Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Determine the percentage of water of crystallisation, iron , sulphur and oxygen in pure ferrous sulphate (FeSO_(4).7H_(2)O).

Answer»

Solution :The formula mass of of FERROUS SULPHATE
=At. Mass of Fe+At. Mass of S+4X At. Mass of oxygen
+7X Mol. Mass of `H_(2)O`
`=56.0+32.0+4xx16.0+7xx18.0=278.0`
So, % of water of crystallisation `=(126)/(278)xx100=45.32`
% of iron `=(56)/(278)xx100=20.14`
% of SULPHUR=`(32)/(278)xx100=11.51`
% of oxygen`=(64)/(278)xx100=23.02`
Discussion
2.

Determine the oxidation states of central metal ions in the following complex compounds : (i) (a) K_(4)[Fe(CN)_(6)] (b) [Co(NH_(3))_(6)]Cl_(3) (ii) (a) K_(3)[Fe(CN)_(6)] (b) [PtCl_(4)]^(2-) (c ) [CoBr_(2)(CN)_(2)]^(+) (iii) (a) [Pt(NH_(3))_(3)Cl_(3)] ion (b) Na_(4)[Ni(CN)_(4)] (c ) [Zn(H_(2)O)_(3)OH]^(+)

Answer»

Solution :(i) (a) + 2 (B) + 3 (ii) (a) + 3 (b) + 2 (c ) + 5 (III) (a) + 2 (b) 0 (c ) + 2
Discussion
3.

Determine the percentage composition (by mass) ofmixture of anhydrous sodium carbonate and sodium bicarbonate from the following datam: wt. of the mixture taken = 2g Loss in weight on heating = 0.124 gm. Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

Answer»


Answer :`NaHCO_(3)=16.8%;Na_(2)CO_(3)=83.2%`
Discussion
4.

Determine the per cent concentration of a solution obtained by mixing 300g of a 25% and 400g of a 40% solution.

Answer»


ANSWER :`33.6%`
Discussion
5.

Determine the packing efficiency of the unit cell in hexagonal close-packed (hcp) structure of particles.

Answer»

Solution :The unit cell in an hcp structure of particles is a hexagonal unit cell as shown in the given figure. The number of particles in the hexagonal unit cell = 6. If radius of each the

particles MAKING up the unit cell is = r, then total volume for 6 particles = `6xx4/3pir^(3)`.
The area of the BASE of a hexagonal unit cell = `=6xxsqrt3/4xx4r^(2)=6sqrt3r^(2)`.
The height of this unit cell = `4rsqrt(2/3)`
Therefore, the volume of unit cell
= area of the base `xx` height = `6sqrt3r^(2)xx4rsqrt(2/3)=24sqrt2r^(3)`
`therefore" "` Packing efficiency of the hexagonal unit cell
`=("Volume of 6 particles")/("Volume of unit cell")xx100=(6xx4/3pir^(3))/(24R^(3)sqrt2)xx100`
= 74.01%
Discussion
6.

Determine the osmotic pressure of a solution prepared by dissolving 2.5xx10^(-2)"g of "K_(2)SO_(4) in 2 L of water at 25^(@)C, assuming that it is completely dissociated. ("R = 0.0821 L atm K"^(-1)"mol"^(-1)," Molar mass of "K_(2)SO_(4)="174 g mol"^(-1))

Answer»


Solution :`K_(2)SO_(4)rarr2K^(+)+SO_(4)^(2-) thereforei=3,""C=(2.5xx10^(-2))/(174)xx(1)/(2)=7.18xx10^(-5)"mol L"^(-1)`
`pi="i CRT,"pi=(3)(7.18xx10^(-5)" mol L"^(-1))("0.0821 L atm K"^(-1)"mol"^(-1))(298K)=5.27xx10^(-3)"atm"`
Discussion
7.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K_2SO_4in 2 litre of water at 25^@C , assuming that it is completely dissociated.

Answer»

SOLUTION :` K_2SO_4` dissolved = 25 mg = 0.025 g`T = 25^@C = 298K`
Molar mass of `K_2SO_4 = 2 xx 39 + 32 + 4xx16 = 78 + 32 + 64 = 174 g "mol"^(-1)`
As `K_2SO_4` dissociates completely as `K_2SO_4 to2K^(+) + SO_4^(2-)`
i.e., ions produced = 3 ` therefore i=3`
Applying the following equation
`pi = iCRT = i n/V RT=i xx w/W xx 1/V RT`
Substituting the values, we get
`pi = 3 xx (0.025 g)/(174 g "mol"^(-1)) xx (1)/(2L)xx 0.0821 L atm K^(-1) "mol"^(-1) xx 298 K`
`= 5.27 xx 10^(-3) atm`
Discussion
8.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K_(2)SO_(4) in 2 litre of water at 25^(@)C, assuming that it is completely dissociated.

Answer»

Solution :`K_(2)SO_(4)to2K^(+)+SO_(4)^(-2)`
Total number of ions produced = 3
Thereforem i = 3
Given,
w = 25 mg = 0.025 g
V = 2 L
`T=250^(@)C=(25+273)K`
= 298 K
Also, we know that :
`R=0.0821" L atm K"^(-1)mol^(-1)`
`M=(2xx39)+(1xx32)+(4xx16)`
`= 174 mol^(-1)`
Appling the following relation,
`pi=i(n)/(V)RT`
`= i(w)/(M)(1)/(v)RT`
`=(0.025)/(174)XX(1)/(2)xx0.0821xx298`
`=5.27xx10^(-3)` atm.
Discussion
9.

Determine the osmotic pressure of a solution prepared by dissolving "25 mg of "K_(2)SO_(4) in 2 litre of water at 25^(@)C, assuming that it is completely dissociated.

Answer»

Solution :`K_(2)SO_(4)" dissolved = 25 mg = 0.025 g"`
`"Molar MASS of "K_(2)SO_(4)=2xx39+32+4xx16="174 g mol"^(-1)`
As `K_(2)SO_(4)` dissociates completely as `K_(2)SO_(4)rarr 2K^(+)+SO_(4)^(2-)`, i.e., ions produced = 3, `therefore i=3`
`thereforepi=iCRT=i(n)/(V)RT=ixx(w)/(M)xx(1)/(V)RT=3xx("0.025 g")/("174 g mol"^(-1))xx(1)/(2K)xx"0.0821 L atm K"^(-1)"mol"^(-1)xx"298 K"`
`=5.27xx10^(-3)" atm."`
Discussion
10.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K_2 SO_4 in 2 litre of water 25^(@)C assuming that it is completely dissociated .

Answer»

Solution :If `K_(2) SO_(4)` is completely dissociated , `K_(2) SO_(4) to 2 K^(+) + SO_(4)^(2-)`
` t = 3`
Mol mass of `K_(2) SO_(4) = 2 xx 39 + 32 + 4 xx 16 = 174 g mol^(-1)`
`pi = iCRT`
`= i(W_(B) xx RT)/(M_(B) xx V)`
`= (3 xx 25 xx 10^(-3) xx 0.082 xx 298)/(174 xx 2.0)`
`= 5.27 xx 10^(-3)` atm
Discussion
11.

Determine the osmotic pressure of a solution prepa.red by diHolving 2.5xx 10^(-2) g of a solution 2I. Of water at 25^(@) C, assuming that it is completely dissociated. (R = 0.0821 " L atm K"^(-1) mol^(-1) , " molar mass of" K_(2)SO_(4) = 174 g " mol"^(-1))

Answer»

Solution :When `K_(2) SO_(4)` is dissolved in water, IONS are produced.
Total number of ions produced = 3
i= 3
`pi = I"CRT " = (I xx N xx R xx T)/(V)`
`pi = 3 xx (25 xx 10^(-2)g)/(174 g "MOL"^(-1)) xx (1 xx 0.0821 " L ATM " xx 298)/(2I)
= 0.00176 " " K^(-1) mol^(-1) xx 298` K
`pi5.27 xx 10^(-3) ` atm
Discussion
12.

Determine the order of stability of the following resonating structure.

Answer»

`igtiigtiii`
`iiigtiigti`
`igtiiigtii`
`iigtigtiii`

ANSWER :A
Discussion
13.

Determine the order of reaction : Step 1. 2NO+H_2 to N_2+H_2O_2 Slow Step 2. H_2O_2 + H_2 to 2H_2O Fast

Answer»

Solution :Rate =`k[NO]^2[H_2]`
ORDER =2+1
=3
Discussion
14.

Determine the number of optical isomers in the following compounds: (a) CH_(3)CH(OH)COOH (b) HOOC(CHOH)_(2)COOH (c) HOOC(CHOH)_(3)COOH (d) CH_(3)(CHOH)_(2)COOH (e) CH_(3)CH(Br)CH(Br)COOH.

Answer»

Solution :(a)2,(b)3,(C)4,(d)4,(E)4.
Discussion
15.

Determine the order of reaction and also determine the units of rateconstant

Answer»

SOLUTION :First order REACTION
`K=MOL^(1-n)L^(n-1)S^(-1)`
n=1
`K=(mol)^(1-1)L^(1-1) S^(-1)`
`=S^(-1)`
Discussion
16.

Determine the number of moles of AgI which may be dissolved in 1.0 litre of 1.0 M CN solution. K_(sp)for Agl and K_f for [Ag(CN)_2]^- are 1.2 xx 10^(-17) M^2and 7.1 xx 10^19 M^2respectively.

Answer»

SOLUTION :
Discussion
17.

Determine the number of formula units of NaCl in the unit cell. NaCl is a face centred cubic crystal. See the diagram of the NaCl unit cell in the text.

Answer»

SOLUTION :Either of `Na^+` or `CI^(-)` IONS may be chosen as the lattice points. Choosing `CL^(-)` ions (as shown in the text), the CUBE contains a `Cl^(-)`ion at the centre of each face as well as at the corners of the unit cell. There are `12 Na^+` ions at the centres of 12 edges and 1 `Na^(+)`ion at the centre of the unit cell.
Discussion
18.

Determine the molecular formula of an oxide of iron in which the mass percent of iron oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.8 g mol^(-1) (Atomic mass : Fe = 55.85, O = 16.00 amu)

Answer»

Solution :Expirical formula mass of `Fe_(2)O_(3)=2xx55.85+3xx16.00=159.7gmol^(-1)`
`n=("Molar mass")/("EMPIRICAL formula mass")=(159.8)/(159.7)=1`
HENCE, molecular formula is same as empirical formula, VIZ., `Fe_(2)O_(3)`.
Discussion
19.

Determine the number of formula units of NaCl in one unit cell. NaCl is face centred cubic.

Answer»

Solution :In the fcc arrangement, there are Eight corners `8 XX 1/8 =Cl^(-)` ion
At the six faces `6 xx 1/2 = 3 Cl^(-)` ions
Along the 12 edges `12 xx 1/4 = 3Na^(+)` ion
At the cube centre `(1 xx 1 = 1Na^(+)" ion")/(" Total 4 " Cl^(-)" ions")`
Hence, the unit cell contains FOUR NaCl UNITS.
Discussion
20.

Determine the molarity of an antifreeze solution containing 250 g water mixed with 222 g ethylene glycol (C_2H_6O_2). The density of this solution is 1.07 g/mL.

Answer»

SOLUTION :Mass of water = 250 g
Mass of ethylene GLYCOL = 222 g
Total mass = 472 g
Density of the solution = 1.07 g/mL
Volume of the solution = 472/1.07 = 441 mL = 0.441 L
No. of moles of ethylene glycol = `222/62 = 3.58`
Molarity = No. of moles / Volume of the solution in litres
` = (3.58)/(0.441) = 8.12 M`
Discussion
21.

Determine the molarity of an antifreeze solution containing 250 g water mixed with 222 g ethylene glycol (C_(2)H_(6)O_(2)). The density of the solution is 1.07 g/mL.

Answer»


Solution :Total mass of solution = `250+222g = 472g" ,Total volume of solution "=(472)/(1.07)mL=441.1 mL`
`"MOLARITY "=(222g)/("62 G mol"^(-1))xx(1)/("441.1 mL")xx"1000 mL"="8.12 M."`
Discussion
22.

Determine the molaarity of an antifreeze soliution sontaining 250 g water mixed with 222 g of ethylene glylene glycol (HOCH_(2)CH_(2)OH). The dnesity of solution is 1.07 g/mL.

Answer»


Solution :Mass of ethylene glycol = 222 g
Mass of water = 250 g
Mass of solution = (222+250)=472 g
Density of solution = 1.07 g `mL^(-1)`
`"Volume of solution (M) "=("Mass os ethylene glycol/molar mass")/("Volume os solution in litres")`
`=((222g))/((62" g MOL"^(-1))XX(0.4412L))`
=8.12 mol `L^(-1)`=8.12 M
Discussion
23.

Determine the hybridization at the carbon atom indicated in each species.

Answer»


SOLUTION :N/A
Discussion
24.

Determine the equilibrium constant of the following reaction at 298 K : 2Fe^(3+)+Sn^(2+) rarr 2Fe^(2+) +Sn^(4+) ("Given: "E_(Sn^(4+)//Sn^(2+))^(@)=0.15" volt, "E_(Fe^(3+)//Fe^(2+))^(@)=0.771" volt" )

Answer»


Solution :CALCULATE `E_(CELL)^(@)`. The value is 0.21 VOLT.
Apply `E_(cell)^(@)=0.0591/2 LOG K`
Discussion
25.

Determine the equilibrium constant for the reaction : HF(aq)+NH_3(aq)toNH_4^(+)(aq)+F^(-)(aq) K_a=6.9xx10^(-4) NH_3(aq)+H_2O(l)toNH_4^(+)(aq)+OH^(-)(aq) K_b=1.8xx10^(-5) 2H_2O(l)toH_3O^(+)(aq)+OH^(-)(aq) K_w=1.0xx10^(-14)

Answer»

`1.2xx10^(-8)`
`1.2xx10^(6)`
`8.1xx10^(7)`
`3.8xx10^(15)`

Answer :B
Discussion
26.

Determine the feasibility of the reaction 2Al_((s))+3Sn_((aq))^(4+)rarr 2Al^(3+)+3Sn_((aq))^(2+)

Answer»

SOLUTION :The given reaction consists of the FOLLOWING half REACTIONS
`{:("Anode",:,2Al_((g))rarr2Al^(3+)+6e^(-),),("Cathode",:,3SN^(4+)+6e^(-)rarr 3Sn^(2+),E^(@)=-1.66V),(,,E_("cell")^(@)=0.15-(-1.66),E^(@)=+0.15V):}`
`=1.81V`
Since `E_("cell")^(@)` is positive, the reaction is FEASIBLE.
Discussion
27.

Determine the difference between exhangeable pair for high spin and low spin d^(6) ions in an octahedral complex.

Answer»


SOLUTION :High SPIN : `t_(2g)^(2,1,1)eg^(1,1)`
Low spin :`t_(2g)^(2,2,2)eg^(0,0)`
2 EXTRA pairs generated in low spin.
Discussion
28.

Determine the density of CsCI which crystallizes in a bcc type structure with an edge length 412.1 pm.

Answer»

SOLUTION :`"Molar mass of cscl"="168.5 g/MOL"`
Number atoms present in PER unit cel for bcc (cscl)
`N=1`
`"Edge length (a) = 412.1 pm"`
`"Density "(rho)=?`
`rho=(nM)/(a^(2)N_(A))`
`=(1xx168.5" g mol"^(-1))/((412.1xx10^(-10)cm)^(3)xx(6.023xx10^(23)"mol"^(-1))`
`=(168.5gcm^(-3))/(6.998xx10^(-23)xx6.023xx10^(23))`
`=(168.5)/(42.148)gcm^(-3)`
`rho=3.997gcm^(-3)`
Discussion
29.

Determinethe densityof cesiumchloride whichcrystallisesin a bcctypestructurewith theedgelength412. 1 pm .Theatomicmasses ofCs and C1 are 133and 35.5respectively.

Answer»

SOLUTION :Densityof CsC1 CRYSTAL = 4.0 `g CM^(-3)`
Discussion
30.

Determine the correct order of stability :-

Answer»

`AL^(+1) gt Al^(+3)`
`Sn^(+2) gt Sn^(+4)`
`Pb^(+2) gt Pb^(+4)`
`Bi^(+3) lt Bi^(+5)`

Answer :C
Discussion
31.

Determine the bond order & formal charge on each oxygen atom in HCO_(2)^(-) respectively ?

Answer»

`1.5,-0.5`
`2,-0.5`
`1.33,-1.5`
`1.5,-1.33`

ANSWER :A
Discussion
32.

Determine the amount of CaCl_2 (i = 2.47) dissolved in 2.5 litre of water such that its osmoticpressure is 0.75 atm at 27^@C .

Answer»

Solution :For a solute undergoing dissociation, the following EQUATION may be used
`pi = ICRT = i n/V RT ` (i is van.t Hoff.s factor)
`pi = (pi xx V)/(i xx R xx T)`
Substituting the values, we GET
`n = (0.75 atm xx 2.5L)/(2.47 xx 0.0821 L atm K^(-1) "mol"^(-1) xx 300 K) = 0.0308` mole
Molar mass of `CaCl_2 = 40 + 2 xx 35.5 = 111 g "mol"^(-1)`
` therefore ` Amount dissolved = 0.0308 x 111 g = 3.42 g.
Discussion
33.

Determine the amount of CaCl_(2)(i=2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at27^(@)C.

Answer»

Solution :`pi=iCRT=i(n)/(V)RT"or"n=(pixxV)/(ixxRxxT)=("0.75 ATM"xx"2.5 L")/(2.47xx"0.0821 L atm K"^(-1)"mol"^(-1)xx"300 K")="0.0308 mole"`
`"Molar mass of "CaCl_(2)=40+2xx35.5="111 g mol"^(-1) therefore"Amount DISSOLVED "=0.0308xx"111 g = 3.42 g."`
Discussion
34.

Determine the amount of CaCl_(2)dissolved in 2.5L at 27^(@)C such that its osmotic pressure is 0.75 atm at 27^(@)C. (i for CaCl_(2) = 2.47)

Answer»

Solution :For `CaCl_(2) , "" i= 2.47`
`pi = ICRT`
`=i(n_(B))/(V) XX RT`
`0.75 = (2.47 xx n_(B) xx 0.082 xx 300)/(2.5)`
`n_(B) = (0.75 xx 2.5)/(2.47 xx 0.082 xx 300)`
`n_(B) = 0.0308` MOL
Amount = `0.0308` mol `xx` 111 g `mol^(-1)`
`= 3.148` g
Discussion
35.

Determine the amount of CaCl_(2)(i=2.47) dissolved in 2.5 litre of waster such that its osmotic pressure is 0.75 atm 27^(@)C at.

Answer»

SOLUTION :`pi = (n)/(V)RT`
`pi=i(W)/(MV)RT`
`k=(pi MV)/(iRT)`
`pi-0.75` atm
V = 2.5 L
`i=2.47`
`T=(27+273)K`
= 300 K
Here,
`R=0.0821 " atm K"^(-1)MOL^(-1)`
`M=1xx40+2xx35.5`
`= 111 G mol^(-1)`
Therefore,
`w=(0.75xx111xx2.5)/(2.47xx0.0821xx300)`
= 3.42 gram.
Discussion
36.

Determine the average life of U^(238) having t^(1//2) = 140 days.

Answer»

SOLUTION :`t^(1//2)=(0.693)/(LAMBDA)`
`TAU=(t^(1//2))/(0.693)`
=202.02 DAYS
Discussion
37.

Determine the absolute configuratins of the labeled carbons (a and b):

Answer»

a = R,b=R
a = R,b=S
a = S, b=R
a = S, b=S

Solution :
Discussion
38.

Determine DeltaH for the following reaction at 500 K and constant pressure : CO(g)+H_(2)O (g) rarr CO_(2) (g) +H_(2)(g) Use the following data : {:("Substance",C_(P)" (J/mol K)",Delta_(f)H (298 K) "(kJ/mol)"),(CO,29.12,-110.5),(H_(2)O,33.58,-241.8),(CO_(2),37.11,-393.5),(H_(2),29.89,0.0):}

Answer»


SOLUTION :Apply `DeltaH_(2)-DeltaH_(1)=DeltaC_(P) [T_(2)-T_(1)] RARR 'DeltaH_(2)=-40.3" kJ mol"^(-1)`
Discussion
39.

Determine pH of the solution that results from addition of 20 ml of 0.01 M Ca(OH)_(2) to 30 ml of 0.01 M HCl.

Answer»

`11.30`
`10.53`
`2.70`
`8.35`

Solution :`Ca(OH)_(2)+2HCl rarr CaCl_(2)+2H_(2)O`
Moles of `Ca(OH)_(2)` in 20 ml solution
`=(0.01)/(1000)xx20=2xx10^(-4)` mole
Moles of HCl in 30 ml solution
`=(0.01)/(1000)xx30`
`=3.0xx10^(-4)`mole.
2 moles of HCl react with 1 mole of `Ca.(OH)_(2)`.
Hence `3xx10^(-4)`moles of HCl react with 1.5 moles of `Ca(OH)_(2)`
Moles of `Ca(OH)_(2)` left
`=2xx10^(-4)-1.5xx10^(-4)=0.5xx10^(-4)`
`=5xx10^(-5)` moles
Moles of `OH^(-)=2xx5xx10^(-5)=10^(-4)`moles.
`[OH^(-)]=(10^(-4))/(50)xx1000=2xx10^(-3)M`
`pOH=-LOG[OH^(-)]=-log(2xx10^(-3))`
`=2.699`
`pH=14-pOH=11.301`.
Discussion
40.

Determine final pressure after the valve is left opened for along time in the apparatus represented in figure. Assume that the temperature is fixed tat 300K. Under the given conditions assume no reaction of CO & O_(2).

Answer»


ANSWER :3.284atm
Discussion
41.

DetermineDelta S_(" total ") and showwhetherthe followingreactionisspontaneousis 25^ (@) C Fe_(2) O_(3(s))+ 3CO_((g)) to 2Fe_((s))+ 3CO_(2(g)) Delta H^(@) =- 24.8 kJ , Delta S^(-1)= 15 JK^(-1)

Answer»

SOLUTION :`DELTA S_("TOTAL ") = 98.2 JK^(-1)`
Discussion
42.

Determine A_(i) and A_( c) for the reversible isothermal evaporation of 90.0 g of water at 100^(@) C. Assume that water vapours behaves as an ideal gas and heat of evaporation of water is 540 cal g^(-1) (R =2 "cal" "mol"^(-1)K^(-1))

Answer»

SOLUTION :44870 CAL
Discussion
43.

Determination of specific conductance and molar conductivity of electrolyte solution.

Answer»

SOLUTION :(1) Specific conductance, `kappa=(1)/(R)XX(l)/(R)`
(2) Molar conductivity, `Lambda_(m)=kappaxx(1000)/(M)`
Discussion
44.

T/FDetergents obtained from linear alkyl benzene are non-biodegradable.

Answer»


ANSWER :F
Discussion
45.

Detergents can be made biodegradable and pollution free by :

Answer»

Using CYCLIC HYDROCARBON chain
Shortening the hydrocarbon chain
Using unbranched hydrocarbon chain
HYDROCARBONS with more branching.

Answer :C
Discussion
46.

Detergents are usually made from products obtained by cracking of petroleum like :

Answer»

Chloroalkanes
Sulphur compounds of BENZENE
`H_2S`
POLYETHYLENE derivative

Answer :B
Discussion
47.

Detergents are sodium of potassium salts of sulphonic acids. Phosphates are added to commercial detergents because they form complex with metal ion that contributes water hardness and keep them dissolved. They also control acidity and micelle formation. Phosphates arediscouraged because there is overgrowth of vegetation and decay of dead plants reduceamount of dissolved oxygen. 1 mole of [AgI]Ag^(+) sol is coagulated by

Answer»

DETERGENTS having LINEAR ALKYL chain 
Detergents having maximum branching 
Both (a) and (b) 
NEITHER (a) nor (b) 

Solution :Linear alkyl chain detergents are biodegradable
Discussion
48.

Detergents are sodium of potassium salts of sulphonic acids. Phosphates are added to commercial detergents because they form complex with metal ion that contributes water hardness and keep them dissolved. They also control acidity and micelle formation. Phosphates arediscouraged because there is overgrowth of vegetation and decay of dead plants reduceamount of dissolved oxygen. Which of the following are cationic detergents ?

Answer»

`[CH_3(CH_2)_(15)N(CH_3)_3]CL`
`CH_3(CH_2)_16COO(CH_2CH_2O)_(11)CH_2CH_2OH`
`CH_3(CH_2)_(16)OSO_3Na`
`C_(17)H_(35)COONA`

Solution :`[CH_3(CH_2)_15N(CH_3)_3]^(+)`
Discussion
49.

Detergents are prepared by the action of H_2SO_4 on which of the following?

Answer»

Cholesterol
Lauryl alcohol
Cyclohexanol
p-Nitrophenol

Answer :B
Discussion
50.

Detergents are prepared by the action of H_2SO_4 followed by neutralization by starting with

Answer»

CHOLESTEROL
Lauryl alcohol
Cyclohexanol
p-Nitrophenol

Answer :B
Discussion