.50 mL of 0.0980N NaOH was required. What is the molarconcentration of the (NH
4)
2SO
4 ? How many g
(NH4)
2 SO
4 are in 1 L solution ?
CS– 13
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
6. A solution contained the mixture of Na2CO
3 and
NaCl. 25 ml of the solution required 20.4 ml of 0.095N-HCl to convert the carbonate into the chloride.25 ml of the later solution (containing chlorideconverted from carbonate) required 38.76 ml ofN/10-AgNO
3 to ppt, the chloride completely.
Calculate the strength of Na2CO
3 and NaCl in one
litre of the original solution.7. 10 g of a mixture of Cu
2S and CuS was treated with
200 mL of 0.75 M MnO4
– in acidic solutionproducting SO
2, Cu2+ and Mn2+ . SO
2 was bubbled
off and excess of MnO4– was titrated with 175 mL
of 1.00 M Fe2+. Calculate the CuS in the mixture.8. Determine the volume of dilute nitric acid (density
1.11 g/mL, 19.0% HNO3 by weight) that can be
prepared by diluting with water 50 mL of theconcentrated acid (density 1.42 g/mL, 69.8% HNO
3
by weight). Calculate the molar concentrations andmolalities of the concentrated and dilute acids.
9. A 1.00 g sample of KClO3 was heated under such
conditions that a part of it decomposed accordingto the equation(i) 2KClO
3 = 2KCl + 3O
2
and the remaining underwent change according tothe equation(ii) 4KClO
3 = 2KClO
4 + KCl
If the amount of oxygen evolved was 146.8 ml atS.T.P., calculate percentage by weight of KClO
4 in
the residue (K = 39.1, Cl = 35.5)10. To a 25 ml H
2O
2 solution, excess of acidified
solution of potassium iodide was added. The iodineliberated required 20 ml of 0.3 N sodiumthiosulphate solution. Calculate the volume strengthof H
2O
2 solution.
11. A solution contains Na2CO
3 and NaHCO
3. 20 cm3
of this solution requires 5.0 cm3 of 0.1 M H2SO
4
solution for neutralization using phenophthalein asthe indicator. Methyl orange is then added andfurther 5.0 cm3 of 0.2 M H
2SO
4 was required.
Calculate the masses of Na2CO
3 and NaHCO
3 in
1 L of this solution.12. A 1.2 g of a mixture containing H
2C
2O
4.2H
2O and
KHC2O
4.H
2O and impurities of a neutral salt,
consumed 18.9 ml of 0.5 N NaOH forneutralization. On titrating with KMnO
4 solution
0.4 g of the same substance needed 21.55 ml of 0.25N KMnO
4. Calculate the percentage composition
of the substance.13. A mixture of FeO and Fe
2O
3 is reacted with
acidified KMnO4 solution having a concentration
of 0.2278 M, 100 ml of which was used. Thesolution then was treated with Zn dust whichconverted the Fe3+ of the solution to Fe2+ . The Fe2+
required 1000 ml of 0.13 M K2Cr
2O
7 solution. Find
the mass % of FeO and Fe2O
3.
14. 10 g of a mixture of anhydrous nitrates of two met-als A and B were heated to a constant weight andgave 5.531 g of a mixture of the corresponding ox-ides. The equivalent weights of A and B are 103.6and 31.8 respectively. What was the percentage ofA in the mixture ? (N = 14, O = 16).
15. Equal weights of mercury and iodine are allowedto react completely to form a mixture of mercurousand mercuric iodides. Calculate the ratio of theweights of mercurous and mercuric iodides formed.(I = 127, Hg = 201).
16. 5 ml of 8 N nitric acid, 4.8 ml of 5 N HCl acid and acertain volume of 17 M sulphuric acid are mixedtogether and made up to 2 litre. Thirty ml of thisacid mixture exactly neutralise 42.9 ml of sodiumcarbonate solution containing one gram of Na
2CO
3.
10H2O in 100 ml of water. Calculate the mass in
gram of the sulphate ions in solution.17. 200 ml of a solution of mixture of NaOH and
Na2CO
3 was first titrated with phenolphthalein and
N/10 HCl. 17.5 ml of HCl was required for the endpoint. After this methyl orange was added and 2.5ml of same HCl was again required for next endpoint. Find out amounts of NaOH and Na
2CO
3 in
mixture.18. 50 ml of solution, containing 1g each of Na
2CO
3,
NaHCO3 and NaOH, was titrated with N HCl.
What will be the titre readings if(a) only phenolphthalien is used as
indicator ?(b) only methyl orange is used as indicator
from the very beginning ?(c) methyl orange is added after the first end
point with phenolpthalein ?19. A mixture of calcium carbonate and sodium
chloride weighing 3.20 g was added to 100 ml of1.02 N HCl. After the reaction had ceased theliquid was filtered, the residue washed and thefiltrate was made up to 200 ml. 20 ml of this dilutesolution required 25 ml of N/5 NaOH forneutralization. Calculate the % of CaCO
3 in the
mixture. (Ca = 40).20. A solution contains a mixture of sulphuric acid and
oxalic acid, 25 ml of the solution require 35.5 ml ofN/10-NaOH for neutralization and 23.45 ml ofN/10 - KMnO
4 for oxidation. Calculate (a) the
Normality of the solution with regard to sulphuricacid and oxalic acid (b) the number of g of each ofthese substances present in one litre of the solution.
CS – 14
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
1. n = 2 2. 333 mL of 1m HCl 3. 0.80 m4. 0.56 M OH–, 0.640 Cl–, 1.2MNa+ 5. 0.0667 M6. 335 mL 7. 86.9 mL 8. 1.62 gm, 0.286 M9. 4.01 % 10. 65.5 % BaO11. (a) 0.1251 N (b) 0.5 N 12. .33 L, .667 L13. CaCO
3 = 62.5 %, MgCO
3 = 37.5%
14. (a) 12.5 mL (b) 3.07 gm15. 50g/eq 16. 0.575 gm 17. 27.2 cm3
18. 0.119 N 19. (a) 3.95 gm (b) 1.49 gm20. Na
2CO
3 = 66.25%, NaCl = 33.75% 21. 0.038
22. 2.27 m23. (a) 3.30 m (b) 3.78 m 24. 0.54 kg C
6H
6, 1.85 m
25. Al% = 54.87%, Mg% = 45.13%26. 66.66% propane, 33.33 % butane 27. (a) 27.24 gm
(b) 2.73 gm28. 916.2 gm29. CaCO
3 = 54.35%, MgCO
3 = 45.65 % 30. KCl = 39.6 %
KI = 60.4%31. 2 32. 70.96 × 106 g/mol 33. (a) 1008 g (b) 8.097 mL34. 80.96 35. 20.03, 40.06
ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1. 6.57 % 2. L4.10V,L302V23 NOHNO
3. % of FeO = 20.25%
% of Fe3O
4 = 79.75 % 4. 8.5% 5. 0.145 M, 19.1 g/L
6. Na2CO
3 = 4.109 gm, NaCl = 4.535 gm 7. CuS = 57.4 %
8. 15.7 M, 3.35 M, 36.8 m, 3.73 m 9. 49.83%
10. 1.344 11. Na2CO
3 = 5.3 gm, NaHCO
3 = 4.2 gm
12. KHC2O
4.H
2O = 80.9 %, H
2C
2O
4.2H
2O = 14.7 %
13. % FeO = 13.34 %, % Fe2O
3 = 86.66 % 14. 32.28
15. 0.513 : 1 16. 6.528 gm 17. WNaOH
= 0.06gm
gm0265.0W32CONa
18. (a) 34.4 mL (b) 55.8 mL (c) 21.3 mL
19. 81.25% 20. H2SO
4 = 0.0482N, 2.36 g/L
H2C
2O
4 = 0.0938N, 4.225 g/L