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51251.

2H_(2(g))+O_(2(g))to2H_(2)(O)_((l)) This reaction is ____ .

Answer»

Redox
Decomposition
Oxidation
Reduction

Solution :
`therefore` Oxidation
In the REACTION Oxidation number of .H. ATOM increases and at the same time - DECREASES of .O. atom.
`therefore` This is a redox reaction.
Discussion
51252.

2H_(2(g))+CO_(2(g))hArrCH_(3)OH_((g)),DeltaH=-92.2kJ. Which of the following condition will shift the equilibrium in the forward direction ?

Answer»

TEMPERATURE of the SYSTEM is increased
CO is removed
`CH_(3)OH` is ADDED
The PRESSURE of the system is increased

Answer :D
Discussion
51253.

2H_(2) O_(2) to 2 H_(2)O_ + O_(2) . This reaction is

Answer»

Decomposition
Combination
DISPROPORTIONATION reactions
Both 1) and 3)

ANSWER :C
Discussion
51254.

2H_(2(g)) + O_(2(g)) rarr 2H_(2)O_((l)) , DeltaH= -ve and DeltaG= -ve. Then the reaction is

Answer»

SPONTANEOUS and endothermic
Spontaneous and slow
NON spontaneous and slow
Non spontaneous ans slow

ANSWER :C
Discussion
51255.

2g of H_2 and 17g of NH_3 are placed in a 8.21 litre flask at 27^@C. The total pressure of the gas mixture is?

Answer»

<P>4 atm
5 atm
6 atm
3 atm.

Solution :`m=n_(H_2)+n_(NH_3)=2/2+17/17=2`
`P=(NRT)/V=(2xx0.0.0821xx300)/8.21=6`atm
Discussion
51256.

Two gram of hydrogen diffuse from a container in 10 minutes. How many grams of oxygen would diffuse through the same container in the same time under similar conditions?

Answer»

0.5g 
4G 
6G 
8g

Answer :D
Discussion
51257.

2g of Al is treated separately with excess dilute H_2SO_4 andexcess NaOH.The ratio of volumes of Hydrogen evolved under similar condition is x y. Find x/y

Answer»

SOLUTION :`2Al+3H_(2)SO_(4) to Al_(2)(SO_(4))_(3)+3H_(2)`
`2Al+2NaOH+2H_(2) 2NaAlO_(2)+3H_(2)`
Ratio is 1:1
Discussion
51258.

2g of a non electrolyte solute dissolved in 75g of benzene lowered the freezing point of benzene by 0.20 k. The freezing point depression constant of benzene is 5.12 K Kg mol ^(-1). Find the molar mass of the solute.

Answer»

Solution :`W_(2) =2g, W_(1) =75G, Delta T _(f)=0.2 K, k _(f) =5.12 kg mol ^(-1) , M _(2)` = ?
`M _(2) = (K _(f) XX W _(2) xx 1000)/( Delta T_(f) xx W_(1)) = (5.12 xx 2 xx 1000)/( 0.2 xx 75) = 682.66 g mol ^(-1)`
Discussion
51259.

2CuIrarrCu+CuI_(2), the reaction is

Answer»

DISPROPORTIONATION reaction
Neutralisation reaction
Oxidation reaction
Reduction reaction

Answer :A
Discussion
51260.

2CuI rarr Cu + CuI_2the reaction is

Answer»

disproportionation
Neutralisation
OXISATION
REDUCTION

Solution :
Same element undergo OXIDATION and reduction , so it is disproportionation REACTION .
Discussion
51261.

Justify that the reaction : 2Cu_(2) O(s) + Cu_(2)S (s) rarr 6 Cu (s) + SO_(2) (g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.

Answer»

Solution :`OVERSET(+1)(2Cu_(2))overset(-2)(O_((s)))+overset(+1)(Cu_(2))overset(-2)(S_((s)))rarroverset(0)(6Cu_((s)))+overset(+4-2)(SO_(2))`
In this reaction copper is reduced from `+ 1` state to zero oxidation state and SULPHUR is oxidised from `- 2` state to `+ 4` state. The above reaction is thus a redox reaction.
Further, `Cu_(2)S` helps sulphur in `Cu_(2)S` to INCREASE its oxidation number,therefore `Cu(I)` is an oxidant, and sulphur of `Cu_(2)S` helps copper both in `Cu_(2)S` itself and `Cu_(2)O` to decrease its oxidation number, therefore, sulphur of `Cu_(2)S` is reductant.
Discussion
51262.

2C_((s))+ 1/2O_(2(gg) rarr CO_(2(g)) , DeltaH = -787 KJ H_(2(g)) + 1/2O_(2(g)) rarr H_2O_((l)) , DeltaH = -286 KJ C_2H_(2(g)) + 2 1/2O_(2(g)) rarr 2CO_(2(g)) + H_2O_((l)) , DeltaH = -1301 KJ. Heat of formation of acetylene in KJ is

Answer»

`-1802`
`+1802`
`-800`
`+228`

ANSWER :D
Discussion
51263.

2CO(g) + O_(2)(g) hArr 2CO_(2)(g) at 1000 K . What is theK_(C) for this reaction ? Predict the extent of this reaction.

Answer»

SOLUTION :`2CO(g) + O_(2)(g) hArr 2CO_(2)(g) at 1000 K `
`K_(C) = 2.2 xx 10^(22)`
`K_(C) gt 10^(3).` So [products ] >> [Reactants]
Reaction nearly goes to completion and FORWARD reaction is favoured.
Discussion
51264.

2ClCH_(2)CH_(2)OH("excess") underset(415K)overset(Conc. H_(2)SO_(4))rarr X underset("Fuse")overset(KOH)rarr Y The compound Y is

Answer»

Ethylene glycol
Ethanol
Divinyl ether
Vinyl chloride

Solution :`2ClCH_(2)CH_(2)OH ("EXCESS") underset(Delta//4.15 K)overset(CONC. H_(2)SO_(4))rarr ClCH_(2)underset((X))(CH_(2)-)O-CH_(2)CH_(2)CL overset(KOH//"FUSE")rarr underset("Divinyl ether (Y)")(CH_(2)=CH-O-CH=CH_(2))`
Discussion
51265.

2CaSO_4(s)hArr2CaO(s)+2SO_2(g)+O_2(g),DeltaH gt 0 Above equilibrium is established by taking some amount of CaSO_4(s) in a closed container at 1600K. Then which of the following may be correct option?

Answer»

Moles of CaSO(s) will increase with increase in temperature.
If the volume of the container is doubled at equilibrium then partical PRESSURE of `SO_2(g)` will change at new equilibrium
If the volume of the container is halved pressure of `O_2(g)` at new equilibrium will remain same.
If TWO moles of the He gas is added at constant pressure then the moles of CAO(s) will increase.

Answer :A::C::D
Discussion
51266.

2CaSO_(4)(s)hArr22CaO(s)+2SO_(2)(g)+O_(2)(g), DeltaHgt0 Above equilibrium is established by taking some amout of CaSO_(4)(s) in a closed container at 1600K Then which of the following may be correct option.

Answer»

Moles of `CAO(s)` will increase with the increase in temperature
If the voulme of the container is doubled at equilibrium then partial pressure of `SO_(2)(g)` will CHANGE at new equilibrium
If the volume of the container is halved partial pressure of `O_(2)(g)` at new equilibrium will remain same
If two moles of the He gas is added at constant pressure then the moles of `CaO(s)` will increase.

Solution :(A) As reaction is endothermic therefore it will GO in the forward direction hence moles of `CaO` will increase.
(B) With the increase or decrease of volume partial pressure of the gases will remain same.
(C) Due to the ADDITION of inert gas at constant pressure reaction will proceed in the direction in which more number of GASEOUS moles are formed.
Discussion
51267.

2B(OH)_(3) +2NaOHhArr NaBO_(2) + Na[B(OH)_(4)]+2H_(2)O How can this reaction be made to proceed in forwards direction ?

Answer»

ADDITION of cis-1, 2-diol
addition of borax
addition of trans-1, 2-diol
addition of `Na_(2)HPO_(4)`

ANSWER :A
Discussion
51268.

298 K of N_2O_(4(g)) hArr 2NO_(2(g))reaction is in equilibrium K_p = 0.14 atm. Calculate of K_c (R=0.082 L atm K^(-1) "mol"^(-1))

Answer»

SOLUTION :`5.73xx10^(-9)`M
Discussion
51269.

29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The exess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is

Answer»

29.5
`59.0`
47.4
23.7

Solution :Weight of substance TAKEN `= 29.5 xx 10^(-3) g`
Volume of ACID taken 20 mL of 0.1 N HCl
`:.` Volume of acid used = 20 - 15 = 5 mL of 0.1 N HCl
`%N= (1.4 xx "Volume" xx "Normality")/("Weight of substance taken") = (1.4 xx 5 xx 0.1)/(29.5 xx 10^(-3))`
`= 23.73`
Discussion
51270.

293 K = .......... ""^(@)F.

Answer»

273
68
293
77

Solution :`293K= 293 - 273 = 20 ""^(@)C`
`F = 9/5xx(C^(@))+32= 9/5xx(20)+32=68""^(@)F`
Discussion
51271.

2.9 g of a gas at 95^(@)C occupied the same volume as 0.184 g of hydrogen at 17^(@)C at the same pressure. What is the molar mass of the gas ?

Answer»

Solution :As `"" P_(1)=P_(2)`
and `"" V_(1)=V_(2)`, But PV=nRT
`:. "" P_(1)V_(1)=P_(2)V_(2)`, i.e., `n_(1)RT_(1)=n_(2)RT_(2)`
`:. "" n_(1)T_(1)= n_(2)T_(2) "or" (w_(1))/(M_(1))T_(1)=(w_(2))/(M_(2))T_(2)`
i.e, `"" (2.9)/(M_(X))XX(95+273)=(0.184)/(2)xx(17+273)`
or `"" M_(X)=(2.9xx368xx2)/(0.184xx290)=40" G mol"^(-1)`
Discussion
51272.

2.9 g of a gas at 95^@C occupied the same volume as 0.184 g of dihydrogen at 17^@C, at the same pressure. What is the molar mass of the gas?

Answer»

Solution :Suppose the molar mass of the GAS is M. According to the gas equation , `PV = NRT`. Since P and V are the same, we have
For the gas :
`PxxV = (2.9)/M xxRxx(273 +95)` ...(i)
For `H_2 : PxxV= (0.184)/2 xx R xx (273+17)` ...(ii)
From EQUATIONS (i) and (ii), we have
`(2.9)/M xx368 = (0.184)/2 xx 290`
or `M = (2.9 xx 368 xx 2)/(0.184xx290)= 40 g MOL(-1)`
Discussion
51273.

2.9 g of a gas at 95^(@)C occupied the same volume as 0.184 g of dihydrogen at 17^(@)C, at the same pressure. What is the molar of the gas ?

Answer»

SOLUTION :According to ideal gas equations,
`pV=nRT=(wRT)/(M) "" therefore M=(wRT)/(pV)`
`therefore (pV)/(R )=(WT)/(M)` where `(pV)/(R )=` constant of number
for 1 gas
`w=w_(1)=2.9` gram
R = Gas constant
`T=T_(1)=95^(@)C`
`= (95+273)=368 K`
p = p bar
V = V L
`M = M_(1)=M_(1)g mol^(-1)`
For `H_(2)` gas :
`w_(2)=0.184` gram
R = Gas constant
`T_(2)=(17+273)K=290 K`
p = p bar
V = V L
`M=M_(2)=2.0 g mol^(-1)`
`therefore (pV)/(R )=(w_(1)T_(1))/(M_(1))=(w_(2)T_(2))/(M_(2))=` constant
`therefore M_(1)=(w_(1)T_(1))((M_(2))/(w_(2)T_(2)))=((2.9g)(368 K)(2g mol^(-1)))/((0.184 g)(290 K))`
`= 40 g mol^(-1)`
Discussion
51274.

2.84 g of methyl iodide was completely converted into methyl magnesium iodide and was decomposed by excess of ethanol. The volume of the gaseous hydrocarbon proudced at NTP will be

Answer»

22.4 litre
224 mL
0.448 litre
0.224 litre

Solution :`UNDERSET("one MOLE")(CH_(3)I)overset(Mg)tounderset("onemole"(CH_(3)Mgl)overset(ETOH)tounderset("one mole")(CH_(4))`
Discussion
51275.

2.82g of glucose is dissolved in 30g of water. The mole fraction of glucose in the solution is

Answer»

0.01
`0.99`
`0.52`
1.66

SOLUTION :No. of moles of glucose `=(2.82)/(180)=0.01567`
No. of moles of WATER `=30/18=1.667`
Total of moles of solution =0.01567+1.667=1.683
Moles fraction glucose `=(0.01567)/(1.683)=0.0093=0.01`
Discussion
51276.

2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water.

Answer»

Solution :MASS of glucose `= 2. 82 G`
No. of moles of glucose `= (2. 82)/( 180) = (30)/(18) = 1. 67`
` X _(H _(2)O)= (1. 67)/( 1. 67 + 0.016) = (1. 67)/( 1. 686) =0.99`
`THEREFORE x _(H _(2)O) + x _("glucose") =1`
` 0.99 + x _("glucose")=1`
`x _("glucose")=1-0.99 =0.01`
Discussion
51277.

28 NO_(3)^(-) + 3As_(2)S_(3) + 4H_(2)O to 6AsO_(4)^(3-)+28 NO+ 9SO_(4)^(2-) + 8H^(+). What will be the equivalent mass of As_(2)S_(3) in above reaction ?

Answer»

`(M. WT.)/(2)`
`(M. wt.)/(4)`
`(M. wt.)/(24)`
`(M. wt.)/(28)`

SOLUTION :N//A
Discussion
51278.

28 gm KOH is dissolved in 90 g water. Find mole Fraction of KOH and water respectively.

Answer»




ANSWER :MOLE FRACTION of `KOH=0.0909`,
Mole fraction of `H_(2)O=0.909`
Discussion
51279.

28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17 g NH_(3) was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.

Answer»

SOLUTION :GIVEN`m_(N_(2))=28G,m_(H_(2))=6g,`
`V=1L`
`(n_(N_(2)))_("INITIAL")=(28)/(28)="1 mol"`
`(n_(H_(2)))_("Initial")=6/2=3" mol"`
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`

`[NH_(3)]=((17)/(17))="1 mol"`
Weight of `N_(2)` = (no. of moles of `N_(2)` ) `xx` molar mass of `N_(2)`
`=0.5xx28=14g`
Weight of `H_(2)` = (no. of moles of `H_(2)` ) `xx` molar mass of `H_(2)`
`=1.5xx2=3g`
Discussion
51280.

28 g of nitrogen and 6 g hydrogen were mixed in a1 litre closed container. At equilibrim 17 g NH_3 was produced. Calculate the weight of weight of nitrogen, hydrogen at equilibrium.

Answer»

Solution :GIVEN `m_(N_2) = 28G, m_(H_2) = 6 g , V = 1L`
`(n_(N_2))_("initial")= 28/28 = 1mol`
`(n_(H_2))_("initial")= 6/2 = 3mol`
`N_2(g)+ 3H_2(g) hArr 2NH_3(g)`

`[NH_3] = (17/17) = 1mol = 2x`
`rArrx = 0.5` mol
At equilibrium, `[N_2] = 1 - x = 0.5 mol`
`[H_2] = 3 -3x =3-3 (0.5) = 1.5 mol`
WEIGHT of`N_2` = (no. of moles of `N_2`)` xx`molar MASS of `N_2 = 0.5 xx 28 = 14g`
Weight of `H_2` = (no. of molesof `H_2`) `xx` molar mass of `H_2= 1.5 xx 2 = 3g`
Discussion
51281.

28 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absored in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. Then the percentage of 'N' is the compound is

Answer»

50
30
25
40

Solution :Excess ACID `= 15xx0.1xx1` m. eq.
Acid taken initially `= 20xx0.1xx1` m.eq.
Acid used for neutralisation
`=(20xx0.1xx1)-(15xx0.1xx1)=0.5` m. eq.
`% N =(NH_(3)" liberated"xx1.4)/("wt.of ORG. COMPOUND")=(0.5xx1.4)/(0.028)=25%`.
Discussion
51282.

28 g of each of the following gases are taken at 27^(@)Cand 600 mm pressures. Which of these will have the least volume?

Answer»

<P>HBr
HCl
HF
HI

Solution :PV=NRT or `V=(nRT)/(P)`. As P and T are same for all given gases, `V prop n`. As `n=(w)/(M)`, n will be LEAST for which M is maximum viz. HI.
Discussion
51283.

27g of Al was treated with NaOHsolution when a whitegelatinous precipitate was obtainedwhichuponstrongheatinggave an oxide. The amount ofoxide (in g ) is

Answer»


SOLUTION :`UNDERSET(54 g)(2Al)overset(NaOH)RARR 2A(OH)_(3)overset(Delta)rarrunderset(102 g)(Al_(2)O_(3))`
`54 g` of Al produce `Al_(2)O_(3) = 102//2 = 51.0 g`
Discussion
51284.

2.79 g of an organic compound when heated in Carius tube with cone. HNO_3 " and "H_3PO_4 formed converted into MgNH_4.PO_4 ppt. The ppt. on heating gave 1.332 g of Mg_2P_2O_7.The percentage of P in the compound is

Answer»

<P>23.33
13.33
33.33
26.66

Solution :`% of P = (62 XX 1.332 xx 100)/(222xx2.79)=13.33`
Discussion
51285.

2.76 g of silver carbonate on strong ignition leaves a residue weighing

Answer»

2.48g
2.16g
2.32g
2.84g

Solution :`2Ag_(2)CO_(3)overset(DELTA)rarrunderset("Residue")(4Ag)+2CO_(2)uarr+O_(2)uarr`
Discussion
51286.

2.76 g of Ag_2CO_3 on being heated yields a residue weighing

Answer»

A)2.16 g
B)2.32 g
C)2.48 g
D)2.64 g

SOLUTION :`underset(2 XX 276 g)(2Ag_(2)CO_(3)) to underset(4 xx 108 g)(4 Ag) + 2CO_(2) uarr + O_(2) uarr`
`therefore` RESIDUE given by 2.76 g of
`Ag_(2)CO_(3) =(4 xx 108)/(2 xx 276)xx 2.76 = 2.16 g`
Discussion
51287.

2.746 g of a compound on analysis gave 1.94 g of silver, 0.268 g of sulphur and 0.538 g of oxygen. Find the empirical formula of the compound.

Answer»

Solution :The PERCENT composition of elements can be calculated as FOLLOWS :
% Ag `=1.94/2.746 XX 100 = 70.6 %`
`5 S = 0.268/2.746 xx 100 = 9.76 %`
`% O = 0.538/2.746 xx 100 = 19.6 %`
Calculation of empirical FORMULA:

`therefore` The empirical formula of the given compound is `Ag_(2)SO_(4)`.
Discussion
51288.

273 mL of a gas at STP was taken to 27^(@)C and 600 mm pressure. The final volume of the gas would be :

Answer»

273 mL
300 mL
380 mL
586 mL

Answer :C
Discussion
51289.

273 K is equal to _______ degree centigrade.

Answer»

0
100
373
1

Answer :A
Discussion
51290.

27 g of Al (at mass=27) will react completely with oxygen equal to

Answer»

24 g
8g
40 g
10 g

Solution :a) The equation of the reaction of aluminium and oxygen, is as
`4Al + 3O_(2) to 2Al_(2)O_(3)`
therefore 108 g of AI reacts with 96g of `O_(2)`.
therefore 27G of AI with react with `=(96 XX 27)/(108)` = 24 g of `CO_(2)`
Discussion
51291.

26c.c of CO_(2) is passed over red hot coke. The volume of CO evolved is (under the same condition)

Answer»

15 c.c
10 c.c
32 c.c
52 c.c

Answer :D
Discussion
51292.

26.8g of Na_(2)SO_(4). nH_(2)O contains 12.6 g of water. The value of n is:

Answer»

1
10
6
7

Answer :D
Discussion
51293.

2.68 xx10^(-3) moles of a solutioncontaing an ion A^(n+) require 1.61 xx10^(-3) moles MnO_(4)^(-) for the oxidation of A^(n+) to AO_(3)^(-) in acid medium what is the value of n ?

Answer»

Solution :STEP 1 To write the reduction and oxidation half reaction
Reducton : `Mno_(4)^(-) + 8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
oxidation `A^(n+)+3H_(2)OrarrAO_(3)^(-)+6H^(+)+(5-n)E^(-)`
Step 2 To FIND out the VALUE of n
SINCE in a REDOX reaction number of electrons lost = number of electrons gained therefore multiple oxidant of Eq (i) i.e `MnO_(4)^(-)` by (5-n) and reductant of Eq (ii) i.e `A^(n+)` by 5 adn equate we have (5-n) `MnO_(4)^(-)=5 A^(n+) i.e (5-n)` moles of `MnO_(4)^(-)` will oxidise `An^(n+)` = moles
Equationg the values of moles of `A^(n+)` actually oxidised =`2.68xx10^(-3)` moles
Equating the value of Eq (iii) and (iv) we have `(5)/(5-n) xx1.61xx10^(-3)=2.68xx10^(-3) or5xxx1.61=(5-n)xx2.68 or 2.68 n = 5(2.68-1.61)=5xx1.07=5.35 or n =(5.35)/(2.68)=2`
Discussion
51294.

26.8 g of Na_2SO_4.xH_2O gave 12.6 g of water on heating. Determine the value of x in the compound using mole concept.

Answer»

Solution :Weight of hydrated salt `Na_2SO_4.xH_2O = 26.8 g`
Weight of water given by it = 12.6 g
`therefore` Weight of anhydrous salt `Na_2SO_4 = 26.8 - 12.6 = 14.2g`
Gram molecular mass of `Na_2SO_4 = (2 xx 22.99) + 32.06 + (4 xx 16.0) = 142.04 g`
`therefore` Number of moles of `Na_2SO_4 = 142 04/(142.04) = 0.0999`
and number of moles of `H_(2)O`associated with it `12.6/18.016`
`therefore 0.0999` moles of `Na_2SO_4` ATTACH moles of water = `0.699 , therefore` 1 MOLE of `Na_2SO_4` will attach moles of water
`=0.699/0.0999 xx 1`
= 6.99
=7, Hence, x=7
Discussion
51295.

2.650 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 500 mL. On titration 50 mL of this solution neutralises 50 mL of a solution of sulphuric acid. How much water should be added to 450 mL of this acid as to make it exactly N//12 ?

Answer»


SOLUTION :N//A
Discussion
51296.

2.65 g of diacidic base was dissolved in water and made up to 500 mL. 20 of this solutioncompletely neutralised 12 mL of N//6 HCl. Find out the equivalent mass and molecular mass of the base.

Answer»


SOLUTION :N//A
Discussion
51297.

2.5mLof 2//5M weak mono-acidic base (K_(b) = 1 xx 10^(-12) at 25^(@)C) is titrated with 2//15M HCI in water at 25^(@)C. Find the concentration of H^(o+) ions at equivalence point. (K_(w) = 1 xx 10^(-14)at 25^(@)C) a. 3.7 xx 10^(-13)M b. 3.2 xx 10^(-7)M c. 3.2 xx 10^(-2)M d. 2.7 xx 10^(-2)M

Answer»

Solution :a. First find the VOLUME of `HCI` re2uired to reach the eqivalent point.
mEq of base = mEq of HCI
`rArr (2.5 xx (2)/(5) xx 1) = ((2)/(15)xx1)V_(HCI)`
`rArr V_(HCI) = 7.5mL`
The net volume of the solution at the equivalent point
`= V_("base") +V_(HCI) = 2.5 +7.5 = 10mL`
`rArr ["salt"] = c (2//5 xx 2.5)/(10) = 0.1M`
`ph` of an aqueous solution of such a salt is given by :
`pH = 7 - (1)/(2) (pK_(b) +logC)`
`= 7-(1)/(2) (12 + log 0.1) = 1.5`
`rArr [H^(o+)] = "Antilog" (-1.5)`
`= "Antilog" [(-1-0.5 +1-1) = bar(2).5]`
`= 3.2 xx 10^(-2)M`
or
`[H^(o+)] = 10^(-15) = (1)/(10sqrt(10)) = 3.2 xx 10^(-2)M` (Answer 'c')
Please note that answer (c) of the above solution is incorrect. (Why ?)
Actually, `pH = 7 - (1)/(2) (pK_(b) + logC)` is valid only when
`1-h = 1`.
To CHECK, claculate `h` using :
`h = sqrt((K_(h))/(c)) = sqrt((K_(w))/(K_(b)C)) = sqrt((10^(-14))/(10^(-12)xx0.1)) = sqrt(0.1)`
`rArr 1 - sqrt(0.1) != 1`
So, we have to solve from basics as follows:
`{:(B^(oplus),+,H_(2)O,hArr,BOH + ,H^(oplus)),(C-Ch,,,,Ch,Ch):}`
`rArr K_(h) = ([BOH][H^(o+)])/([B^(o+)]) = ((Ch)(Ch))/(C-Ch)`
`K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(10^(-12)) = 10^(-2) = 0.01`
Also,
`K_(h) = (Ch^(2))/((1-h))` (Not neglecting `h` and solving, we get)
`0.01 = (0.1 xxh^(2))/(1-h)`
or `h~~ 0.271`
`[H^(o+)] = Ch = 0.27 xx 0.1 = 0.0271 = 2.71 xx 10^(-2) M`
Hence answer is (d).
Note: In such problem, the answer cna be obtained DIRECTLY. The answer (c) `(3.2 xx 10^(-2)M)` is calculated by using the formula of salt of `W_(B)//S_(A)`. The concentration of salt is `0.1M` (which is very high). So `(1-h)` cannot be taken equal to `1`, since the two answer (c) and (d) are very close to each other. THEREFORE, `[H^(o+)]` would slightly less than the calculated value.
`:. [H^(o+)]` calculated by using formula `= 3.2 xx 10^(-2)M`
`[H^(o+)]` observed (by not using approximation) `= 2.7 xx 10^(-2)M`
In such problem, if the `pH` is asked, calculated it by using formula will be slightly less than the observed `pH` (by not using approcimation) e.g., `pH` (calculated by using formula) `=- log (3.2 xx 10^(-2))`
`=- log [(2)^(5) x 10^(-3)]`
`=-5 log 2+3 = 1.5`
`pH` (observed) (by not usingapproximation
`=- log (27 xx 10^(-3))`
`=- 3log 3+3 = 1.56`
Discussion
51298.

25mL of a solution of ferric alum Fe_(2)(SO_(4))_(3) (NH_(4))_(2) (SO_(4))//24H_(2)O containing1.25g of the salt was boiled with iron when the reaction Fe + Fe_(2) (SO_(4))_(3) rarr 3FeSO_(4)occurredtreated with 0.107N KMnO_(4) in acid medium. What is titre value? If Cu has been used in place of Fe what would have been titre value?

Answer»


ANSWER :`36.36mL, 24.24 ML`
Discussion
51299.

25mLof decimolar solution of a stable divalent transition metal cation is oxidised by 50mL of 0.02M acidified permanganate solution. What is the final oxidation state of the metal?

Answer»

Solution :5 moles of `M^(2+)` ion =x moles of permanganate , where x is the increase in oxidation number of the metal ion, `M^(2+)`
Using the MOLARITY EQUATION for the TITRATION
`(V_(1)M_(1))/(5)=(V_(2)M_(2))/(x)`
The final oxidation state of metal `=+2-(-2)=+4`
Discussion
51300.

25mL of a solutioncontainingFe^(2+) and Fe^(3+) sulphate acidifed with H_(2)SO_(4) is reducedby 3g of metallliczinc. The dsolution required 34.25mL of N//10 solution of K_(2)Cr_(2)O_(7) for oxidation. Beforereduction with zinc, 25mL of the same solution. Calculate the strengthof FeSO_(4) and Fe_(2) (SO_(4))_(3) in solution.

Answer»


ANSWER :`FeSO_(4) = 13.64g//"LITRE'`,
`Fe_(2)(SO_(4))_(3) = 9.45g//"litre"`
Discussion