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 antibiotic contains nitro group attached to aromatic nucleus. It is

Answer»

Tetracycline
Streptomycin
Penicillin
CHLORAMPHENICOL

Solution :Chloramphenicol is
2.

An antibiotic with a broad spectrum

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KILLS the antibodies
Acts on a specific antigen
Acts on DIFFERENT antigens
Acts on both the antigens and antibodies

Solution :Broad SPECTRUM antibiotics act on different antigens.
3.

An antibiotic contains nitro group attached to aromatic nucleus . It is :

Answer»

Penicillin
Streptomycin
Tetracycline
Chloramphencol

Answer :D
4.

…………………..an anti cancer agent used to treat stomach and colon cancer.

Answer»


ANSWER :MITOMYCIN C
5.

An anti-freeze solution is prepared from 222.6 g of ethylene glycol C_(2) H_(4) (OH)_(2) and 200 g of water . Calculate themolality of the solution. If the density of this solution be 1.072 g mL^(-1) , what will be the molarity of the solution ?

Answer»

Solution :Mg of `C_(2) H_(4) (OH)_(2) = 62 g mol^(-1)`
Molality = `(n_(B))/(W_(A))xx 1000 = (W_(B))/(M_(B) xx W_(B)) xx 1000 = (222.6 xx 1000)/(62 xx 200) = 17.95 m `
DENSITY= `("Mass")/("Volume")`
So , Volume = `("Mass")/("Density") = (422.6)/(1.072) = 394.22` ml
`M = (n_(B))/(V) xx 1000`
`= (222.6)/(394.22 xx 62) xx 1000 = 9.11` M
6.

An antibiotic contains nitro group attached to aromatic nucleus in its structure. It is

Answer»

PENCILLIN
sulphadiazine
tetracycline
CHLORAMPHENICOL

ANSWER :D
7.

An analytical balance has uncertainity in measurement equal to pm1 mg. Then the result in terms of persentage would be if the weight of a compound is 10 g

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`10pm0.1%`
`10pm0.01%`
`10pm1%`
`10pm0.001%`

ANSWER :B
8.

An anhybride and an ester can be distinguished with

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`C_2H_5OH`
`NH_3`
water
aqueous NaOH

Answer :C
9.

An aniline on nitration gives

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Both a and C

ANSWER :D
10.

An analyst starts a first order chemica reaction at 8.00 A.M. in the morning at the laboratory temperature of 27^(@) C At 1.00 P.M. he discovered that only 10% of the reaction was complete by that time To speed-up the reaction he increased the temperature to 127^(@) C At 4.00 P.M. he found that only 50% of the reaction was complete Any how he did not want to stay in laboratory beyond 5.00 P.M. but he could not leave the laboratory until the reaction was 90% complete Fortunately he found a suitable catayst adding which at 4.00 P.M. at 127^(@) C he could meet the target of 5.00 P.M. and 90% Answer the following questions based on theabove observation (Use l n 10/9 = 0.1 l n 9/5=0.6, l n10=2.3 l n5=1.6,l n8=2) What was the activation energy of the catalyzed pathway?

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`4.92 kcal//mol`
`3.92 kcal//mol`
`4.72 K cal//mol`
`9.84 kcal//mol`

ANSWER :B
11.

An analysis of the hydrolysis products of salmine, a polypeptide from salmon gave following result of weights of amino acidsd in gram per 109.66g of salmine{:(Ie :,"Isoleusine : 1.31","Alanine : 0.89","Serine : 7.35","Argemine : 86.40"),(,"valine : 3.51","Glycine : 3.0","Proline : 6.90",):} {:("molecular weight of salmine is : 10, 966","molecular weight of Serine : 105"),("molecular weight of alanine is : 89","molecular weight of Argenine : 174"),("molecular weight of Isoleucine : 131","molecular weight of Protine : 115"),("molecular weight of Valine : 117","molecular weight of Glycine : 75"):}Salmine is also known as

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`Ala. Arg_(50).Gly_(4).Ile Pro_(6). Ser_(7)Val_(3)`
`Ala. Arg_(50).Gly_(14).Ile_(3).Pro_(6).SER.Val`
`Ala.Arg_(50).Gly_(3).Ile_(4).Pro_(6).Ser_(7).Val`
`Ala.Arg_(50).Gly_(6).Ile.Pro_(4).Ser_(7).Val_(3)`

Solution :Salmine composition is `Ala, Arg_(5)-Gly_(4)-Ile-Pro_(6)-Ser_(7)-Val_(3)`
12.

An analyst starts a first order chemica reaction at 8.00 A.M. in the morning at the laboratory temperature of 27^(@) C At 1.00 P.M. he discovered that only 10% of the reaction was complete by that time To speed-up the reaction he increased the temperature to 127^(@) C At 4.00 P.M. he found that only 50% of the reaction was complete Any how he did not want to stay in laboratory beyond 5.00 P.M. but he could not leave the laboratory until the reaction was 90% complete Fortunately he found a suitable catayst adding which at 4.00 P.M. at 127^(@) C he could meet the target of 5.00 P.M. and 90% Answer the following questions based on theabove observation (Use l n 10/9 = 0.1 l n 9/5=0.6, l n10=2.3 l n5=1.6,l n8=2) What was the activation energy of the original pathway?

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`10.14 kcal//mol`
`5.52 kcal//mol`
`2.64 kcal//mol`
`7.92 kcal//mol`

ANSWER :B
13.

An analgesic is used to

Answer»

KILL bacteria
arrest growth
relieve pain
alll of these

ANSWER :C
14.

An analysis of pyrex glass showed 12.9%B_(2)O_(3),2,2% Al_(2)O_(3).3.8% Na_(2)O,0.4% K_(2)O and remaining is SiO_(2). What is the ratio of silicon to boron atoms in the glass?

Answer»

Solution :PERCENTAGE composition of `B_(2)O_(3)=12.9%`
Percentage composition of
`SiO_(2)=100-[12.9+2.2+3.8+0.4]`
=80.7%
Number of moles of `B_(2)O_(3)=("MASS")/("Molar mass")=(12.9)/(70)=0.184`
Number of moles of boron atoms `=2xx0.184`
Number of moles of `SiO_(2)=("Mass")/("Molar mass")=(80.7)/(60)=1.345`
Number of moles of silicon atoms =1.345
`("Number of atoms of silicon")/("Number of atoms of boron")=(N_(A)xx1.345)/(N_(A)xx0.184)=(7.3)/(1)`
Where `N_(A)=` AVOGADRO's number.
15.

Choose the most appropriate answer from the following options : Identify the metallic oxide which is amphoteric in nature : Calcium oxide Barium oxide Zinc oxide Copper(ll)oxide

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`NA2O`
`SO_2`
`B_2O_3`
ZnO

Answer :D
16.

An amphoteric oxide out of the following is :

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`NA2O`
`SO_2`
`B_2O_3`
ZnO

Answer :D
17.

An amount of solid NH_(4)HSis placed in a flask already containing ammonia gas at a certain temperature and 1.0 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH_(3)and H_(2)Sgases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 2 atm. What will be the equilibrium constant for NH_4HS decomposition at this temperature?

Answer»

<P>

SOLUTION :`underset("initial At. EQN")(NH_(4)HS(s)) X atm)(H_(2)S)(g)`
Then, `1.0 + x + x=2x+1.0=2.0` (given)
`rArr x=0.5` atm
`P_(NH-(3)) =1.0 + 1.5 1.5 atm, p-(H_(2)S) = 0.5 atm^(2)`
`0.75 atm^(2)`
18.

An amount of solid NGH_(4)HS is placed ina flask already containg ammonia gas at a certain temperaturee and 0.50 atm. Pressue. Ammounium hydrogen sulphide decomposes to yield H_(3)and H_(2)S gases in the flask. When the decomposition reaction reaches equlibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH_(4)HS decompositions at this temperaturee is

Answer»

`0.30`
`0.18`
`0.17`
`1.11`

SOLUTION :`{:(NH_(4)HShArrNH_(3(g))+H_(2)S_((g))),("""a""0.5atm"),(""a-X ""0.5+x ""x):}`
Total PRESSURE `=0.5+2x=0.84 i.e., x=0.17`
`K_(p)=P_(NH_(3)).P_(H_(2)S)=(0.67).(0.17)=0.1139`
19.

An amorphous solid (X) burns in air to form a gas ( Y) which turns lime water milky. This gas decolourises aqueous solution of acidified KMnO_(4). Gas (Y) reacts with oxygen to give another gas (Z) which is responsible for acid rain. X, Y and Z are

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X-C, Y-CO, Z-`CO_2`
X-S, Y-`SO_2` , Z-`SO_3`
X-P , Y-`P_2O_3` ,Z - `P_2O_5`
X-S , Y-`SO_3` ,Z-`H_2SO_4`

Solution :`{:(S,+O_2,toSO_2),("(X)",,"(Y)"):}`
`{:(SO_2+,Ca(OH)_2to , CaSO_3 darr , + H_2O),("(Y)",,"white ppt.",):}`
`{:(2KMnO_4+,5SO_2+,2H_2OtoK_2SO_4+,2MnSO_4 +,2H_2SO_4),("Purple","(Y)",,"colourless",):}`
`{:(2SO_2+,O_2to,2SO_3),("(Y)",,"(Z)"):}`
`underset"(Z)"(SO_3) + H_2O to H_2SO_4` (responsible for ACID RAIN )
20.

An amorphous solid 'A' which has a crown shaped structure, burns in air to form a gas 'B' which turns lime water milky. 'B' is also produced by roasting of sulphide ores. 'B' undergoes oxidation in the presence of V_(2)O_(5) to give 'C' and to carry out this oxidation low temperature and high pressure is mandatory to get a good yeild of 'C'. 'C' is then absorbed in H_(2)SO_(4) to give 'D'. 'D' is then diluted to give a very important. 'E' in concentrated form, when combined with Cu metal, gives compound 'F'. From this description Give two important functions of 'E' in the chemical industry.

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SOLUTION :(i) MANUFACTURE of FERTILISERS LIKE ammonium sulphate.
(ii) In detergent inductry.
21.

An amorphous solid 'A' which has a crown shaped structure, burns in air to form a gas 'B' which turns lime water milky. 'B' is also produced by roasting of sulphide ores. 'B' undergoes oxidation in the presence of V_(2)O_(5) to give 'C' and to carry out this oxidation low temperature and high pressure is mandatory to get a good yeild of 'C'. 'C' is then absorbed in H_(2)SO_(4) to give 'D'. 'D' is then diluted to give a very important. 'E' in concentrated form, when combined with Cu metal, gives compound 'F'. From this description Give a balanced chemical equation for the conversion of 'E' to 'F'.

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SOLUTION :`Cu+2H_(2)SO("CONC.")toCuSO_(4)+SO_(2)+2H_(2)O`
22.

An amorphous solid 'A' which has a crown shaped structure, burns in air to form a gas 'B' which turns lime water milky. 'B' is also produced by roasting of sulphide ores. 'B' undergoes oxidation in the presence of V_(2)O_(5) to give 'C' and to carry out this oxidation low temperature and high pressure is mandatory to get a good yeild of 'C'. 'C' is then absorbed in H_(2)SO_(4) to give 'D'. 'D' is then diluted to give a very important. 'E' in concentrated form, when combined with Cu metal, gives compound 'F'. From this description Elucidate the structure of 'A' to 'F'.

Answer»

SOLUTION :
23.

An amorphous solid "A" burns in air to form a gas "B" which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO_4 solution and reduces Fe^(3+) to Fe^(2+)Identify the solid "A" and the gas "B" and write the reactions involved.

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Solution :(A) `S_(8)`, (B) `SO_(2)`
with OXYGEN (air), `S_(8)` oxidises to `SO_(2)`
`S_(8) + 8O_(2) overset(Delta)to 8SO_(2)`
`SO_(2) + Ca(OH)_(2) to CaSO_(3) + H_(2)O`
`5SO_(2) + underset("violet")(2MnO_(4)^(-)) + 2H_(2)O to 5SO_(4)^(2-) + 4H^(+) + underset("COLOURLESS") (2MN^(2+))`
`2Fe^(3+) + SO_(2) + 2H_(2)O to underset("green")(2Fe^(2+)) + SO_(4)^(2-) + 4H^(+)`
24.

An amorphous solid "A" burns in air to form a gas "B" which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO_4 solution and reduces Fe^(3+) to Fe^(2+). Identify the solid "A" and the gas "B" and write the reactions involved.

Answer»

Solution :The solid A is sulphur , `S_8` and the gas B is `SO_2`.
The reactions are given as under :
`S_8 + 8O_2 OVERSET(Delta)to 8SO_2`
`2MnO_4^(-) + 5SO_2 + 2H_2O to 5SO_4^(2-) + 4H^(+) + underset("(Colourless)")(2MN^(2+))`
`2Fe^(3+) + SO_2 + 2H_2O to 2Fe^(2+) + SO_4^(2-) + 4H^(+)`
`SO_2` is produced as a by - product during roasting of sulphide ore. The sulphides are converted into oxideswith the EVOLUTION of `SO_2`.
25.

An amorphous solid (A) burns in air to form a gas (B) which turns lime water milky . The gas is also produced as a byproductduring roasting of sulphide ore. This gas decolourises acidifiedaqueousKMnO_(4) solution and reduces Fe^(3+) to Fe^(2+). Identify the solid 'A' and the gas 'B' and write the reactions involved.

Answer»


SOLUTION :(i) Since the byproduct of roasting to sulphide ore is `SO_(2)`. It turns lime water milky. Therefore, gas 'B'must be `SO_(2)`.
(ii) As the gas 'B' is obtained when amorphous solid 'A' burns in air therefore, amorphous solid 'A' must be sulphur `s_(8)`.
`underset("(A)")(S_(8))+8O_(2) overset(Delta)RARR underset("(B)")(8SO_(2))`
Gas (B) reduces acidified aqueous KMnO4 solution and reduces `Fe^(3+)` to `Fe^(2+)` salts as shown below:
`underset("(yellow)")(2MnO_(4)^(-))+underset("(B)")(SO_(2))+2H_(2)O rarr underset("(GREEN)")(2Fe^(2+)+SO_(4)^(2-)+4H^(+))`
(IV) Thus, solid 'A' is `S_(8)` and gas 'B' is `SO_(2)`.
26.

An amorphous solidA burnsin airto form a gas B whichturns limewatermilky. Thegasis also producedduringroasting of sulphide ore. This gasdecolourises acidifedaqueousKMnO_(4) solutionand reduces Fe^(3+) to Fe^(2+). Identify the solid 'A' and thegas B and write thereactionsinvolved .

Answer»

Solution :`A to S_(8) , B to SO_(2) , S_(8) + 8O_(2) OVERSET(Delta)to 8SO_(2)`
`underset("(VIOLET)")(2MnO_(4)^(-)) + 5SO_(2) + 2H_(2)O to5SO_(4)^(2-) +4H^(+) + underset("(colouless)")(2Mn^(2+))`
`2Fe^(3+) + SO_(2) + 2H_(2)Oto 2Fe^(2+) + SO_(4)^(2-) + 4H^(+)`
27.

An amorphous solid 'A' burns in air to form a gas 'B' which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO_(4) solution and reduces Fe^(3+) to Fe^(2+). Identify the solid "A" and the gas "B" and write the reaction involved.

Answer»

Solution :(i) Since gas 'B' is obtained as a by-product during roasting of sulphide, therefore, gas 'B' must be `SO_(2)`.
`2 ZnS + 3 O_(2) rarr 2 ZnO + 2 SO_(2) (B)`
(ii) Since gas 'B' is obtained when AMORPHOUS SOLID 'A' burns in air, therefore, amorphous solid 'A' must be SULPHUR, `S_(8)`
`S_(8) (A) + 8 O_(2) rarr 8 SO_(2) (B)`
(iii) Gas 'B' reduces ACIDIFIED `KMnO_(4)` solution and reduces `Fe^(3+)` to `Fe^(2+)` salts as shown below :
`{:(underset(("Purple"))(2MnO_(4)^(-)) + underset((B))(5SO_(2))+ 2H_(2)O rarr underset(("COLOURLESS"))(2Mn^(2+))+5SO_(4)^(2-) + 4H^(+)),(underset(("Yellow"))(2Fe^(3+))+underset((B))(SO_(2))+2H_(2)O rarr underset(("Green"))(2Fe6(2+)) + SO_(4)^(2-) + 4 H^(+)):}`
Thus, solid 'A' = `S_(8)` and gas 'B' = `SO_(2)`.
28.

An amorphous solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aq. KMnO_(4) solution. Identify the solid ‘A’ and the gas ‘B’ and write the reaction involved. Write its any two uses.

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SOLUTION :It is USED as BLEACHING AGENT and DISINFECTANT.
29.

An amorphous solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aq. KMnO_(4) solution. Identify the solid ‘A’ and the gas ‘B’ and write the reaction involved. What happens when SO_(2) is passed through water and reacts withNaOH ? Write balanced equation.

Answer»

Solution :`2NaOH+SO_(2)(G)rarrNa_(2)SO_(3)(aq)+H_(2)O`
`Na_(2)SO_(3)(aq)+SO_(2)+H_(2)Orarr2NaHSO_(3)(aq)`
30.

An amorphous solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aq. KMnO_(4) solution. Identify the solid ‘A’ and the gas ‘B’ and write the reaction involved. How is SO_(2) prepared in laboratory ?

Answer»

Solution :`Na_(2)SO_(3)(s)+H_(2)SO_(4)(aq)rarrSO_(2)(g)+Na_(2)SO_(4)(aq)+H_(2)O(L)`
31.

An amorphous solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aq. KMnO_(4) solution. Identify the solid ‘A’ and the gas ‘B’ and write the reaction involved.

Answer»

SOLUTION :`A = S_(8), B = SO_(2) (G)`
32.

An ammonia solution is 9.9% ammonia by mass and has a density of 0.99 g/mL. Calculate the pH of the solution. K_b (NH_4OH) = 1.7 xx 10^(-5)

Answer»


ANSWER :`12.0`
33.

An amino acid with a phenolic hydroxyl group is

Answer»

alanine
tyrosine
valine
phenyl glycine

Answer :B
34.

An amine (X) reacts with benzene sulphonyl chloride and the product thus obtained is soluble in KOH . The amine (X) is

Answer»

`1^(@)`amine
`2^(@)` amine
`3^(@)` amine
NONE of thesen

Solution :Since an ainine (X) REACTS with benzene sulphonyl CHLORIDE to form a product which is soluble in KOH, therefore, amine (X) must be a `1^(@)` amine.
35.

An amine reacts with CH_(3)SO_(2)Cl and the product is soluble in alkali, amine is

Answer»

`1^(@)` AMINE
`2^(@)` amine
`3^(@)` amine
All of these

Answer :1
36.

An amino acid is characterized by two pKa values the one correspondingto the more acidic site is designated as pKa_1 and the other corresponding to the less acidic site is designated as pKa_2 The isoelectric point also called isoionic point (pI) is the pH at which concentration of zwitter ion is maximum . pI is the average of pKa_1 and pKa_2.Generally the value of pI is slightly less than 7. some amino acids have side chain with acidic or basic groups.These amino acids have pKa_3 value also for the side chain.Acidic amino acid have acidic side chains and basic amino acids have basic side chains . pI for acidic amino acid is average of pKa_1 and pKa_3.pI for basic amino acid is the average of pKa_2 and pKa_3. {:(S.NO,"Amino acid",p^(Ka1),p^(Ka2),p^(Ka3)("side chain")),(I,"Aspartic acid",1.88,9.6,3.65),(II,"Glutamic acid",2.19,9.67,4.25),(III,"Lysine",2.18,8.95,10.53),(IV,"Arginine",2.17,9.04,12.48):} In the table given above the acidic amino acids are

Answer»

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

Solution :`P_3^(KA)` value of side chain DETERMINES the NATURE of amino acid
37.

An amine reacts with C_6H_5SO_2Cl and the productis soluble in alkali amine is

Answer»

`3^(@)`
`1^(@)`
`2^(@)`
All

ANSWER :B
38.

An amine reacts with C_6H_5SO_2Cl and the product is soluble in alkali, amine is :

Answer»

`1^@`
`2^@`
`3^@`
All

Answer :A
39.

An amine reacts with benzenesulphonyl chloride to form a white precipitate which is insoluble in aq. NaOH. The amine is

Answer»




SOLUTION :`2^(@)` amines REACT with benzenesulphonyl chloride to give the corresponding benzenesulphonamide which due to the absence of acidic H on N does not dissolve in aq. NaOH.
40.

An amine hormone is

Answer»

Cortisone
ADRENALINE
Insulin
Estrone

Solution :Adrenaline is AMINE HORMONE.
41.

An amine A reacts with benzene sulphonyl chloride and the product, thus, formed is soluble in KOH. The correct representation of the functional group of amine is

Answer»

`-NH`
`-NH_(2)`
`-underset(|)overset(|)(N)-`
both A and B.

Solution :Only a `1^(@)` amines (with `-NH_(2)` group) REACT with benzene sulphonyl CHLORIDE to give a product which is soluble in `KOH`.

On the other HAND `2^(@)` amines react with benzene sulphonyl chloride to give a product which is insoluble in `KOH` whereas `3^(@)` amines do not react with benzene sulphonyl chloride.
42.

An amine C_(3)H_(9)N reacts with benzene sulphonyl chloride to fonn a white precipitate which is insoluble in aq. NaOH. The amine is

Answer»




Solution :`2^(@)`amine reacts with benzene sulphonyl CHLORIDE and the product FORMED is insoluble in NaOH. Therefore, the amine should have only ONE H-atom attached to nitrogen atom.
43.

An ambident nycleophile is

Answer»

AMMONIA
Ammonium ion
Chloride ion
NITRITE ion

SOLUTION :Nitrite ion
44.

An alumina-silica clay called bentonite is dropped from aeroplanes in the slurry form for :

Answer»

FERTILIZING the soil
Spreading WATER over fires
Cooling the soil
Fumigation

Answer :D
45.

An alpha- paticle approaches the target nucleus of copper (Z = 29) is such a way that the vlaue of impact parameter is zero. The distance of closest approach will be

Answer»

`(2pi epsilon_(0)(K.E.)_(alpha))/(29e^(2))`
`(29e^(2))/(2pi epsilon_(0)(K.E.)_(alpha))`
`(4pi epsilon_(0)(K.E.)_(alpha))/(29e^(2))`
`(K.E.)_(alpha)`

ANSWER :B
46.

An alpha-helix is a structural feature of

Answer»

Sucrose.
Polypeptides.
Nucleotides.
Starch.

Answer :B
47.

An alpha-amino acid exists as, overset(+)(NH_3)-underset(R)underset(|)CH-COOH at (pH=2) and its isoelectric point is 6. The amino acid at pH 10.97 will exist as :

Answer»

`OVERSET(+)(NH_3)-UNDERSET(R)underset(|)CH-COO^-`
`NH_2-underset(R)underset(|)CH-COO^-`
`NH_2-underset(R)underset(|)CH-COOH`
`underset(+)(NH_2)-underset(R)underset(|)overset(-)CH-COOH`

ANSWER :B
48.

An alloy which does not contain copper is :

Answer»

Solder
Bronze
Anode mud
Electrolyte

Answer :A
49.

Which of the following alloys does not contain copper?

Answer»

SOLDER
Bronze
Brass
Bell METAL

ANSWER :A
50.

An alloy weighing 1.05g of Pb-Ag was dissolved in desired amount of HNO_(3) and the volume was made 350 mL. An Ag electrode was dipped in solution and E_(cell) of the cell underset(1 atm)(Pt H_(2)|)underset(1 M)(H^(+))"||"Ag^(+)|Ag was 0.503V at 298K. calculate the percentage of lead in alloy. ("given," E_(Ag^(+)//Ag)^(@)=080V)

Answer»


ANSWER :`99.967% ;`