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.

An aromatic compound 'A' (C_(7)H_(9)N) on reacting with NaNO_(2)//HCl at 0^(@)C forms benzyl alcohol and nitrogen gas. The number of isomer possible for the compound 'A' is

Answer»

3
5
6
7

Answer :B
Discussion
2.

An aromatic compound A (C_(2)H_(6)O_(2)) liberates hydrogen with metallic sodium. Compound A when heated with anhydrous zinc chloride ultimately gives B (C_(2)H_(4)O) whereas, when heated with conc. Phosphoric acid gives C (C_(4)H_(10)O_(3)). A on oxidation with acidified K_(2)Cr_(2)O_(7) gives compound D (CH_(2)O_(2)). Identify A, B, C and D. Explain the reactions involved.

Answer»

Solution :(i) An organic compound A `(C_(2)H_(6)O_(2))` liberates hydrogen with metallic sodium.
`underset("Ethylene glycol compound"(A))underset(CH_(2)OH)overset(CH_(2)OH)|overset(NA)underset(323K)to underset("MONOSODIUM glycolate")underset(CH_(2)OH)overset(CH_(2)ONa)|overset(Na)underset(433 K) to underset("disodium glycolate")underset(CH_(3)ONa)overset(CH_(3)ONa)|+H_(2)uarr`
(ii) Compound A when heated with anhydrous zinc chloride ultimately gives B `(C_(2)H_(4)O)`.

(iii) Compound A when heated with conc. phosphoric acid gives `C (C_(4)H_(10)O_(3))`.

(iv) Compound A on OXIDATION with acidified `K_(2)Cr_(2)O_(7)` gives compound D `(CH_(2)O_(2))`.
`underset(CH_(2)OH)overset(CH_(2)OH)|overset(3(O))underset("acidified " K_(2)Cr_(2)O_(7))to underset("HCOOH formic acid")overset(HCOOH)+ + H_(2)O`
Discussion
3.

An aromatic compound (A), C_(7)H_(5)NO_(2)CI_(2) on reduction with Sn//HCI gives (G), which on reaction with NaNO_(2)//HCI gives (B). Compound (B) does not form a dye with beta-naphthol. However, (B) gives red colour with ceric ammonium nitrate and on oxidation gives an acid (D) of equivalent weight 191. Decarboxylation of (D) gives (E), which forms a mono nitro derivative (F) on nitrotion. Give the structures of (A) to (F) with proper reasoning.

Answer»

SOLUTION :`DU in (A) `= ((2n_(C) + 2) - (n_(H) + n_(X) - n_(N)))/(2)`
`= (16-(5+2-1))/(2)=5^(@)`
`5^(@)` DU is due benzene (4 DU) and ONE `(---NO_(2))` group (1 DU)`.

In (A), `(---NO_(2))` group cannot be with benzene ring, since reduction changes to `(---NH_(2))` ggroup, which on reaction with `(NaNO_(2) + HCI)` changes to alcohol. So it is in the side chain.

Reactions:
Discussion
4.

An aromatic compound 'A'(C_7H_9N)on reacting with NaNO_2 // HClat 0^(@)C forms benzyl alcohol andnitrogen gas. The number of isomers possible for the compound 'A' is

Answer»

5
7
3
6

Solution :
Discussion
5.

An organic amine (X) was treated with alcoholic potast and another compound (Y),a foul smelling gas ws formed with formula C_(6)H_(5)NC,(Y) was formed by reacting a compound (Z) with Cl_(2) in the presence of slaked lime. The compound (Z) is:

Answer»

`C_(6)H_(5)NH_(2)`
`C_(2)H_(5)OH`
`CH_(3)OCH_(3)`
`CHCl_(3)`

Answer :B
Discussion
6.

An aromatic among other things should have a pi-electron cloud containing electrons where n can't be

Answer»

`1//2`
3
2
1

Solution :According to Huckel's rule, all aromatic compounds must have `(4n+2)PI` electrons where n is an integer,i.e., n = 0, 1, 2, 3,.... And possesses unusual STABILITY due to the complete DELOCALISATION of `pi`-electrons.
Discussion
7.

An aromatic aldehyde (A) of molecular C_7H_6Oreacts with acidified KMnO_4 to give (B) of molecular formula C_7H_6O_2 , calcium salt' of compound (B) on dry distillation gives (C) of molecular formula C_13H_10 O. Compound (C) on Clemmenson's reduction, gives (D) of formula C_13H_12O.Identify A,B,C,D and explain the reaction involved.

Answer»

Solution : (i)From the MOLECULAR FORMULA (A) is identified as Benzaldehyde `C_6H_5CHO` .Benzaldehyde on reaction with acidified `KMnO_4`undergoes oxidation to give BENZOIC acid as compound (B)
Discussion
8.

An aquesous solution of urea [6%w/v] is isotonic with NaCI solution then mass-volume percentage (%w/v) of NaCI solution then mass- volume precentage (%w/v) of NaCI solution will be :-

Answer»

`1.46%`
`5.85%`
`2.92%`
`11.7%`

Solution :`pi_(1)=pi_(2)`
`:.i_(1)c_(1)=i_(2)c_(2)`
`1XX(6)/(60)xx10=2xx(W)/(58.5)xx10`
`:.W=(58.5)(20)=2.925g`
Discussion
9.

An arbitrary combound P_2Q decomposes according to reaction : 2P_2Q(g)hArr 2P_2(g)+Q_2(g) If one starts the decomposition reaction with 4 moles of P_2Q and value of equilibrium constant K_P is numerically equal to total pressure at equilibrium.Then which option is/are correct at equilibrium :

Answer»

mole of `.^nP_2Q=.^nQ_2`
MOLES of `.^nP_2=(8//3)`
DEGREE of dissociation is `alpha =(2//3)`
total number of moles of products (`P_2` and `Q_2`) at equilibrium is 4

Solution :`{:(2P_2Q,hArr,2P_2+,Q_2),(t=0,4,0,0),(t=eq,4(1-alpha),4alpha,2alpha):}`
`K_p=(((4alpha)/(4+2alpha)XXP)^2((2alpha)/(4+2alpha)xxP))/(((4(1-alpha))/(4+2alpha)xxP)^2)=(2alpha^3)/((1-alpha)^2(4+2alpha))xxP`
But `K_p=P` (given) `:. 2alpha^3=(1-alpha)^2(4+2alpha)`
`:. -6alpha+4=0 "" alpha=(2//3)`
Total moles at eq `=4+2alpha=(16//3)`
`n_(P_2Q)=n_(Q_2)=(4//3),n_(P_2)=(8//3)`
`:.` Total number of moles of products `=8/3+4/3=12/3=4`
Discussion
10.

An aqueous which is 20% (w/w) in NaOH. What will be mass fraction of water in 30ml of such solution.

Answer»

Solution :100gm solution contain `w_(NAOH)=20gm`
100 ML solution contain `w_(NaOH)=30gm`
30 ml solution contain `w_(NaOH)=(30)/(100)xx30=9gm`
20 gm NaOH present in volume of solution `=(100)/(30)xx20ml`
`d_("solution")=(100)/(100xx20)=(3)/(2)gm//ml=(w_("solution"))/(30ML)rArrw_("solution")=45gm`
`w_(NaOH)=(9)/(45)=(1)/(5)=0.2`
`w_(NaOH)=9gm`
`x_(NaOH)=(9)/(45)=(1)/(5)=0.2`
`x_(H_(2)O)=0.8`
0.8xx10=8
Discussion
11.

An aqueus soution of an organic compound containing 0.6 of it dissolved in 21.7 g of water, freezes at 272.187 K. If the value of K_(f) is 1.86 K mol^(-1) for water which freezes at 273 K, calculate the molecular mass of organic compund.

Answer»

SOLUTION :`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Mass of compound`(W_(B))=0.6 g`
Mass of water `(W_(A))=21.7 g= 0.0 217 KG`
`(DeltaT_(f))=(273-272.187)=0.813 K`
Molal depression constant `(K_(f))=1.86 K kg MOL^(-1))` ltbtgt `M_(B)=((1.86 K kg mol^(-1))xx(0.6 g))/((0.813 K)xx(0.0217 kg))=63.26 g mol^(-1)`
Discussion
12.

An aqueous solution whose pH = 0 is

Answer»

Alkaline
ACIDIC
Neutral
Amphoteric

Solution :pH = 0 MEANS `[H^(+)] = 10^(@) = 1M`. HENCE solution is strongly acidic.
Discussion
13.

An aqueous solution of urea is found to boil at 100.52^(@)C. Given K_(b) for water is 0.52 K kg mol^(-1), the mole fraction of urea in the solution is

Answer»

1
0.5
0.018
0.25

Solution :`DeltaT_(b)=K_(b)xxm`.
Hence, molality, `m=(DeltaT_(b))/(K_(b))=(0.52)/(0.52)=1`
Molality = 1 means 1 mole of solute in 1000 G of solvent.
But 1000 g of solvent (water)
`=(1000)/(18)" moles "="55.55 moles"`
`therefore"Mole fraction of urea"=(1)/(1+55.55)=0.018`.
Discussion
14.

An aqueous solution of urea is:

Answer»

ACIDIC
BASIC
neutral
amphoteric

ANSWER :C
Discussion
15.

An aqueous solution of urea has freezingpoint of -0.604^(@)C &At 27^(@)C . The osmatic pressure of the same solution is ____________atm . (assume molality and molarity are same)

Answer»


ANSWER :8
Discussion
16.

An aqueous solution of urea had a freezing point of -0.52^(@)C. Predict the osmotic pressure of the same solution at 37^(@)C. Assume that the molar concentration and the molality are numerically equal. (K_(f)=1.86)

Answer»

Solution :MOLALITY `=(DeltaT_(F))/(K_(f))=(0.52)/(1.86)=` MOLARITY (given)
Osmotic pressure `=CRT`
`=(0.52)/(1.86)xx0.0821xx(273+37)`
`=7.1` atm
Discussion
17.

An aqueous solution of titanium chloride, when subjected to magnetic measurement, meaured zero magnetic moment. Assuming the octahedral complex in aqueous solution, the formulae of the complex is:

Answer»

`[TI(H_(2)O)_(6)]Cl_(2)`
`[Ti(H_(2)O)_(6)]Cl_(4)`
`[TiCl_(3)(H_(2)O)_(3)]`
`[TiCl_(2)(H_(2)O)_(4)]`

Solution :`[Ti(H_(2)O)_(6)]Cl_(4)`
Coordination number 6`implies`octahedral complex
Ti is in +4 oxidations state `implies`no unpaired ELECTRONS
`implies`MAGNETIC moment=0 B.M.
Discussion
18.

An aqueous solution of urea containing 6 g in 500 ml has a density equal to 1.05. If the molar mass of urea is 60, then the molality of the solution is :

Answer»

0.193 M
0.190 M
0.20 M
0.10 M

Solution :Mass of solution `=500 ML xx (1.05 "g ml"^(-1))`
`=525g`
Mass of solvent `=525-6=519g`
Urea`=6g=6/60`MOLE
MOLALITY `=(6xx1000)/(60xx519)=0.193`m .
Discussion
19.

An aqueous solution of [Ti(H_2O)_6]^(+3) appears

Answer»

greenish-yellow in colour
BLUE in colour
VIOLET in colour
PURPLE in colour

Solution :Purple colour DUE to d-d transition
Discussion
20.

An aqueous solution of sucrose,C_(12) H_(2) O_(11) containing 34.2 g/L has an osmotic pressure of 2.38 atmospheres at 70^(o) C.for an aquesous solution of glucose, C_(6)H_(12)O_(6) to be istonic with this solution ,it would have :

Answer»

34.2g/L
17.1 G/L
18.0 g/L
36.0 g/L of glucose

Solution :`34.2 g L ^(-1) "SUCROSE" = 0.1 M,`
`18 g L^(-1) "glucose" = 0.1 M`
Discussion
21.

An aqueous solution of sodium chloride is marked 10% (w/w) on the bottle. The density of the solution is "1.071 g mL"^(-1). What is the molality and molarity? Also what is the mole fraction of each component in the solution?

Answer»

SOLUTION :1.83 M, 1.90 m, X (NACL) = 0.03 , x `(H_(2)O)`= 0.97
Discussion
22.

An aqueous solution of sodium carbonate is alkaline because sodium carbonate is a salt of

Answer»

Weak ACID and weak BASE
Strong acid and weak base
weak acid and strong base
Strong acid and strong base

Solution :`Na_(2)CO_(3) + 2H_(2) hArr 2NAOH + H_(2)CO_(3)`. It is a strong base and weak acid so it is a basic.
Discussion
23.

An aqueous solution of sodium carbonate has a pH greater than 7 because

Answer»

It contains more CARBONATE ION than `H_(2)O` molecules
Contains more hydroxide IONS than carbonate ions
`Na^(+)` ions REACT with water
Carbonate ions react with `H_(2)O`

Solution :`pH gt 7` = Basic
It means contain more hydroxide ion than carbonate ions.
Discussion
24.

An aqueous solution of salt 'R' when treated with dil. HCl, a colourless gas is given out. The gas so evolved when passed through acidified KMnO_(4) decolourisesKMnO_(4) solution. The salt 'R' is

Answer»

`Na_(2)CO_(3)`
`NaClO_(3)`
`NaNO_(2)`
`Na_(2)SO_(3)`

Solution :Since .R. gives a COLOURLESS gas on reaction with dil. HCl, so it CONTAINS `CO_(3)^(2-) or SO_(3)^(2-)` as anion (i.e. `CO_(2) or SO_(2)` is evolved)
Since the gas decolourises acidified `KMnO_(4)` solution so it is `SO_(2)` and thus the anion present is `SO_(3)^(2-)` i.e., the salt .R. is `Na_(2)SO_(3)`.
Discussion
25.

An aqueous solution of salt is alkaline. This show that the salt is made from as:

Answer»

STRONG acid and strong BASE
strong acid and week base
weak acid and week base
weak acid and strong base

ANSWER :D
Discussion
26.

An aqueous solution of salt gives white precipitate with AgNO_3 solution as well as with dil. H_2 SO_4. It may be

Answer»

`Pb(NO_(3))_(2)`
`Ba(NO_(3))_(2)`
`BaCl_(2)`
`CuCl_(2)`

Solution :`BaCl_(2) +2AgNO_(3) to UNDERSET("WHITE ppt.")(2AgCl DARR) + BaNO_(3)`
`BaCl_(2)+H_(2)SO_(4) to underset("White ppt.")(BaSO_(4) darr)+2HCL`
Discussion
27.

An aqueous solution of nitrous acid (HNO_(2)) , free of salts , can be obtained from the reaction

Answer»

`Ba(NO_(2))_(2)+H_(2)SO_(4) to `
`NaNO_(2) + H_(2)SO_(4) OVERSET("Cold ") to `
`NH_(4)NO_(2) + H_(2)SO_(4) to `
`KNO_(3) +HNO_(3) to `

Solution :Nitrous ACID can be obtained by the REACTION of sulphuric acid with `Ba(NO_(2))_(2)`.
Discussion
28.

An aqueous solution of NaCl on electrolysis gives H_(2)(g), Cl_(2)(g) and NaOH according to the reaction. 2Cl^(-)(aq) 2H_(2)O 2OH^(-) (aq) + H_(2) (g) + Cl_(2)(g)

Answer»

SOLUTION :At ANODE`2C1^(-) to C1_2+2e^(-)`
At cathode
`2H_2O+2e^(-) toH_2+2oH^(-)`
` 1kg Cl_2=100/35.5 ` equivalent of `Cl_2=28.17` equivalent `implies` Thereforeelectricity requiremnt `=28.17 F`
Q EFFICENCY is only 62%
`THEREFORE` Electricity requirement (experimental)`=(28.17xx10)/(62)=F=45.44f`
`implies 45.44 xx96500=25t` (in second) `implies t=48.72 H`
Also, gram equivalent of `HO^(-)` produced `=28.17`
`implies` Molarity of `HO^(-)=28.17/20=1.4085 M`
Discussion
29.

An aqueous solution of NaCL on electrolysis gives H_2(g),CL_2(g) and NaOH according to the equation 2Cl^(-)(aq)+2H_2O

Answer»

48.71 HR,1041M
2880 MIN, 1041M
17.54 hr, 2M
170.54 min, 2M

Answer :A
Discussion
30.

An aqueous solution of NaCl on electrolysis gives H_2(g), Cl_2(g) and NaOH according to the reaction:2Cl^(-)(aq) + 2H_2O = 2OH^(-) (aq) + H_2(g) + Cl_2(g)A direct current of 25 amp with a current efficiency of 62% is passed through 20 litres of NaCl solution (20% by weight). Write down the reactions taking place at the anode and at the cathode. How long will it take to produce 1 kg of CI_2? What will be the molarity of the solution with respect to hydroxide ions? (Assume no loss due to evaporation.)

Answer»

SOLUTION :49 H, 1.408 M
Discussion
31.

An aqueous solution of NaCl shall boil at

Answer»

`100^(@)C`
below `100^(@)C`
above `100^(@)C`
`99.9^(@)C`

ANSWER :C
Discussion
32.

An aqueoussolutionof NaCI on electrolysisgivesH_(2)(g),CI_(2)(g) and NaOH accordingto reaction 2CI^(-)(Aq)+2H_(2)O rarr 2OH^(-)(Aq)+H_(2)(g)+CI_(2)(g) A direct currentof 25 Awitha current efficiencyof 62% is passed through 20 L of NaCI solution(20%)by weight Howlong will it take to produce 1 kgof CI_(2)

Answer»

`30.20h`
`12.17 h`
`48.71 h`
`14.61 h`

SOLUTION :`(W)/("EQUIVALENT WEIGHT")=("it")/(96500)`
Given `w_(C 12)=10^(3) g` equivalent weight `w_(CI_(2))=35.5` and
`i=(25xx62)/(100) =15.5 A`
HENCE`t=(10^(3)xx93500)/(35.5 xx15.5)=175374.83 s=48.71 h`
Discussion
33.

An aqueous solution of NaCl containing 5.85 g/L of NaCl was electrolysed by using platinum electrodes. Hydrogen and chlorine gases evolved at the cathode and anode respectively. Calculate the pH of the solution after electrolysis assuming complete electrolysis of NaCl.

Answer»


ANSWER :13
Discussion
34.

An aqueous solution of NaCI is marked 10% W/W on the borrle. The density of soution is 1.071 g mL^(-1). Calculate its molarity.

Answer»


Solution :`"Mass of solution"=100g," Mass of NACI"=10 g`
`"Density of soluion "1.071" g ML"^(01)`
`"VOLUME of 100 g of solution"= ("Mass of solution")/("Density of solution")=((100g))/((1.071" g mL"^(-1)))=93.37mL`
`"Molarity of solution (M)"=("Mass of solute/Molar mass of solute")/("Volume of solution in litres")`
`=((10G))/((58.5"g mol"^(-1))xx0.0934L)=1.83" mol L"^(-1) or 1.83 M`
Discussion
35.

An aqueous solution of methanol on water has vapour pressure:

Answer»

EQUAL to that of WATER
Equal to that of methanol
More than that of water
LESS than that of water

ANSWER :C
Discussion
36.

An aqueous solution of methanol in water has vapour pressure

Answer»

equal to that of water
equal to that of methanol
more than that of water
less than that of water

Solution :Methanol being more VOLATILE than water, the solution of methanol in water will have more VAPOUR PRESSURE than water. Alternatively, `CH_(3)OH-H_(2)O` forms a non-ideal solution showing `+ve` deviation from Raoults' law. Hence, the solution has higher vapour pressure than that of PURE COMPONENTS.
Discussion
37.

An aqueous solution of metal ion M_(1) reacts separately with reagents Q and R in excess to gie tetrahedral and sqaure planar complexes, respectively. An aqueous solution of another metal ion M_(2) always forms tetrahedral complexes with these reagents. Aqueous solution of M_(2) on reaction with reagent S gives white precipitate which dissolves in excess of S. The reaction are summarised in the scheme given below Scheme M_(1), Q and R, respectively are

Answer»

`Zn^(2+), KCN` and `HCl`
`Ni^(2+), HCl` and `KCN`
`CD^(2+), KCN` and `HCl`
`CO^(2+), HCl` and `KCN`

ANSWER :B
Discussion
38.

An aqueous solution of metal ion M_(1) reacts separately with reagents Q and R in excess to gie tetrahedral and sqaure planar complexes, respectively. An aqueous solution of another metal ion M_(2) always forms tetrahedral complexes with these reagents. Aqueous solution of M_(2) on reaction with reagent S gives white precipitate which dissolves in excess of S. The reaction are summarised in the scheme given below Scheme Reagent S is

Answer»

`K_(4)[Fe(CN)_(6)]`
`Na_(2)HPO_(4)`
`K_(2)CrO_(4)`
`KOH`

Answer :D
Discussion
39.

An aqueous solution of KI is 1.0 molal in KI. Which change will cause the vapour pressure of the solution to increases ?

Answer»

Addition of NaCI
Addition of `Na_(2)SO_(4)`
Addition of 1.0 molaol KI
Addition of WATER

Solution :Addition of water decreases the concentration of the KI solution. Therefore, its vapour its vapour PRESSURE increases. Allother solutes being IONIC in nature, increases the concentration of the solution. Therefore, the vapour pressure of the soution gets LOWERD.
Discussion
40.

An aqueous solution of mannitol in water has a vapour pressure of 17.504 mm at 20^@C , at which temperature, the vapour pressure of pure water is 17.535 mm. What is the f.p. depression for this solution? K_f(H_2O) = 1.86.

Answer»

SOLUTION :`0.183^@C`
Discussion
41.

An aqueous solution of KI does not give a precipitate with :

Answer»

`Mg^(2+)`
`Pb^(2+)`
`HG^(2+)`
`CU^(2+)`

ANSWER :A
Discussion
42.

An aqueous solution of KBr is treated with each of the following. In which case bromine will be liberated ?

Answer»

`Cl_(2)`
HI
`SO_(2)`
`I_(2)`

SOLUTION :`Cl_(2)` being a stronger oxidising AGENT that `Br_(2)`, liberates `Br_(2)` from KBR.
`Cl_(2) + 2KBr RARR 2 KCl + Br_(2)`.
Discussion
43.

An aqueous solution of hydrogen sulphide shows the equilibrium, H_2S ⇌H^+ + HS^- if dilute hydrochloric acid is added to an aqueous solution of hydrogensulphide without any change in temperature:

Answer»

The equilibrium CONSTANT will change
The CONCENTRATION `HS^-` will increase
The concentration of undissociated hydrogen SULPHIDE will decrease
The concentration of `HS^-`will decrease.

Answer :D
Discussion
44.

An aqueous solution of hydrogen peroxide is:

Answer»

Alkaline
Neutral
Strongly acidic
Weakly acidic

Answer :D
Discussion
45.

an aqueous solution of HNO_2 (nitrous acid), free of salt can be obtained from the reaction

Answer»

`BA(NO_(2))_(2)+H_(2)SO_(4) to `
`NaNO_(2) +HNO_(4) overset("Cold")to`
`NH_(4)NO_(2)+H_(2)SO_(4) to`
`KNO_(3) +H_(2)SO_(4) to `

Answer :A
Discussion
46.

An aqueous solution of glucose containing 60 g glucose (C_(6)H_(12)O_(6)) per litre has an osmotic pressure of 5.2 bar at 300 K. The concentration of the glucose solution having osmotic pressure 0of 1.3 bar at the same temperature is :

Answer»

`1/10M`
`1/5M`
`/220M`
`1/12M`

Solution :Osmotic PRESSURE `(PI)alphaM`
`pi_(1)/pi_(2)=M_(1)/M_(2)` (for the same solution at the same temperature)
`(5.2)/(1.3)=(1/3)/M_(2)or M_(2)=1/3xx1/4=1/12M`
Discussion
47.

An aqueous solution of glucose boils at 100.01^(@)C. The molal elevationconstant for water is 0.5 K kg mol^(-1). What is the number of glucose molecules in the solution containing 100 g of water ?

Answer»

Solution :`Delta T_(B)=K_(b)XX m , 0.01 = 0.5 xx m` which gives m = 0.02.
Thus, 1000 G of water contain glucose = 0.02 mole
`therefore` 100 g of water contain glucose = 0.002 mole `= 0.002 xx 6.02xx10^(23)` MOLECULES `= 1.204xx10^(21)` molecules.
Discussion
48.

An aqueous solution of gas (X) shows the following reactions :- On addition of FeCl_(3) solution a brown ppt. soluble in dilute nitric acid is obtained. Identify (X) and give equations for the reactions at step (ii) & (iii)

Answer»

Solution :`X - NH_(3)`
`FE(OH)_(3)+3HNO_(3) underset("Soluble")(rarrFe(NO_(3))_(3))+3H_(2)O`
Discussion
49.

An aqueous solution of gas (X) shows the following reactions : (i) It turns red litmus blue. (ii) When added in excess to a copper sulphate solution, a deep blue colour is obtained. (iii) On addition to a ferric chloride solution, a brown precipitate soluble in dilute nitric acid, is obtained. Identify (X) and give equations for the reactions at steps (ii) and (iii).

Answer»

Solution :(i) (a) SINCE gas (X) turns red litmus blue, it must be basic in nature. The only important gas which is basic in character is ammonia `(NH_(3))`. Thus, (X) may be ammonia. This is supported by the observation that it gives a deep blue colour with copper sulphate solution.
(b) Having identified that the gas (X) is ammonia, the reactions at steps (ii) and (iii) may be explained as follows:
(ii) With copper sulphate solution, it gives a deep blue colour due to the formation of tetrammine copper (II) COMPLEX.
`CuSO_(4) + underset(underset((X))("Ammonia"))(4NH_(3)) rarrunderset(underset("(Deep blue colour)")("Tetramminecopper (II) sulphate"))([Cu(NH_(3))_(4)] SO_(4))`
(iii) With FERRIC chloride solution, (X) gives a brown PPT. of ferric hydroxide which dissolves in dilute nitric acid to form ferric nitrate.
`NH_(3) + H_(2)O rarr NH_(4)^(+) + OH^(-)`
`FeCl_(3) + 3 OH^(-) rarr underset(underset(("Brown ppt."))("Ferric hydroxide"))(Fe(OH)_(3)) + 3 Cl^(-)`
Discussion
50.

An aqueous solution of formaldehyde in water is sold under the name_____.

Answer»

SOLUTION :FORMALIN.
Discussion