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.

K_(sp) (BaSO_(4)) is1.1 xx 10^(10). In which case is BaSO_(4) precipitated ?

Answer»

`100 mL of 4xx 10^(-3)M of BaCl _2 +300 mL of 6.0 XX 10^(-4)M of Na_2SO_4`
` 100 mLof 4xx 10^(-4)M of BaCl _2+ 300 mLof 6xx 10^(-8)M of Na_2SO_4`
` 300 mL of 4xx 10^(-4)M of BaCl_2 +100 mLof 6.0 xx 10 ^(-8)M of Na_2SO_4`
in all cases

Solution :Before MIXING ` [Ba^(+2) ]= 4 xx 10 ^(_3)M rArr [SO_4^(-2)] = 6 xx 10 ^(-4)M `
after mixing,
` 4 xx 10 ^(-3)xx 100 =M_2xx 400`
` [ Ba^(+2) ] = 10 ^(-3) M`
` 6 xx 10 ^(-4) xx 300 = M_2 xx 400`
` [SO_4^(-2) ]=4.5 xx 10^(-4)M`
` IP=[Ba^(+2) ][SO_4^(-2) ]= 4.5 xx 10 ^(-7),KsPlt I.P.`
Discussion
2.

Kp value for 2SO_(2(g))O_(2(g)) hArr 2SO_(3(g))is 5.0 atm^(-1). What is the cquilibrium partial pressure of O_(2) ifthe equilibrium pressures of SO_(2) and SO_(3) are equal ?

Answer»

<P>0.2
0.3
0.4
0.1 atm

Solution :`2SO_(2(g))+O_(2(g)) HARR 2SO_(3(g))`
`K_(P)=(P_(SO_(3))^(2))/(P_(SO_(2))^(2) xx P_(O_(2)))=(1)/(P_(O_(2)))` (`:. P_(SO_(3))=P_(SO_(2))`)
`PO_(2)=(1)/(K_(P))=(1)/(5)=0.2` atm
Discussion
3.

K_p of the following reaction at 700 K is 1.3 xx 10^(-3) atm ^(-1) . The K_Cat same tempertaure for the reaction 2SO_(2)(g)+O_(2)(g) rArr 2SO_(3)(g) will be..........

Answer»


Solution :`K_(p) = K_(C)(RT)^(An)g` R = GAS constant
`K_(c)= (K_p)/(RT^(Deltans)) = (1.3 xx 10^(-3))/((0.0821 xx 700)^(-1))=7.4 xx 10^(-2)`
Discussion
4.

K_p//K_e for N_2+3H_2 harr 2NH_3 (gaseous phase) at 400 K is

Answer»

400 R
`(400R)^2`
`(400R)^(-2)`
`(127)^(-2)`

SOLUTION :`(K_(P))/(K_(C))=(RT)^(DELTA ng)=(400R)^(-2)`
Discussion
5.

K_(p)//K_(c) " for the reaction " CO(g) + 1/2 O_(2) (g) hArr CO_(2) (g)is

Answer»

1
RT
`1//sqrt (RT)`
`(RT)^(1//2)`

ANSWER :A::B::C::D
Discussion
6.

K_(p) "is" atm^(2) for the reaction: LiCI.3NH_(3)(s)hArrLiCI.NH_(3)(s)+2NH_(3)(g) "at" 40^(@)C. How many moles of ammonia must be added at this temperature to a 5 litre flask containing 0.1 mole of LiCI. NH_(3) in order to completely convert the solid to LiCI.3NH_(3)?

Answer»


SOLUTION :N//A
Discussion
7.

K_(p) is how many times equal to K_(c) for the given reaction ? N_(2(g))+3H_(2(g))hArr2NH_(3(g))

Answer»

<P>`(1)/(R^(2)T^(2))`
`R^(2)T^(2)`
`R/T`
`RT`

Solution :`K_(p)=K_(c)(RT)^(Deltang)`
`K_(p)=K(RT)^(-2)`
`:.K_(p)=(K_(c))/(R^(2)T^(2))`
Discussion
8.

K_(p) is always equal to K_(c) " if"Delta n_(g) is ………..

Answer»


ANSWER :ZERO
Discussion
9.

K_(p) for the reaction, NH_(4)HS(s) Leftrightarrow NH_(3) (g)+H_(2)S(g), at certain temperature is 4"bar"^(2). Calculate the equilibrium perssure.

Answer»

<P>

Solution :Equilibrium constant,
`K_(P)=P_(NH_(3)). P_(H_(2)S)=4"bar"^(2)`
PARTIAL pressure are given as,
`NH_(3)=P_(H_(2)S)=SQRT(K_(p))=2"bar"`
`P_(NH_(3))+P_(H_(2)S)=2"bar" +2 "bar"=4 "bar"`
Discussion
10.

K_(p) for the reaction N_(2)O_(4)(g) hArr 2NO_(2) (g) is 0.157 atm at 27^(@)C and 1 atm pressure. Calculate K_(c) for the reaction.

Answer»

<P>

Solution :`DELTA N = n_(p) - n_(r) = 2 - 1 = 1`,
`K_(c) = K_(p)//(RT)^(Delta n) = (0.157 " atm)//(0.0821 L "atm" K^(-1) mol^(-1) xx 300 K)^(1) = 6.37 xx 10^(-3) mol L^(-1)`.
Discussion
11.

K_(p) " for the reaction " N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at "400^(@)C " is " 1*64 xx10^(-4) atm^(-2)." Find " K_(c). " Alsocalculate " DeltaG^(@) " using " K_(p) and K_(c) " values and interpret the difference ".

Answer»

Solution :`N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)`
`Delta n= 2-4 = -2`
`K_(p) = K_(c) (RT)^(Delta n)`
`1*64 XX 10^(-4) " atm"^(-2) = K_(c) (0*0821) L " atm "K^(-1) mol^(-1) xx 673 K ) ^ (-2) ""(T=400 + 273 K = 673 K)`
or `K_(c) = (1*64 xx 10^(-4) "atm"^(-2))/((0*821 xx 673 " atm"^(-1) mol^(-1))^(2))=0*5372 " mol"^(2)L^(-2)`
Now,`Delta G^(@) = - 2*303" RT " log K`
If `K=K_(p)`
` Delta G^(@) = - 2*303 xx (8*314 " JK"^(-1) mol^(-1)) (673 K) xx log (1*64 xx 10^(-4))`
`=-2*303 xx 8*314 xx 673 xx (-3* 7852) " J "mol^(-1) = +48*78 " kj "mol^(-1)`
If `K=K_(c)` ,
`Delta G^(@)= - 2*303 xx (8*314 xx 673 xx (-0*27) " J "mol^(-1) = + 3479 " J "mol^(-1)`
INTERPRETING the difference .`DeltaG^(@)` is the FREE energy CHANGE when all the reactants and products are in their standard state. In case of `K_(p)`, standardstate pressure is used which is 1 atm (or now 1 bar) wheras in case of `K_(c)` , standard concentration is used which is 1 mol `L^(-1)` (as already explained on page 7/22).
Discussion
12.

K_(p) for the reaction N_(2) + 3H_(2) hArr 2NH_(3) , is 1.6 xx 10^(-4) atm^(-2) at 400^(@) C. What will be K_(p) at 500^(@)C ? Heat of reaction in this temperature range is-25.14 Kcal.

Answer»

`1.43 xx 10^(-5) ATM^(-2)`
`1.43 xx 10^(5) atm^(-2)`
`1.43 xx 10^(-5) atm^(2)`
`1.43 xx 10^(5) atm^(2)`

SOLUTION :`log(K_(P2))/(K_(P1))=(Delta H)/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
Discussion
13.

Kp for the reaction 3/2O_(2(g)) overset(rarr)(larr)O_(3(g)) at 25^@C is 2.47 xx 10^(-29). Then DeltaG^@ for the conversion is

Answer»

`163 kJ mol^(-1)`
` 264 kJ mol^(-1)`
`376 kJ mol^(-1)`
`488 kJ mol^(-1)`

ANSWER :A
Discussion
14.

K_(p) " for the following reaction will be equal to " 3 Fe (s) + 4 H_(2)O (g) hArrFe_(3)O_(4) (s) + 4 H_(2) (g)

Answer»

<P>`(p_(H_(2)))^(4)(p_(Fe_(3)O_(4)))`
`p_(H_(2))/(p_(H_(2)O))`
`(p_(H_(2)))^(4)/((p_(H_(2)O))^(4))`
` ((p_(H_(2))) (p_(Fe_(3)O_(4))))/(p_(F_(e)))`

Answer :C
Discussion
15.

K_p for the reaction2SO_3(g) +O_2(g) leftrightarrow 2SO_3(g) is 2.5 times 10^20 atm^-1 at 500K. Calculate the K_p for the reaction SO_3(g) leftrightarrow SO_2(g)+1/2 O_2(g).

Answer»

SOLUTION :`6.3 TIMES 10^-6 atm^0.5`
Discussion
16.

K_p for the reaction : 2SO_2(g) + O_2(g) hArr 2SO_3(g) is 2.5xx10^10 Calculate the K_p values for the following reactions : (i)SO_2(g) + 1/2O_2(g) hArr SO_3(g) (ii)SO_3(g) hArr SO_2(g) +1/2O_2(g) (iii) 3SO_2(g) + 3/2 O_2(g) hArr 3SO_3(g)

Answer»

Solution :`2SO_2(G) +O_2(g) hArr 2SO_3(g) "" K_p = 2.5xx10^10`
(i)`SO_2(g) +1/2O_2(g) hArr SO_3(g)`
`K_i=(K_p)^(1//2)`
`=(2.5xx10^10)^(1//2) =1.58xx10^5`
(ii)`SO_3(g)hArr SO_2(g) +1/2O_2(g)`
`K_(ii)=1/K_i`
`=1/(1.58xx10^5)=6.3xx10^(-6)`
(III)`3SO_2(g) +3/2O_2(g) hArr 3SO_3(g)`
`K_(iii)=(K_p)^(3//2)`
`=(2.5xx10^10)^(3//2)=3.95xx10^15`
Discussion
17.

Kp for Lane's process is :

Answer»

`(P_(H_(2))^(4))/(P_(H_(2)O)^(4))`
`([Fe_(3)O_(4)])/([Fe])`
`(P_(H_(2)O)^(4))/(P_(H_(2))^(4))`
`(P_(H_(2)))/(P_(H_(2)O)) xx ([Fe_(3)O_(4)])/([Fe])`

Solution :LANES process is ONE of the methods of PREPARATION of `H(2)`
`4Fe(s)+4H_(2)O(g) harr Fe_(3)O_(4)(s)+4H_(2)(g)`
Discussion
18.

K_(p) for a reaction at 25^(@)C is 10 atm. The activation energy for forward and reverse reactions are 12 and 20 kJ mol^(-1) respectively. The K_(c) for the reaction at 40^(@)C will be:

Answer»

`4.33xx10^(-1) M`
`3.33xx10^(-2)M`
`3.33xx10^(-1)M`
`4.33xx10^(-2)M`

ANSWER :C
Discussion
19.

K_p= 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C_2H_6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium ? C_2H_(6(g)) hArr C_2H_(4(g)) + H_(2(g))

Answer»

Solution :`{:("Reaction equilibrium:",C_2H_(6(g)) hArr, C_2H_(4(g)) +, H_(2(g))),("Initial partial pressure :","4.0 ATM","0 atm","0 atm"),("CHANGE in pressure","-p atm","+p atm","+p atm"),("Partial pressure at equilibrium :", "(4-p)atm","p atm","p atm"):}`
`K_p=((p_(C_2H_4))(p_(H_2)))/((p_(C_2H_6)))`
`therefore 0.04 = ((p)(p))/((4-p))`
`therefore p^2+ 0.04 p - 0.16 =0`
In this quadratic equation put B= +0.04 , a=1 and c=-0.16
So, `p=(-bpmsqrt(b^2-4ac))/(2a)`
`=(-(0.04)pmsqrt((+0.04)^2-4(1)(-0.16)))/2`
`=(-(0.04)PM sqrt(0.0016 +0.64))/2`
`=(-(0.04) pm sqrt0.6416)/2`
`=(-0.04)pm0.8009)/2` OR `(-0.4-0.80009)/2`
p = 0.3805 atm = 0.38 atm OR - 0.6005 atm
it is possible OR impossible
`therefore p_(C_2H_6)`=(4-p)=(4-0.3805)=3.62 atm
Discussion
20.

K_(p) = 0*04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C_(2)H_(6) when it is placed in a flask at 4*0 atm pressure and allowed to came in equilibrium ?C_(2) H_(6) (g) hArr C_(2) H_(4) (g) + H_(2) (g)

Answer»

<P>

Solution :` {:(,C_(2)H_(6)(g),hArr,C_(2)H_(4)(g),+,H_(2)(g)),("Intial pressure",4*0"atm",,0,,0),("At eqm".,4-p,,p,,p):}`
`K_(p) = (p_(C_(2)H_(4))xx p_(H_(2)))/(p_(C_(2)H_(6))):. 0*04 = p^(2)/(4-p) or p^(2) = 0*16 - 0*04 p `
or ` p^(2) + 0*04 p - 0*16 = 0:.p = (-0*04 PM sqrt(0*0016 - 4(-0*16)))/2 = (0*04 pm 0*89)/2`
aking positive value, ` p= (0*80)/2 = 0*40 :.[C_(2)H_(6) ]_(eq)= 4 - 0*40 "atm"= 3* 60 "atm " `
Discussion
21.

Kolbe's synthesis of 2,2-dimethylpropanoic acid gives the following major product at anode

Answer»

isoctane
ISOBUTENE & ISOBUTANE
2,2,3,3-tetramethyl BUTANE
HYDROGEN gas

Solution :isobutene & isobutane
Discussion
22.

Kolbe's electrolysis of sodium salt of 2, 2-dimethyl propanoic acid gives the following majorproduct(s) at anode.

Answer»

`CO_2` gas
ISOBUTENE & isobutane
2, 2, 3, 3-tetramethyl butane
hydrogen gas

SOLUTION :
Discussion
23.

Kolbe's electrolysis of _____ produces n-hexane at anode .

Answer»


ANSWER :SODIUM butanonate
Discussion
24.

Koble's electrolysis can be applied

Answer»

`CH_(3)COONa`


`C_(6)H_(5)COONa`

ANSWER :A::B::C
Discussion
25.

KO_(2)(potassium superoxide) is used in oxygen cylinders in space and submarines because it

Answer»

absorbs `CO_(2)` and increases `O_(2)` content
eliminates moisture
absorbs `CO_(2)`
PRODUCES ozone

Solution :`4KO_(2) + 2 CO_(2) to 2 K_(2)CO_(3) + 3 O_(2)`
Although `KO_(2)` also reacts with moisture of the BREATH but does not completely eliminate it .
`4 KO_(2) + 2 H_(2)O to 4 KOH + 3 O_(2)`
or `4 KO_(2) + 4 CO_(2) + 2H_(2)O to 4 KHCO_(3) + 3 O_(2)`
For further EXPLANATION ,
Discussion
26.

Potassium superoxide is used in oxygen cylinders in space and submarines because it

Answer»

absorbs `CO_(2)` and INCREASE `O_(2)` content
eliminate moisture
absorbs `CO_(2)`
Produces `O_(3)`

Solution :`4KO_(2) + 2CO_(2)rarr 2K_(2)CO_(3) + 3O_(2)`
Discussion
27.

KO_(2)(potassium super oxide) is used in oxygen cylinders in space and submarines because it.....

Answer»

ABSORBS `CO_(2)` and increases `O_(2)` content
ELIMINATES moisture
absorbs `CO_(2)`
PRODUCES ozone

Answer :A
Discussion
28.

KO_(2) is used in space crafts and submarines because it

Answer»

absorbs CO2 and MOISTURE to INCREASE O2 concentration
absorbs moisture
absorbs CO2
produces ozone

Solution :`2KO_(2)+CO_(2)rarrK_(2)CO_(3)+3/2O_(2)`
`2KO_(2)+H_(2)O rarr2KOH+3/2O_(2)`
Discussion
29.

Knowing the properties of H_(2)O and D_(2)O , do you think that D_(2)O can be used for drinking purpose ?

Answer»

Solution :HEAVY water is injurious to human BEINGS, plants and animals SINCE it slows down the rates of reactions occuring in them. Thus, heavy water does not support life so WELL as does ordinary water.
Discussion
30.

Knowing the properties of H_2O and D_2O. Do you think that D_2O can be used for drinking purpose ?

Answer»

Solution :HEAVY water is injurious to human BEINGS, plants and animals since it SLOWS down the rates of reactions occurring in them. Thus heavy water does not support life so it is not used for drinking
Heavy water `(D_2O)` is a oxide of deuterium, the isotope of Hydrogen.
Heavy water is extensively used as a moderator in nuclear reactors and in exchange reactions for the study of reaction mechanisms.
It can be prepared by EXHAUSTIVE electrolysis of water or as a by PRODUCT in some fertilizer industries.
`CaC_2+2D_2O to C_2D_2+Ca(OD)_2`
`SO_3 + D_2O to D_2SO_4`
`Al_4C_3 + 12D_2O to 3CD_4+4Al(OD)_3`
Discussion
31.

Known sources of pollution is known as ........

Answer»

SOLUTION :POINT SOURCE
Discussion
32.

Knowing the properties of H_2O and D_2O . Do you think that D_2O can be used for drinking purpose?

Answer»

SOLUTION :`D_2O` dose not SUPPORT life because it slow down the REACTIONS in plants and animals.
Discussion
33.

KMnO_(4)underset(R.A.)overset(H^(+))toMn^(x) KMnO_(4)underset(R.A.)overset(OH)toMn^(y)

Answer»

`K_(2)Cr_(2)O_(7)overset(OH)toCr^(Z)`
`x+y+z` is

Answer :A::B
Discussion
34.

KMnO_(4) solution is to be standardised by titration against As_(2)O_(3) (s). A 0.1097 g sample of As_(2)O_(3) requires 26.10 mL of the KMnO_(4) solution for its titration. What are the molarity and normality of the KMnO_(4) solution (Mol. Wt. of As=75)

Answer»

Solution :`Mn^(7+)+5e to Mn^(2+) ("REDUCTION")`
`As_(2)^(3+) to 2As^(5+) +4e ("reduction")`
Meq. Of `As_(2)O_(3)="Meq. Of KMnO_(4)`
`[0.1097//(198//4)] xx 1000=26.10 xx N`
`[E_(A) S_(2)O_(3))=M//4]`
`N_(KMnO_(4))=0.085, M_(KMnO_(4)=0.085//5=0.017` by MOLE concept method
`As_(2)OVERSET(+3)(O_(3)) to 2 overset(+5)(As)`
n-factor =2|5-3|=+4
`[overset(+3)(As_(2)) to 2 overset(+5)(As)+4e^(-)] xx 5`
`[overset(+7)(Mn)+5e^(-) to overset(+2) (Mn)]xx 4`
`5 overset(+3)(As_(2))+4 overset(+7)(Mn) to 10 overset(+5)(As)+4 overset(+2)(Mn)`
5 mole 4 mole
5 mole of `As_(2)^(+3) "require to "4 mole of "KMnO_(4)`
thus, 1 mole require to `4/5" mole of KMnO"_(4)`
thus, `(0.1097)/(198) "mole require "to " mole "KMnO_(4)`
Thus, molarity of `KMnO_(4)= 4/5 xx (0.1097)/(198 xx (26.10//1000))=0.017M`
i,e, `N=M xx n-"factor"`
`N=0.017 xx 5=0.085N`
Discussion
35.

KMnO_(4) solution is to be standaridised by titration against AsO_(3)(s). A 0.1097gsample of As_(2)O_(3) requires26.10mL of the solution for its titration. What are the molarity and normally of the KMnO_(4) solution?

Answer»


ANSWER :`0.085N, 0.017M`
Discussion
36.

KMnO_(4) reacts with oxalic acid according to the equation, 2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)to 2Mn^(2+)+10CO_(2)+8H_(2)O Here, 20 mL of 0.1 M KMnO_(4) is equivalent to :

Answer»

120 mL of 0.25M `H_(2)C_(2)O_(4)`
`150 mL " of" 0.1 M H_(2)C_(2)O_(4)`
`50 mL " of" 0.1 M H_(2)C_(2)O_(4)`
`50 mL " of" 0.2 M H_(2)C_(2)O_(4)`

SOLUTION :`(M_(1)V_(1))/(n_(1))(KMnO_(4))=(M_(2)V_(2))/(n_(2))(H_(2)C_(2)O_(4))`
`(0.1xx20)/(2)=(M_(2)V_(2))/(5)`
`M_(2)V_(2)=5`
It is POSSIBLE in the option c
Discussion
37.

KMnO_(4) reacts with oxalic acid according to the reaction : 2KMnO_(4) +5C_(2)O_(4)^(2-)+16 H^(+) to 2Mn^(2+) + 10 CO_(2) + 7H_(2)O Then, 20 mL of 0.1M KMnO_(4) is equivalent to :

Answer»

30 mL of 0.5 M `C_(2)H_(2)O_(4)` (oxalic acid)
50 mL of 0.1 M` C_(2)H_(2)O_(4)`(oxalic acid)
20 mL of `0.5 M C_(2)H_(2)O_(4)` (oxalic acid)
10 mL of `0.1 M C_(2)H_(2)O_(4)` (oxalic acid)

Solution :`(M_(1)V_(1))/(n_(1))(KMnO_(4))=(M_(2)V_(2))/(n_(2))(H_(2)C_(2)O_(4))`
`(0.1xx20)/(2)=(M_(2)V_(2))/(5)`
`M_(2)V_(2)=5`
(It is possible in case of b.)
Discussion
38.

KMnO_4oxidizes X^(n+)to XO_3^(-)and it self changing to Mn+2is acid medium 2.68xx10^(-3) moles of requires mole of. The value of .n. is

Answer»


SOLUTION :`2.68xx10^(-3)XX(5-x)=1.61xx10^(-3)XX5`
Discussion
39.

KMnO_(4) oxidizes X^(n+) ion to XO_(3)^(-) itself changing to Mn^(2+) in acid solution. 2.68xx10^(-3) moel of K_(2)Cr_(2)O_(7) was required 1.61xx10^(-3) mole of MnO_(4)^(-). What is the value of n ? Also calcualtethe atomic mass of X, if the weight of 1g equivalent of XCl_(n) is 56

Answer»


ANSWER :`2,97`
Discussion
40.

KMnO_(4) reacts with KI,in basic medium to form I_(2) and MnO_(2)When 250 mL of 0.1M KI solution is mixed with 250 mL of 0.02 M KMnO_(4), in basic medium, what is the number of moles of I_(2) formed?

Answer»

0.015
0.0075
0.005
0.01

Answer :A
Discussion
41.

KMnO_(4) oxidises NO_(2)^(-) to NO_(3)^(-) in basic medium. How many moles ofNO_(2)^(-) are oxidised by 1 mol of KMnO_(4)?

Answer»


ANSWER :`1.5 MOLE NO_(2)^(-)`
Discussion
42.

KMnO_(4) and K_(2)Cr_(2)O_(7) are replaced in volumetric analysis by

Answer»

LA(III) salts
Ce(III)salts
Ce(IV)salt
Gd(III)salts

Answer :C
Discussion
43.

KMnO_4acts as an oxidising agent in alkaline medium, when alkaline KMnO_4is treated with KI, iodine ion is oxidised to

Answer»

`I_2`
`IO^(-)`
`IO_(3)^(-)`
`IO_(4)^(-)`

SOLUTION :`2KMnO_(4)+H_(2)O+KIrarr 2MnO_(2) +2KOH+KIO_3`
Discussion
44.

KMnO_(4) acts as an oxidising agent in alkaline medium when alkaline KMn_(4) is treated with KI iodide ion is oxidised to

Answer»

`I_(2)`
`IO^(-)`
`IO_(3)^(-)`
`IO_(4)^(-)`

Solution :`2KMnO_(4)+H_(2)O+KIrarr2MnO_(2)+2KOH+KIO_(3)`
or `I^(-)+6OH^(-)rarrIO_(3)^(-)+3H_(2)O+6e^(-)`
Discussion
45.

Kjeldahl's method for estimation of nitrogen is not applicable to:

Answer»

pyridine
hexamethylenediamine
propan-1-amine
2-phenylethanamine

Answer :A
Discussion
46.

Kjeldahl's method cannot be used for the estimationof nitrogen in

Answer»

Pyridine
Nitro COMPOUNDS
AZO compounds
All the THREE above

ANSWER :D
Discussion
47.

Kjeldahl's method cannot be used for estimation of nitrogen in: (I) C_(6)H_(5)CONH_(2) (II) Pyridine (III) C_(6)H_(5)-N=N-C_(6)H_(5) (IV) C_(6)H_(5)NHCOCH_(3)

Answer»

I,II
II,III
III,IV
I,II

Answer :B
Discussion
48.

Kjeldhal.s method cannot be used for the estimation of nitrogen in

Answer»

Nitrobenzene
PYRIDINE
Azobenzene
ACETANILIDE

ANSWER :A::B::C
Discussion
49.

Kjeldahl's method cannot be used for the estimation of nitrogen in

Answer»

Nitrobenzene
PYRIDINE
Azobenzene
ACETANILIDE

SOLUTION :Kjeldahl.s METHOD is not suitable for nitro COMPOUNDS, azo compounds and compounds having N in the ring.
Discussion
50.

Kjeldahl's is the name of

Answer»

A SCIENTIST
A ROUND - BOTTOM FLASK
A sealed CAPILLARY tube
A long - necked round-bottom flask

Solution :Kjeldayl is name of long necked round botton flask.
Discussion