Explore topic-wise InterviewSolutions in .

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

93151.

1.0 g of a non-electolyte solute (molar mass 250 g mol^(-1)) was dissolved in 51.2 g of benzene. If the K_(f) for benzene is 5.12 K kg mol^(-1), the freezing point of benzene will be lowerd by :

Answer»

0.2 K
0.4 K
0.3 K
0.5 K.

Solution :`DeltaT_(f)=(K_(f)xxW_(B))/(M_(B)xxW_(A))`
`=((5.12 KG mol^(-1))XX(1.0g))/((250gmol^(-1))xx(0.0512kg))=0.4 K`
93152.

1.0 g of a non-electrolyte solute (molar mass 250 g "mol"^(-1)) was dissolved in 51.2 g of benzene. If the freezing point depression constant of benzene is 5.12 K kg "mol"^(-1), the lowering in freezing point will be

Answer»

0.5 K
0.4 K
0.2 K
0.3 K

ANSWER :B
93153.

1.0 g of a mixture of carbonates of calcium and magnesium gave 240 cm^(3) of CO_(2) at STP. Calculate the percentage composition of the mixture.

Answer»

Solution :Mass of mixture of carbonates of Ca and Mg TAKEN = 1.0 G
Suppose the mass of `CaCO_(3) = x g "" therefore`Mass of `MgCO_(3)=(1-x)g`
The CHEMICAL equations involved are : `underset(40+12+3xx16=100g)(CaCO_(3))rarr CaO+underset(22400 cm^(3) "at STP")(CO_(2))`....(i)
`underset(24+12+3xx16=84 g)(MgCO_(3))rarr MgO+underset(22400 cm^(3) "at STP")(CO_(2))`....(ii)
Step 1. To calculate the volume of `CO_(2)` evolved at STP from x g of `CaCO_(3)`.
100 g of `CaCO_(3)` evolve `CO_()` at STP `= 22400 cm^(3)`
`therefore x` g of `CaCO_(3)` will evolve `CO_(2)` at STP `= (22400)/(100)xx x cm^(3)`
Step 2. To calculate the volume of `CO_(2)` evolved at STP (1 - x) g of `MgCO_(3)`.
84 g of `MgCO_(3)` evolve `CO_(2)` at STP `= 22400 cm^(3)`
`therefore (1-x)` g of `MgCO_(3)` will evolve `CO_(2)` at STP `=(22400)/(84)xx(1-x)cm^(3)=(800)/(3)(1-x)cm^(3)`
Step 3. To calculate the value of x
Total volume of `CO_(2)` evolved at STP `= 224 x + (800)/(3)(1-x)cm^(3)`
But total volume of `CO_(2)` evolved at STP `= 240 cm^(3)` (Given)
`therefore 224 x+(800)/(3)(1-x)=240` or `672 x + 800-800x = 720` or 128 x = 80
`therefore x= (5)/(8)`
Step 4. To calculate the percentage composition of the mixture.
`therefore` Percentage of `CaCO_(3)=(5)/(8xx1)xx100=62.5`
`therefore` Percentage of `MgCO_(3)=100-62.5=37.5` .
93154.

10 g of a gas at NTP occupies a volume of 2 litres. At what temperature will the volume be double, pressure and amount of the gas remaining same ?

Answer»


ANSWER :B
93155.

10 g of a mixture of BaO and CaO requires 100cm^(3)of 2.5 M HCl to react completely. The percentageof calcium oxide in the mixture is approximately (Given : molar mass of BaO= 153)

Answer»

52.6
55.1
44.9
47.4

Solution :`BaO+2HCl rarr BaCl_(2)+H_(2)O`
`CaO+2HCl rarr CaCl_(2)+H_(2)O`
REACTIONS show that 2 moles of MIX. of BaOand CaO requires 4 moles of HCl to react completely.
HENCE, no. of moles of HCl `=(2.5xx100)/(1000)=0.25` moles
`{:("Now,",HCl,BaO+CaO,),("Initial","4 moles","2 moles",),("Final","0.25 moles",""?,):}`
No. of moles of MIXTURE of BaO and CaO
`=(0.25xx2)/(4)=0.125` moles
Let the mass of CaO be `x` G and BaO is `(10-x)` g
Then, `(x)/(56)+(10-x)/(153)=0.125` [ Mol. mass f CaO `=56, BaO=153`]
`153x+56(10-x)=0.125xx56xx153`
`153x+560-56x=0.125xx56xx153`
`97x=1071-560`
`97x=511`
`x=5.268`
`:.%` of `CaO=(5.268)/(10)xx100=52.68%`
93156.

10 g of a metal gives 14 g of its oxide. The equivalent mass of its oxide and hydroxide will be respectively

Answer»

20 and 37
28 and 37
56 and 74
None of these

Solution :Equivalent MASS of metal `= (10)/((14-10)) xx 8=20 G`
Hence, equivalent mass of metal oxide = Eq mass of metal + equivalent mass of oxygen = 20 +8=28
Equivalent mass of hydroxide = Equivalent mass of metal + Equivalent mass of `OH^(-)` = 20+ 17 = 37
93157.

10 g of a hydrated salt of BaCI_2 was dissolved in one litre water. The solution was then treated with 1.65 litre of 0.05 N AgNO_3 till complete precipitation. The number of water molecules in hydrated salt are :

Answer»

6
10
4
2

Answer :D
93158.

10 g of a gas at NTP occupies a volume of 2 litre. At what temperature will the volume of 2 litre. At what temperature will the volume be double, pressure and amount of the gas remaining same :

Answer»

273 k
546 k
`-273^@C`
`546^@C`

Answer :B
93159.

10 g mixture of KI and NaClO_(3) treated with 200 cc HCl gives a gas which absorbes in 40 ml of (N) /(10) sodium thiosulphate solution. Then the correct statement is//are

Answer»

equivalent weight of IODINE in the reaction is `M//2`
PERCENTAGE of KL in the mixtre is 66.4%
sodium thiosulphate converted into `Na_(2)S_(4)O_(6)`
percentage of `NaClO_(3)` is 66.4%

Answer :AC
93160.

10 g CaCO_(3) were dissolved in 250 ml of 1 M HCl. What volume of 2 M KOH would be required to neutralise excess HCl ?

Answer»


ANSWER :V= 25ML
93161.

10 g CaCO_(3)on heating leaves behind a residue weighing 5.6 g. Carbon dioxide released into the atmosphere at STP will be

Answer»

2.24 L
4.48 L
1.12 L
0.56 L

Solution :`CaCO_(3) to CAO + CO_(2)`
Thus 100G `CaCO_(3)` gives 22.4 L `CO_(2)` at STP. Hence 10 G will give 2.24 L `CO_(2)` at STP.
93162.

10 g atoms of an alpha-active radioactive isotope aredisintegrating in a sealed container. In one hour, helium gas collected at STP is 11.2 "cm"^3. The half-life of the radioactive isotope is

Answer»

138.6 hr
1386 hr
13860 hr
138600 hr

Solution :Gram atoms of HELIUM gas formed in 1 hour`=11.2/22400=5xx10^(-4)`
i.e., Gram atoms of radioactive isotope disintegrated in one hour =`5xx10^(-4)`
RATE of DISINTEGRATION = `lambdaN`
`5xx10^(-4) = lambda xx 10` or `lambda=5xx10^(-5)hr^(-1)`
`t_(1//2) = 0.693/lambda=0.693/(5xx10^(-5))` hr = 13860 hr
93163.

1.0 g an alloy of aluminium and magnesium when treated with excess of dilute HCl forms magnesium chloride and aluminium chloride and hydrogen collected over mercury at 0^(@)C has a volume of 1.20 L at 0.92 atmospheric pressure. Calculate the composition of the alloy.

Answer»

Solution :SUPPOSE Al in the alloy = X g
Then Mg in the alloy `=(1-x)g`
Al and Mg in the alloy will react with HCL acid as follows :
`{:(underset(2xx27=54g)(2Al)+6HCl rarr2AlCl_(3)+underset(3xx22.4"L at S.T.P.")(3H_(2))),(""underset(24g)(Mg)+2HCl RARR MgCl_(2)+underset("22.4 L at S.T.P.")(H_(2))):}`
`H_(2)" priduced from x g of Al"=(3xx22.4)/(54)xx x=(22.4x)/(18)"L at S.T.P.`
`H_(2)" produced from "(1-x)g" of Mg"=(22.4)/(24)xx(1-x)"L at S.T.P."`
`therefore" Total "H_(2)" produced at S.T.P."=(22.4x)/(18)+(22.4(1-x))/(24)"L at S.T.P."`
Let us now convert the actual volumeof `H_(2)` produced to VOLUME at S.T.P.
`{:(P_(1)=0.92atm""P_(2)=1atm),(V_(1)=1.20L""V_(2)=?),(T_(1)=273K""T_(2)=273K),((P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) therefore (0.92xx1.20)/(273)or v_(2)=1.104L):}`
`"or"4xx22.4x+3xx22.4(1-x)=1.104xx72`
`"or"89.6x+67.2-67.2x=79.488`
`"or"22.4x=12.888 or x=0.5486g`
`therefore""%" of Al"=54.86 and %" of Mg"=100-54.86=49.14.`
93164.

10 dm^(3)of N_(2)gas and 10 dm^(3)of gas X at the same temperature contain the same number of molecules. The gas X is

Answer»

CO
`CO_(2)`
`H_(2)`
NO

SOLUTION :Number of moles of `N_(2)` and X should be equal. This can be so if X has same molecular WEIGHT as `N_(2)`
93165.

10 dm^(3) of N_(2) gas and 10 dm^(3) of gas X at the same temperature and pressure contain the same number of molecules. The gas X is

Answer»

`CO_(2)`
CO
NO
`H_(2)`

SOLUTION :GAS X MUST have the same MOLECULAR mass
93166.

10 dm^(2) of N_(2) gas and 10 dm^(3) of gas X at the same temperature contain the same number of molecules. The gas X is

Answer»

CO
`CO_(2)`
`H_(2)`
`NO`

SOLUTION :No. of MOLES of `N_(2)` and X should be equal. This can be so if X has same molecular mass as `N_(2)` viz. CO.
93167.

10 cc of H_(2) O_(2)solution when reacted withKIsolution produced 0 . 5 g of iodine . Calculate the percentage purity of H_(2) O_(2)

Answer»

Solution :`H_(2) O_(2) + KI to KOH + I_(2)`
First, applyingPOACfor H atoms to calculate moles of KOH, thenapplying POAC for Katoms to calculate moles of KI, andthenfinallyapplying POACfor Iatoms tocalculate moles ofiodine.
POACfor H atoms :
molesof H in `H_(2)O_(2) ` = moles of H in KOH
`2 xx ` moles of `H_(2)O_(2) = 1 xx ` moles of KOH. . . (i)
Applying POAC for K atons,
molesof K atoms in KI= moles o F K atoms in KOH
`1 xx` molesof KI = ` 1 xx ` moles of KOH
`=2 xx ` molesof `H_(2) O_(2)`
[from the equ. (i)]
Applying POACfor I atoms,
molesof I atoms in KI = moles of I atoms in` I_(2)`
or moles of `I_(2) = (1)/(2) xx ` moles of KI
`= (1)/(2) xx 2 ` molesof `H_(2) O_(2)`[from the eqn. (ii)]
Now, `(wt. of I_(2))/( mol. wt. of I_(2)) = (wt. of H_(2) O_(2))/(mol . wt. of H_(2)O_(2))`
Suppose x is thewt. of `H_(2)O_(2)` . Then,
`(0. 5)/( 254) = (x)/( 34)`
`x = (34 xx 0 . 5)/( 254) = 0.0669 G` .
`% of H_(2)O_(2) = (0.0669)/(10) xx 100 = 0 . 669 %`
93168.

10 cc of a liquid A were mixed with 10 cc of liquid B. The volume of the resulting solution was found to be 19.9 cc. What do you conclude?

Answer»

Solution :DECREASE in volume means stronger intermolecular forces of ATTRACTION on MIXING. This implies that the solution SHOWS negative deviation from RAOULT's law.
93169.

10^(-6)M NaOH is diluted to 100 times. The pH of the diluted base is

Answer»

between 5 and 6
between 6 and 7
between 10 and 11
between 7 and 8

Solution :`[OH^(-)]` of `10^(-6)` M NaOH `=10^(-6)`
when 100 times DILUTED
`[OH^(-)]=(10^(-6))/(10^(2))=10^(-8)`
As it is a WEAK base, autoprotolysis of water will also take place. So, thecontributaioin of `OH^(-)` ions from water MUST also be considered.
`:.` Total `[OH^(-)]=10^(-8)+[OH^(-)]_("water")`
93170.

An acid solution of pH = 6 is diluted 1000 times, the pH of the final solution becomes :

Answer»

between 5 and 6
between 6 and 7
between 10 and 11
between 7 and 8

Solution :`[OH^(-)]=10^(-6)XX(1)/(100)+10^(-7)=10^(-8)+10^(-7)`
`=1.1xx10^(-7)`
`pOH=-LOG[OH^(-)]=-log(1.1xx10^(-7))`
`=6.96`
`pH=14-6.96=7.04`
93171.

10^-6 M HCl is diluted to 100 times. Find its pH value.

Answer»

6
8
6.95
9.5

Answer :C
93172.

10^(-6) M NaOH is diluted 100 times. The pH of the diluted base is

Answer»

Between 5 and 6
Between 6 and 7
Between 10 and 11
Between 7 and 8

Solution :`[OH^(-)]` of `10^(-6)M NaOH = 10^(-6)`.
When 100 TIMES diluted,
`[OH^(-)] = (10^(-6))/(10^(2)) = 10^(-8)`
As it is weak base, but proteolysis of water will also TAKE place. So, the contribution of `OH^(-)` ions from water must also be considered.
`:.` Total `[OH^(-)] = 10^(-8) + 10^(-7)`.
93173.

10^(-4)g ofgelatin is required to be added to 100 cm^(3)of a standard glodsolutionof justpreventits precipitation by theaddition of 1 cm^(3) of 10% NaCl solutionto it . Hence , the glod numberof gelation in mg is .

Answer»

10
0.01
0.001
0.1

Solution :Gelatin added to `100 cm^(3)` of astandard goldsolutionto preventprecipiation by` 1 cm^(3)` of 10% NaClsolution` = 10^(-4) G = 10^(-1) mg `
Gelation required to be added to `10 cm^(3)` of gold solutionto completelypreventcoagulationby `1 cm^(3)` OF10% NaCl solution` = 10^(-2) mg`
`therefore` By DEFINITION gold numberofgelatin ` = 10^(-2) mg `
` = 0.01 mg`
93174.

10^(-4)g of gelation is required to be added to 100 cm^(3) of a standard gold solution to just prevent its precipitation by addition of 1 cm^(3) of 10% NaCI solution to it . Hence the gold number of gelation in mg is

Answer»

10
`1.0`
`0.1`
`0.01`

SOLUTION :By defination 100 ML of gold sol REQUIRES `10^(-4)` GM 10 ml of gold sol requires
`=(10)/(100)xx10^(-4) =10^(-5)` gms
93175.

10^(-2)F charge can remove entrie Cl_2 from 100 ml of HCl solution. What is the P^(H) of the solution initially

Answer»


Solution :`1F to ` remove1 MOLE ` Cl^(Ɵ)` to ` (1)/(2)Cl_2 , 10^(-2) F to ` remove ` 10^(-2)` mole `Cl^(Ɵ)` to `((10^(-2))/(2))Cl_2`
`(H^(OPLUS)) = [HCl]=(n)/(v) = (10^(-2)" mole")/((100//1000)) = 10^(-2) xx 10 = 10^(-1) , P^(H) =1`
93176.

10^-2 g atom of Ag can be oxidised to Ag^+ during the electrolsis of AgNO_3 solution using silver electrode by:

Answer»

965 coulomb
96500 coulomb
9650 coulomb
96.500 coulomb

Answer :A
93177.

1 yoctometer is…………………..m whereas 1 yottametre is……………..m.

Answer»

SOLUTION :`10^(-24),10^(24)`
93178.

(1) Write down the IUPAC nomenclature of the following:(a)NH_(4)[Cr(NCS)_(4)(NH_(3))_(2)](b)K_(3)[Cr(C_(2)O_(4))_(3)] (c) Fe(C_(2)O_(4))_(2)N_(2)O_(2)]^(+) (ii)(a) K_(2)[OsCI_(5)N](b) [Zn(NCS)_(4)]^(2-) (c) [Cd(SCN)_(4)]^(3-)(d) [Cr(en)_(3)]CI_(3)]Cl_(3) (e) K_(4)[Fe(NC)_(6)] [Co (NH_(3))_(5) ONO]Cl_(2) (g) K_(3)[Cr(CN)_(6)](h) [Co(NH_(3))_(5)NO_(2)(NO)_(3))_(2)](i)[Co(en)(NH_(3))_(2)CIBr]NO_(3)(j) Na_(2) [Fe(CN)_(5)NO] (iii) Write the formula for each of the following complexes. (a) Chlorodiammineplatinum (II) ion (b) Dichlorobis (ethylenediaminie) cobalt(II) ion (c) Dichlorotetramminecobalt (iii) ion (d) Potassium pentacyanonitrosylcobaltate (III) (e) Sodium ethylenediaminetetraacetato chromate (ll) (f) Carbonylchlorobis(triphenylphosphine)iridium(I) (g) Ammoniumtrisoxalatocobaltate(iii) ( Puntaaquahydroxoelumimumf(lll)sniphate (i) Hexaaquavanadium(II) chloride (j)Tetraamminedichlorocobalt(II) hexacyanochromate(V)

Answer»

Solution :Ammoniumdiamminetetrakisthiocyanato(-N)chromate(III0
(b) Potassiumtrisoxalatochromate(III)
(c)Potassiumdinitrogenbisoxalatodixoygenferrate(III)
(Potassiumpentachloronitridoosmate(VI)
(b) Tetrathiocyanato-N-zinc(ll)
(c) Tetrathiocyanato-s-cadmium(I)
(d) Tris(ethylenediamine)chromium(III) chloride
(E) Potassium hexacyano-N-ferrate(II)
(f) Pentaamminenitritocobalt(III) chloride (9) Potassium hexacyano-C-chromate(III)
(h) Pentaamminenitrocobalt(III) nitrate (i)-Diamminebromochloroethylenediaminecobalt(III) nitrate
(J) Sodiumpentacyanonitrosylferrate(III)
(a) `[PtCI(NH_(3))_(2)]^(+)`,br>(b) `[CoCl_(2)(en)_(2)]^(+)`
(c) `[CoCl_(2)(NH_(3))_(4)^(+)]`
(d) `K_(2)[CO(CN)_(5)NO]`
(e) `Na_(2) [Cr(CH_(3)COO)_(4)(en)]`
(f) `[lr(Ph_(3)P)_(2)(CO)CI]`
(g) `(NH_(4))_(3)[Co(C_(2)O_(4))3]`
(h) `[Al(OH)(H_(2)O)_(5)]SO_(4)`
`[V(H_(2)O)_(6)]Cl_(3))`
`[CO(NH_(3)).Cl_(2)][Cr(CN)_(6)]`
93179.

1 xx 10^(-3) mole of HCl is added to a buffer solution made up of 0.01 M acetic acid and 0.01 M sodium acetate. The final pH of the buffer will be (given pK_(a) of acetic acid is 4.75 at 25^(@)C)

Answer»

`4.60`
4.66
4.75
4.8

Solution :`{:(CH_(3)COO^(-),+,H^(+),RARR,CH_(3)COOH),(" "0.01,,0.001,,""0.01),(0.01-0.001,,,,0.01+0.001),(= 0.009,,,,= 0.011):}`
`pH = pKa + log.(["SALT"])/(["ACID"]) = 4.75 + log. (0.009)/(0.011) = 4.66`.
93180.

1 xx10^(-3)mole of HCI is added to a buffer solution made up of 0.01 M acetic acid and 0.01 M sodium acetate. The final pH of the buffer will be given, PK of acetic acid is 4.75 at 25^@ C)

Answer»

`4.60 `
`4.66`
`4.75`
`4.8`

SOLUTION :
`PH =pK_a + LOG(["SALT"])/(["ACID"])=4.75 + log( 0.009 )/( 0.011 ) = 4.66`
93181.

1% (wt./vol) solution of urea is isotonic with 3% (wt./vol.) solution of a non electrolytic solute . The mol. Wt. of solute is

Answer»

240
120
180
60

Answer :C
93182.

1. Which of the following reaction will give 1-methoxy-4-nitrobenzene. (a) 4-nitro-1-bromobenzene + sodiummethoxide. (b) 4-nitrosodium phenoxide + bromomethane

Answer»

Solution :(a) SET - (a) :

(b) Set - (b) :

Chemically both sets equally possible. In Set-1, the Br-atom is ACTIVATED by electron withdrawing effect of `-NO_2` group. Therefore nucleophilic attack by `CH_3ONa` followed by elimination of NaBr gives the desired ether.

In set-2, nucleophilic attack by 4-nitrosodium phenoxide ION on methyl bromide gives the desiredether

SINCE alkyl halides `(CH_3Br)` are more REACTIVE thanaryl halides in nucleophilic substitution reactions, therefore set-2 reactants are preferTed.
93183.

(1) We metallurgical process is used for pyrites ores of lower grade. (2) 2Cl^(-) + 2H_(2)O to 2OH^(-) + H_(2) + Cl_(2) . The cell potential is -2.186 V. This reaction will take place in forward direction. (3) Pure Zn metal is called spleter. (4) The abundance of Al is highest . Its place is third and is about 9.3 % by weight .

Answer»

TFFT
TTTT
FTFT
FFFT

Answer :A
93184.

1 volume of a gaseous compound consisting C,H,O on complete combustion in presence of 2.5 volume of O_(2) gives 2 vol. of steam and 2 volt. Of CO_(2), what is the formula of the compound if all measurements are made at NTP?

Answer»


ANSWER :`C_(2)H_(4)O`
93185.

(1) underset((d)) overset(n-BuLi)toA+B (B does not contain pi bond) (2) Aoverset(Cl-CH_(2)OMe)(to)C Which is/are true for above reactions?

Answer»

`B` can be prepared by corey house synthesis
`C` after huydrogenation on TREATMENT with 1 eq cold conc `HI` PRODUCES an alkyl HALIDE with highest density
`C` does not decolorizes `Br_(2)`-water
`A` is organometallic compound

Solution :
93186.

1% solution of phenol 40% aqueous solution or methanal, SO_2 solution Cl_2 solution H_2O_2 solution among them how many chemicals are disinfections.

Answer»


SOLUTION :1% PHENOL , FORMALIN, `SO_2` solution ,`Cl_2 ` solution
93187.

1 solution contains a mixture of Na_(2)CO_(3) and NaOH. Using phenolphthalein as indicator. 25ml. of mixture required 19.5 ml. of 0.995 N HCl for the end point. With methyl orange. 25ml. of solution required 25 ml. of the same. HCl for the point. Calculate grams per litre of each substance in the mixture.

Answer»


ANSWER :23.3,24,4
93188.

1% sollution of Ca(NO_(3))_(2) has freezing point

Answer»

`0^(@)C`
Less than `0^(@)C`
GREATER than `0^(@)C`
None of the above

Solution :Freezing POINT is lowered on ADDITION of solute in it.
93189.

_1 , r_2 , r_3 represents rate of cuclisation. Then relation between r_1, r_2 and r_3 is

Answer»

`r_1 GT r_2 gt r_3`
`r_3 gt r_2 gt r_1`
`r_1 gt r_3 gt r_3`
`r_1 = r_2 = r_3`

Solution :Electron REALISING groups favors ESTERFICATION `r_3 gt r_2 gt r_1`
93190.

1- propanol on oxidation give

Answer»

PROPANAL
propionic ACID
PROPANONE
ETHANAL

Answer :B
93191.

1-propanol is obtained from C_(2)H_(5) MgI and what ?

Answer»

`CH_(3)CHO`
`HCHO`
`HCN`
`C_(2)H_(5)CHO`

Answer :B
93192.

1-propanol is converted into propene, which ofthe following agent is used ?

Answer»

ALC. KOH
dil. NaOH
dil.HCl
conc. `H_2SO_4`

ANSWER :D
93193.

1-Propanol and 2-propanol can be distinguished by

Answer»

oxidation with alkaline KMnO4 followed by reaction with Fehling solution
oxidation with acidic DICHROMATE followed by reaction with Fehling solution
oxidation by heating with copper followed by reaction with Fehling solution
oxidation with CONCENTRATED `H_(2)SO_(4)` followed by reaction with Fehling solution.

Solution :1-Propanol and 2-propanol in presence of `Cu//Delta` undergo dehydrogenation to give acetaldehyde and acetone respectively which can be DISTINGUISHED by Fehling solution since acetaldehyde reduces while acetone does not reduce Fehling solution.
93194.

1-propanol can be converted into 1-chloro propane by HCI in the presence of catalyst,

Answer»

HYDROUS `ZnCl_2`
unhydrous `ZnCl_2`
`H_3PO_4`
`P_2O_5`

ANSWER :B
93195.

1-Propanol and 2- propanol can be distinguished by:

Answer»

a.Oxidation with alkaline `KMnO_(4)` FOLLOWED by reaction with Fehiling's solution.
b.Oxidation with acidic DICHROMATE followed by reaction with Fehling's solution
c.Oxidation by heating with COPPER followed by reaction with Fehling's solution.
d.Oxidation with concentrated `H_(2)SO_(4)` follwed by reaction with Fehling's solution.

Solution :This can be SEEN from the reaction below:
a.
93196.

1-propanol and 2-propanol can be best distinquishedby

Answer»

oxidation with alkaline`KMnO_(4)` followed by reaction with Fehling solution
Oxidation with acidic dichlormate followed by reaction with Fehling solution
Oxiation by HEATING with copper followed by REACTIONWITH Fehling solution
Oxidation with concentration`H_(2)SO_(4)` followed by reaction with Fehling solution

Solution :`underset("1-propanol")(CH_(3)-CH_(2)-CH_(2)OH)underset(Delta300^(@)C)overset(Cu)to CH_(3)-CH_(2)-CHO underset("Fehling solution")overset(2CuO)to underset("red ppt.")(Cu_(2)O)DARR + CH_(3)CH_(2) - COOH `
`underset("2-proponol")(CH_(3)-underset(OH)underset(|)CH-CH_(3))underset(Delta,300^(@)C)overset(Cu)to CH_(3)-CO-CH_(3)underset("Fehling solution")overset(2CuO)to ` No reaction
93197.

1-Propanal and 2-propanol can be best distinguished by

Answer»

oxidation with `KMnO_(4)` followed by reaction with Fehling solution
oxidation with acidic dichromate followed by reaction with Fehling solution
oxidation by HEATING with copper followed by reaction with Fehling solution
oxidation with CONCENTRATED `H_(2)SO_(4)` followed by reaction with Fehling solution.

Answer :C
93198.

1 ppm = ……….. Mass of solute in 1 litre solution.

Answer»

`10^(-3)` gram
`10^(-4)` gram
`10^(-6)` gram
`10^(-2)` gram

Answer :A
93199.

1- Phenylethanol can be prepared by the reaction of benzaldehydewith

Answer»

METHYL BROMIDE
ethyl IODIDE and magnesium
methyl iodidie and magnesium
methyl bromide and ALUMINIUM bromide

Answer :C
93200.

1 - phenyl ethanoyl chloride on hydrolysis gives

Answer»

ethanoicacid
1- phenylethanoicacid
2- phenylethanoicacid
BENZOIC acid

Answer :C