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85551.

After adding non-volatile solute, freezing point of water decreases to -0.186^(@)C. Calculate Delta T_(b) if : K_(f) = 1.86 "K kg mol"^(-1) and K_(b) = 0.521 "K kg mol"^(-1)

Answer»

`0.521`
`0.0521`
`1.86`
`0.0186`

Solution :`DELTA T_(F)=K_(f)xx m rArr 0.186=1.86xx m`
So m = 0.1, Put the VALUE of m in `Delta T_(b)=K_(b)xx m`
`Delta T_(b)=0.521xx(0.1)=0.0521`
85552.

acetyl salicylic acid, analgesic, antipyretic, antibiotics, relieve or decerease pain, antacid, 2-acetoxybenzoic acid, preservatives, synthetic detergent, aspartame, chloroxylenol, a-terpineol, wide spectrum Sodium p-n decylbenzenesulfonate is a……........... .

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SOLUTION :SYNTHETIC DETERGENT.
85553.

acetyl salicylic acid, analgesic, antipyretic, antibiotics, relieve or decerease pain, antacid, 2-acetoxybenzoic acid, preservatives, synthetic detergent, aspartame, chloroxylenol, a-terpineol, wide spectrum Main constituents of dettol are ........ and ............ .

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SOLUTION :CHLOROXYLENOL, a-terpineol
85554.

After adding non-volatile solute freezing point of water decreases to -0.186^@C. Calculate DeltaT_b if K_f = 1.86 K kg mol^(-1)and K_b = 0.521 K kg mol^(-1)

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`0.0521`K
`0.0186` K
`0.521 `K
`1.86` K

SOLUTION :`DeltaT_(f) = T_(f) = T_(f)^(@) - T_(f) = 0 - (-0.186) = 0.186 ^(@)C`
`DeltaT_(f) = K_(f) xx m `
` RARR m = (DeltaT_(f))/(K_(f)) = (0.186)/(1.86) = 0.1 m `
` DeltaT_(b) = K_(b) xx m = 0.521 xx 0.1= 0.0521 K `
85555.

acetyl salicylic acid, analgesic, antipyretic, antibiotics, relieve or decerease pain, antacid, 2-acetoxybenzoic acid, preservatives, synthetic detergent, aspartame, chloroxylenol, a-terpineol, wide spectrum Chloramphenicol is a ............... antibiotic.

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SOLUTION :WIDE SPECTRUM
85556.

acetyl salicylic acid, analgesic, antipyretic, antibiotics, relieve or decerease pain, antacid, 2-acetoxybenzoic acid, preservatives, synthetic detergent, aspartame, chloroxylenol, a-terpineol, wide spectrum Tranquillizers are the chemicals used to treat ............... .

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SOLUTION :MENTAL DISESES
85557.

After a discussion about thestructures of hydrides of group-15 elements, .Neethu wrote the order of bond angles as NH_3ltPH_3ltAsH_3. Also arrange the above hydrides in the increasing order of their thermal stability. Justify your answer.

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SOLUTION :Thermal STABILITY order is `AsH_3ltPH_3ltNH_3` The thermal stability is related to the STRENGTH of E-H BOND(where E=N, Por As).The strength of E-H bond DECREASES from `NH_3`to `AsH_3`.
85558.

acetyl salicylic acid, analgesic, antipyretic, antibiotics, relieve or decerease pain, antacid, 2-acetoxybenzoic acid, preservatives, synthetic detergent, aspartame, chloroxylenol, a-terpineol, wide spectrum Aspirin acts as ………...... and ………….. .

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SOLUTION :ANALGESIC, ANTIPYRETIC
85559.

After a discussion about the structures of hydrides of group-15 elements, .Neethu wrote the order of bond angles as NH_3ltPH_3ltAsH_3. Justify ‘your answer:

Answer»

Solution :The bond angle decreases from `NH_3`to `BiH_3`,i.e., the correct ORDER of bond’ angle is `NH_3gtPH_3gtAsH_3`.As the size of the central ATOM in the hydride INCREASES, its electronegativity. decreases, Consequently the position of the bond PAIR of electrons tend to lie away from the central atom in moving from `NH_3` to `AsH_3` As a result the REPULSION between bond pair of electrons' ' decreases and the bond angle decreases.
85560.

acetyl salicylic acid, analgesic, antipyretic, antibiotics, relieve or decerease pain, antacid, 2-acetoxybenzoic acid, preservatives, synthetic detergent, aspartame, chloroxylenol, a-terpineol, wide spectrum Analgesics are drugs which .................. .

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SOLUTION :RELIEVE or DECREASE PAIN
85561.

After a discussion about the structures of hydrides of group-15 elements, .Neethu wrote the order of bond angles as NH_3ltPH_3ltAsH_3. Give the hybridization and shape of these hydrides.

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SOLUTION :HYBRIDISATION: `sp^3` SHAPE: PYRAMIDAL
85562.

acetyl salicylic acid, analgesic, antipyretic, antibiotics, relieve or decerease pain, antacid, 2-acetoxybenzoic acid, preservatives, synthetic detergent, aspartame, chloroxylenol, a-terpineol, wide spectrum The chemical name of aspirin is ................ .

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SOLUTION :2-Acetoxybenzoic ACID or acetyl-salicylic acid
85563.

After a discussion about the structures of hydrides of group-15 elements, .Neethu wrote the order of bond angles as NH_3ltPH_3ltAsH_3. Is this the correct order?

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SOLUTION :The ORDER WRITTEN is RIOT CORRECT
85564.

acetyl salicylic acid, analgesic, antipyretic, antibiotics, relieve or decerease pain, antacid, 2-acetoxybenzoic acid, preservatives, synthetic detergent, aspartame, chloroxylenol, a-terpineol, wide spectrum Substances produced by one microorganism are toxic to other microorganism are called ................ .

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SOLUTION :ANTIBIOTICS
85565.

After 24 hrs, only 0.125 g out of the initial quantity of 1 g of a radioactive isotope remains behind. What is its half-life period?

Answer»

Solution :The DATA given can be WRITTEN as under:
`a=1g, a-x=0.125g, t=24` hours
Substituting the VALUES in the first order equation
`K=(2.303)/(t)"LOG"(a)/((a-x))`, we have
`k=(2.303)/(24h)xx"log"(1)/(0.125)`
We have, `k=(2.303)/(24h)xx log 8=0.0866 h^(-1)`
`t_(1//2)=(0.693)/(k)=(0.693)/(0.0866)"hours"=8` hours
85566.

Acetyl derivative of which of which carbohydrate is used in sizing of industry:

Answer»

Glucose
Fructose
Lactose
Starch

Answer :B
85567.

Acetyl derivative of which carbohydrate is used in sizing industry:

Answer»

Glucose
Fructose
Lactose
Starch

Answer :B
85568.

Acetyl chloride reacts with ether in the presence of zinc chloride to form:

Answer»

`CH_(3)COOCOCH_(3) + C_(2)H_(5)Cl`
`CH_(3)COOC_(2)H_(5) + CH_(3)Cl`
`CH_(3)COOC_(2)H_(5) + C_(2)H_(5)Cl`
`CH_(3)COOH + C_(2)H_(5)Cl`

ANSWER :A
85569.

After 24 hours, only 0.125 g out of the initial quantity of 1g of a radioisotope remains behind. what is half-life period?

Answer»

Solution :`N_(0)` N=0.125 G ,t=24 HOURS
`therefore lambda =(2.303)/(t)log(N_(0))/(N)`
`=(2.303)/(24) log (1)/(0.25)`
`therefore t^(1//2) =(0.693)/(lambda)=(0.639)/(0.866)=7.99` hours
85570.

Acetyl chloride reacts with ................. to form acetone.

Answer»

SOLUTION :DIMETHYL CADMIUM
85571.

After 24 hour, only 0.125" g" out of the initial quantity of 1 g of a radioactive isotope remains behind. The half-life period of the radioactive isotope in hours is

Answer»

`7.2 h`
`7.99 h`
`6.99 h`
`10.0 h`

Solution :In this problem , 'a'=1.0g and 'a-x'=0.125gand 't'=24h. Therefore,the DISINTEGRATION CONSTANT.
`lambda= (2.303)/(t) "log" (a)/(a-x)`
Or `lambda =(2.303)/(24)"log" (1)/(0.125)=0.0866h^(-1) " "` Or `" " t_(1//2)=(0.693)/(0.0866)=7.99` hours
85572.

Acetyl chloride on treating with ammonia gives :

Answer»

Acetylamine
Aminoacetyl chloride
Acetamide
Amine

Solution :`CH_(3)COCl+2NH_(3)2NH_(3)toCH_(3)CONH_(2)+NH_(4)Cl`
85573.

After 2 hours , a radioactive substance becomes (1/16)^(th) of original amount . Then the half life (in min) is ..........

Answer»

60 minutes
120 minutes
30 minutes
15 minutes

Solution :`1overset(t_1//2)rarr((1)/(2))overset(t_1//2)rarr((1)/(4))overset(t_1//2)rarr((1)/(8))overset(t_1//2)rarr((1)/(16))`
`:. 4 t^(1//2)=2` hours
`t_(1//2)=30` MIN
85574.

Acetyl chloride reacts with aniline a product which on reduction with LiAlH_(4) give

Answer»




SOLUTION :
85575.

After 2 hours, a radioactive substance becomes ((1)/(16))^(th) of original amount. Then the half life (in min) is

Answer»

60 minutes
120 minutes
30 minutes
15 minutes

Solution :`t OVERSET(1//2) to (1/2) overset(t_(1//2)) to (1/4) overset(t_(1//2)) to (1/8) overset(t_(1//2)) to (1/16)`
`THEREFORE 4 t^(1//2) = 2` HOURS `rArr t^(1//2) = 30` min.
85576.

Acetyl chloride is prepared industrially by distilling sodium acetate with:

Answer»

HCL
`SO_(2)Cl_(2)`
`PCl_(5)`
`POCl_(3)`

Answer :B::D
85577.

Acetyl chloride is reduced to acetaldehyde by:

Answer»

`Na+C_2H_5OH`
`LiAlH_4`
`H_2+Pd-BaSO_4`
`H_2+Ni`

ANSWER :C
85578.

After 11.2 g of carbon reacts with oxygen originally occupying 21.2 litres at 18^@Cand 750 mmHg, the cooled gases are passed through 3 litres of 2.50 M NaOH solution. Determine the concentration of NaOH remaining in solution which is

Answer»

SOLUTION : CO does not REACT with NAOH under these conditions.
1.95M
85579.

For a given data : E_(Mg^(2+) // Mg)^(@) = -2.37 V , E_(Cu^(2+)// Cu)^(@) = + 0.34 V Calculate the emf of the cell in which the following reaction takes place . underset((0.0001 M)) ( Mg_((s))) + Cu_((aq))^(2+)underset((0.001M)) to Mg_((aq))^(2+) + Cu_((s))

Answer»

Solution :Limiting molar CONDUCTIVITY of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.If the molar conductivity of cation is denoted by and that of anion `X^@`.
Where `V_(+)` and `V_(-)` are the number of cations and anions per formula unit of an electrolyte.
(ii)Mercury cell is a primary cell. SUITABLE for low CURRENT devices.
Anode: Zinc - Mercury amalgam
Cathode:Paste of HgO and Carbon
Electrolyte: Paste of KOH and ZnO.
Cell reactions:Anode:`Zn(HG) 4- 2OH to ZnO(s) + H_(2)O + 2e^(-)`
Cathode:`HgO + H_(2)O_(4)- 2e^(-) to Hg(Z) 4- 2OH^(-)`
Over all reaction:`Zn(Hg) + HgO(s) to ZnO(s) + Hg(Z)`
Cell potential1.35 V.
85580.

Acetyl chloride fumes when exposed to air. Give reason.

Answer»

Solution : Acetyl chloride emit PALE FUMES of HYDROGEN chloride when exposed to air on account of their reaction with WATER vapour.
`UNDERSET("acetyl chloride")(CH_3COCl) + H_2O to underset("acetic acid")(CH_3COOH) + underset("pale fumes")(HCl uarr)`
85581.

(A)Find the solubility product of a saturated solution of Ag_(2)CrO_(4) in water at 298 K if the emf of the cell Ag|Ag+ (saturated. Ag_(2)CrO_(4) solution) || Ag+ (0.1 M) | Ag is 0.164 V at 298 K.(B) What will be the resultant pH when 200 mL of an aqueous solution of HCl (pH = 2.0) is mixed with 300 mL of an aqueous of NaOH (pH = 12.0) ?

Answer»

Solution :(a) `E = 0.164 = -0.059 log[Ag^(+)]_(anode)/0.10`
`[Ag^(+)]_(anode) = 1.66 xx 10^(-4) Ml`
`[CrO_(4)^(2-)] = [Ag^(+)]/2 = 8.3 xx 10^(-5)M`
`K_(SP) = [Ag^(+)]^(2)[CrO_(4)^(2-)] = (1.66 xx 10^(-4))^(2)(8.3 xx 10^(-5)) = 2.3 xx 10^(-12)`
(b) pH of HCl = 2
`:. [H^(+)] = 10^(-2)M`
Moles of `H^(+)` ions in 200 mL of `10^(-2) M` HCl solution` = 10^(-2)/1000 xx 200 = 2 xx 10^(-3`)
Similarly, pH of NaOH = 12
`:. [H^(+)] = 10^(-12)M or [OH^(-)] = 10^(-2)M[:.[H^(+)][OH^(-)] = 10^(-14)M]`
Moles of `[OH^(-)]` ion in 300 mL of `10^(-2) M` NaOH solution `= 10^(-2)/1000 xx 300 = 3 xx 10^(-3)`
Total volume of solution after mixing = 500 mL
Moles of `OH^(-)` ion left in 500 mL of solution `= (3 xx 10^(-3))-(2 xx 10^(-3)) = 10^(-3)`
Solution `= 10^(-3)/500 xx 1000 = 2 xx 10^(-3) M`
`pOH = - log(2 xx 10^(3)) = -log2 + 3log10 = -0.3 103 = 2.699`
`:. pH = 14 - 2.699 = 11.301`
85582.

Acetyl chloride does not react with

Answer»

DIETHYL ether
Aniline
Phenol
Ethanol

Solution :`CH_(3)COCl+ C_(2)H_(5)O-C_(2)H_(5)to ` No REACTION
85583.

Acetyl chloride cannot be obtained by treating acetic acid with

Answer»

`CHCl_(3)`
`SOCl_(2)`
`PCl_(3)`
`PCl_(5)`

SOLUTION :`CH_(3)COOH+CHCl_(3)to`No REACTION.
85584.

Acetyl chloride can be prepared

Answer»

by the ACTION of `CH_(3) COOH` with CHLOROFORM
BYTHE action of `PCl_(3)` on acetic acid
by the action of `Cl_(2)` on acetic acid
by the action of `C Cl_(4)` on acetic acid

Answer :B
85585.

Acetyl chloride cannot be obtained by the reaction of acetic acid with

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`CHCl_3`
`SOCl_2`
`PCl_3`
`PCL_5`

ANSWER :A
85586.

Acetyl chloride is not obtained from acetic acid by the action of

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`SOCl_2`
`CHCl_3`
`PCl_3 `
`PCl_5`

ANSWER :B
85587.

Acetyl bromide reacts with excess of CH_(3)MGI followed by treatement with a saturated solution of NH_(4)C1 gives:

Answer»

2-METHYL-2-propanol
Acetamide
Acetone
Acetyl iodide

Solution :`UNDERSET("Acetyl BROMIDE ")(CH_(3)-overset(O)overset(|)C-Br)underset((ii)"Saturated " NH_(4)CL)overset((i) "Excess"-CH_(3)MgI)to underset(2-"methyl" 2-"propanol")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH)`
85588.

2-Acetoxy benzoic acid is :

Answer»

ANTISEPTIC
Aspirin
Antibiotic
MORDANT dye

Answer :B
85589.

Acetoxy benzoic acid is :

Answer»

ANTISEPTIC
Aspirin
Antibiotic
Mordant dye

Answer :B
85590.

Acetoximeon reductionand followed by acetylationgives

Answer»

ethyl AMINE
isopropyl amine
monoacetyl isopropyl amine
DIACETYL isopropyl amine

Solution :`CH_(3) - OVERSET(CH_(3))overset(|)(C) = NOH overset(Na + C_(2)H_(5) OH) to `
` CH_(3) - overset(CH_(3))overset(|)(CH) - NH_(2) overset(2CH_(3) COCl) to underset("diacetyl isoproppyl amine")(CH_(3)-overset(CH_(3))overset(|)(CH) - N(COCH_(3))_(2))`
85591.

Acetophenone when reacted with a base, C_(2)H_(5)ONa, yields a stable compound which has the structure

Answer»




SOLUTION :
85592.

Acetoxime on reaction with trifluoroperoxy acetic acid gives

Answer»

1-nitropropane
2-nitropropane
propionic ACID
isobutyric acid

ANSWER :B
85593.

Acetoxime on catalytic reduction gives

Answer»

ACETIC acid
acetic anhydride
ethylamine
isopropylamine

ANSWER :A::D
85594.

Acetophenone when reacted with a base, C_(2)H_(5)ON a, yields a stable compound which has the structure

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ANSWER :C
85595.

Acetophenone when reacted with a base, C_(2)H_(5)O Na, yields a stable compound which has the structure

Answer»




ANSWER :A
85596.

Acetophenone to benzoic acid

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SOLUTION :
85597.

Acetophenone overset(NaNO_2//HCl)to A underset(Delta)overset(AC_2O)to B overset(H_3 O^+)to C . Incorrect statement(s) aboutproduct C is …..

Answer»

PRODUCT 'C' on heating liberates `CO_2` gas.
Product 'C' is more ACIDIC than PICRIC acid
Product 'C' can reduce Tollen's reagent
product 'C' can give TEST with NEURAL `FeCl_2`

Solution :Conceptual
85598.

Acetophenone on reaction with hydroxylamine hydrochloride can produce two isomeric oximes. Write structures of the oximes.

Answer»

SOLUTION :Like alkenes, oximes also SHOW GEOMETRICAL isomerism. For example, ACETOPHENONE exists in the following two geometrical ISOMERIC forms:
85599.

Acetophenone is used in:

Answer»

Toilet soaps
Preparation of HYPNOTIC drug
Perfumery
Phenacyl chloride preparation used in TEAR GAS shells

Answer :C
85600.

Acetophenone is prepared from

Answer»

ROSENMUND reaction
Sandmayer reaction
Wurtz reaction
Friedel CRAFT reaction

Answer :D