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92901.

1.25of a mixture of Na_(2)CO_(3) and Na_(2)SO_(4) was dissolvedin 25o mL of water.25 mL of this solution required20 mL of 0.01 " N " H_(2)SO_(4)solution forexact neutralisation .Calculate the percentageof Na_(2)CO_(3) in the mixture .

Answer»

Solution :In this PROBLEM only `Na_(2)CO_(3)`is NEUTRALISED by `H_(2)SO_(4)`.
Let the AMOUNT of `na_(2)CO_(3)` be x g
` :. " equivalent of " Na_(2)CO_(3) = x/53 "" …(Eqn.4i)`
m.e of `Na_(2)CO_(3) = x/53 XX 1000`
` :." m.e of "Na_(2)CO_(3) ` in 250 mL of the mixture solution = `(1000x)/53 `
` :. ` m.e of `Na_(2)CO_(3) ` in 25 mL of the mixture solution = `(100X)/53`
Now m.e of 25 mL of mixture solution
= m.e of 20 mL of 0.1 N `H_(2)SO_(4)`...(Eqn.2)
` (100x)/53 = 0.1 xx 20 `
` x = 1.06 g `
` :." % of " Na_(2)CO_(3) =(1.06)/(1.25) xx 100 = 84.8 %`
Discussion
92902.

1.25 gm of a chalk sample (CaCO_(3)) is strongly heated, 0.44 gm CO_(2)^(-) gas is produced. Determine the percentage % yield of the reaction :-

Answer»

`40%`
`60%`
`80%`
`90%`

Answer :C
Discussion
92903.

12.5 g of H_(2)SO_(4) are dissolved in water to make 1250ml of solution. The concentration in normality is :

Answer»

0.204
0.102
0.408
0.051

Solution :GRAM EQUIVALENTS of `H_(2)SO_(4)=(12.5)/(49)`
NORMALITY `= (12.5xx1000)/(49xx1250)=0.204 N`
Discussion
92904.

1.25g of a sample chlorine in 25 mL of water and 25 mL of which are treated with KI solution . The iodine so liberatedrequired 12.5 of which are treated with KI solution . The iodine so liberated required 12.5 mL of N/25 hypo solution in titration . Find the percentage of chlorine available from the sample of bleaching powder .

Answer»

Solution :m.e of available CHLORINE in 25 mLbleachingpowder solution
= m.e of iodine liberated
= m.e of hypo solution
` = 1/25 xx 12.5 = 0.5`
` :. ` m.e of available chlorine in 100 mL = `(0.5 xx 100)/25 = 2 `
` :. ` eq of available chlorine ` = 2/100`
wt . of available chlorinein `1.25 ` of bleaching powder
` = (0.2)/1000 xx 35.5 = 0.071` g ( eq. wt of CHLORNE = 35.5)
` :. ` PERCENTAGE of available chlorine ` = (0.071)/(1.25) xx 100`
` = 5.68 % `
Discussion
92905.

1.249 g of a sample of pure BaCO_(3). and impure CaCO_(3) containing some CaO was treated with dil . HCl and it evoled 168 ml of CO_(2)at NTP . From this solution , BaCrO_(4) was precipitated , filtered and washed . The precipitate was dissolved in dilute sulphauric acid and diluted to 100 ml. 10 ml of this solution , when treated with Kl solution , liberated iodine which required exactly 20 ml of 0.05 ml of 0.05 N Na_(2)S_(2)O_(3). Calculate the percentage of CaO in the sample.

Answer»


SOLUTION :N/A
Discussion
92906.

1.245 g of CuSO_(4). xH_(2)O was dissolved in water and H_(2)S gas was passed through it will till CuS was completely precipitated . The H_(2)SO_(4) produced in the filtrate required 100 ml of 0.1 M NaOH solution . Calculate x (approximately)

Answer»

SOLUTION :`CuSO_(4) +H_(2)Sto CuS +H_(2)SO_(4)`
`{:(" m.e of "CuSO_(4).x H_(2)O " solution = m.e of "H_(2)SO_(4)),( "= m.e of 10 mL of N NaOH"),(""1XX10 =10 ):}`
` :. ` number of equivalentof `CuSO_(4).x H_(2)O` solution = `10/1000`
Weight of `CuSO_(4) . x H_(2)O ` = equivalent `xx ` EQ.wt
`= 10/1000xx (159.5 +18x)/2 `
`{ " eq . wt of "CuSO_(4) .x H_(2)O = (159.5 +18x)/2 }`
Thus , `10/1000 xx (159.5 +18x)/2 = 1.245 ` (given)
18 x = 89.5
`x approx 5`
Discussion
92907.

1.249 g of a sample of pure BaCO_(3) and impure CaCO_(3) containing some CaO was treated with dil. HCl and it evolved 168ml of CO_(2) at NTP. From this solution, BaCrO_(4) was precipitated, filtered and washed. The precipitate was dissolved in dilute sulphuric acid and diluted to 100ml. 10ml of this solution, when treated with Kl solution, liberated iodine which required exactly 20ml of 0.05N Na_(2) S_(2) O_(3). Calculate the percentage of CaO in the sample.

Answer»

Solution :`n_(CaCO_(3)) + n _(BaCO_(3)) = n_(CO_(2)) = ( 168)/( 22400) = 7.5 xx 10^(-3) `.....(1)
`2BaCO_(3) rarr 2BaCrO_(4) overset( H^(+))(rarr) BaCr_(2) O_(7) overset( Kl) ( rarr) l_(2) -=Na_(2) S_(2) O_(3)`
eq. of `Na_(2) CO_(3)` = eq. of `I_(2) = ` eq of `BaCr_(2) O_(7) = ( 20 xx 10^(-3) xx 0.05 xx 100)/( 10) = 1 xx 10^(-12)`
Moles of `BaCr_(2) O_(7) = ( 1)/( 6)xx 10^(-2)`
Moles of `BaCrO_(4) = ( 2)/( 6) ( 1 xx 10^(-2))`
Moles of `BaCO_(3) = ( 1)/( 3) xx 10^(-2) = 3.33 xx 10^(-3) `.....(2)
Weight of `BaCO_(3) = 0.650 g m`
From equation (1) and (2) we get
`n_(CaCO_(3)) = 4.17 xx 10^(-3)`
weight of `CaCO_(3) = 100 xx 4.17 xx 10^(-3)`
=0.417g
weigght of `CaO = 1.249 - 0.656 -0.417 = 0.176`
% of CaO `= ( 0.176 )/( 1.249) xx 100`
= 14.09 %
Discussion
92908.

1,2,3-trihydroxybenzene is also known as

Answer»

Phrogallol
Phloroglucinol
Resorcinol
Orcinol

Answer :A
Discussion
92909.

1.23g of a substance dissolved in 10g of water raised the boiling point of water to 100.39^(@)C. Calculate the molecular weight of the substance .(K_(b)=0.52^(@)Cm^(-1))

Answer»


ANSWER :`164`
Discussion
92910.

1.234 gm gold is deposited on electrode when 3 ampere current is pass through solution containing AuCl_(4)^(-) ions. So calculate for how long such current should be passed ? (Atomic weight of Au=197 gm/mol)

Answer»

20 min. 8 sec.
10 min. 4 sec.
30 min. 12 sec.
10 min. 40 sec.

Solution :In `AuCl_(4)^(-) AU=(+3)`
`therefore Au_((aq))^(3+)+3e^(-)toAu_((S))`
So, 3F is REQUIRED for 1 mol Au
So, 3F will gives 197 gm Au.
So, for 1.234 gm GOLD required faraday
`=(1.234xx3)/(197)=(1.234xx3)/(197)xx96500` coulomb
Time=`("coulomb")/("ampere")=(1.234xx96500xx3)/(197xx3)`
=604.47 second
`~~10` minute 4 second.
Discussion
92911.

12.2 g of benzonic acid (molar mass = 122 ) in 100 g of benzene has a depression in freezing point 2.6^(@). If there is 100 percent polymersation, what is the nunber of molecues of benzoic acid in associated state ? (Given K_(f) for benzene =5.2 K kh mol^(-1))

Answer»


Solution :Observed molar mass of benzoic ACID MAY be CALCULATED as FOLLOWS :
`M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
`W_(B)=12.2g, W_(A)=100g=0.1 kg, K_(f)=5.2" K kg mol"^(-1), DeltaT_(f)=2.6 K`
`M_(B)=((5.2" K kg mol"^(-1))xx(12.2 g))/((2.6 K)xx(0.1"kg"))=244.0" g mol"^(-1)`
`"On comparison, observed molar mass "(244" g mol"^(-1)) "of benzoic acid is twice its normal molar mass (122" g mol"^(-1))`.
This means that benzoic acid exists as a dimer in the associated state.
Discussion
92912.

1.22 g C_6H_5COOHis added to two solvents. (i) In 100 g CH_3COCH_3 , Delta T_b = 0.17 , K_b = 1.7 kg K "mol"^(-1)(ii) In 100 g C_6H_6 , Delta T_b = 0.13 , K_b = 2.6 kg K "mol"^(-1) Find out the molecular weight of C_6H_5COOH in both the solvents and interpret the result.

Answer»

Solution :APPLY `Delta T_b = K_b` .m in both CASES
(i) 122 (II) 244, `C_6H_5COOH` dimerises in `C_6H_6`
Discussion
92913.

1.216g of a sample of (NH_(4))_(2)SO_(4) was boiled with excess of NaOH and the ammonia gas so produced was absorbedin 100 mL of N H_(2)SO_(4) solution .The unreacted H_(2)SO_(4) requied 81.6 mL of normal solution of a base for exact neutralisation. Calculatepercentage amount of ammonia in ammonium sulphate .

Answer»

Solution :m.e of unreacted`H_(2)SO_(4)` = m.e of the base
` = 1 xx 81.6 = 81 .6`
m.e of `NH_(3)` = m.e of `H_(2)SO_(4)` reacted with AMMONIA
= m.e of total `H_(2)SO_(4)` - m.e of unreacted `H_(2)SO_(4)`
` = 1xx 100 - 81.6 = 18.4`
` :. ` equivalent of `NH_(3) = (18.4)/(1000) "" ...(Eqn.3)`
Wt. of `NH_(3) = (18.4)/(1000) xx 17 "" ...(Eqn.4i)`
` = 0.3128 g `
` :. " % o f " NH_(3)" in " (NH_(4))_(2)SO_(4) = (0.3128)/(1.216) xx100`
` = 25.72 % `
Discussion
92914.

1.20 gm sample of NaCO_(3) " and " K_(2)CO_(3) was dissolved in water to form 100 ml of a Solution : 20 ml of this solution required 40 ml of 0.1 N HCl for complete neutralization . Calculate the weight of Na_(2)CO_(3) in the mixture. If another 20 ml of this solution is treated with excess of BaCl_(2) what will be the weight of the precipitate ?

Answer»

Solution :Let, weight of `Na_(2)CO_(3) = xgm`
Weight of `K_(2)CO_(3) = y ` gm
` :. X + y = 1.20` gm
For neutralization reaction of 100 ml
Meq. of `Na_(2)CO_(3) + " Meq. of" K _(2)CO_(3) = ` Meq. Of HCl
`RARRX/(106) xx 2 xx 1000+ y/(138) xx 2 xx 1000 = (40 xx 0.1 xx 100)/20 `
` :. 69x + 53Y = 73 .14`
From Eqs. (1) and (2) , we get
`x = 0.5962 ` gm
` y = 0.604` gm
Solution of `Na_(2)CO_(3) " and " K_(2)CO_(3) ` gives ppt. of `BaCO_(3)` with `BaCl_(2)`(Meq. of `Na_(2)CO_(3) + ` Meq. of `K_(2)CO_(3))` in 20 ml = Meq. of `BaCO_(3)`
`rArr ` Meq. of HCl for 20 ml mixture= Meq. of `BaCO_(3)`
`rArr` Meq. of `BaCO_(3) = 40 xx 0.1 = 4`
` (197)/2= W_(BaCO_(3))/E_(BaCO_(3)) xx 1000 = 40 xx 0.1 = 4`
`W_(BaCO_(3))/(197) xx 2 xx 1000= 4`
` :. W_(BaCO_(3)) = 0.394` gm
Discussion
92915.

120 gm of urea are present in 5 litre solution, the active mass of urea is

Answer»

`0.2`
`0.06`
`0.4`
`0.08`

SOLUTION :Active `"MASS"=("MOLES")/("Litre")`
`=("wt. in gm/molecular wt.")/("V in litre")=(120//60)/(5)=2/5=4`
Discussion
92916.

12.0 g urea is dissolved in 1 litre of water and 68.4 g sucrose is dissolved in 1litre of water. The relative lowering of vapour pressure of urea solution is

Answer»

greater than SUCROSE solution
less than sucrose solution
double that of sucorse solution
equal to that of sucrose solution.

Solution :`"12 G urea "(NH_(2)CONH_(2))=(12)/(60)" MOL"=(1)/(5)" mol"="0.2 mol"`
`"68.4 g sucrose "=(68.4)/(342)" mol"=(1)/(5)"mol"="0.2 mol"`
As molar concentrations are equal, relative lowering of VAPOUR pressures will also be equal.
Discussion
92917.

1.2 of a monprotic acid HA, is titrated with 0.222 M NaOH solution. The pH of the solution is monitrored with pH meter. A portion of the titration curve is shown in the diagram. What is the molar mass of HA?

Answer»

180
222
282
None

ANSWER :C
Discussion
92918.

1.2 of a monprotic acid HA, is titrated with 0.222 M NaOH solution. The pH of the solution is monitrored with pH meter. A portion of the titration curve is shown in the diagram. What is the pH of solution at the equivalence point?

Answer»

`3.50`
`7.00`
`8.40`
5

Answer :C
Discussion
92919.

1.2 of a monprotic acid HA, is titrated with 0.222 M NaOH solution. The pH of the solution is monitrored with pH meter. A portion of the titration curve is shown in the diagram. How many mL of NaOH is required to bring about the titration to its equivalence point?

Answer»

`4.00`
`9.00`
`19.00`
NONE

ANSWER :C
Discussion
92920.

1.2 mL acetic acid having density 1.06 g cm^(-3) is dissolved in 1 litre of water. The depression in freezing point observed for this concentration of acid was 0.041""^(@)C. The van't Hoff factor of the acid is ( K_(f) of water = 1.86 K kg mol^(-1) )

Answer»

Solution :Number of moles of acetic acid ,
`n=(1.2xx1.06)/60=0.0212`
Molality of acid solution,
`C=0.0212/2=0.0106 "mol kg"^(-1)`
`DeltaT_f` expected for this strength = `k_f xx m`
=1.86 x 0.016 = 0.0197 K
Van.t HOFF FACTOR (i) =`"observed freezing POINT"/"expected freezing point" ="-0.0205"/"-0.0197"`=1.041
Acetic acid is ionised as, `CH_3COOH hArr CH_3COO^(-) + H^(+)`
If `.alpha.` is the extent of ionisation, total number of particles =`n(1+alpha)`
van.t Hoff factor = i= `(n(1+alpha))/n=1+alpha =1.041, alpha` =0.041
Proton concentration =`C alpha` = 0.0106 x 0.041
`=4.24 xx 10^(-4) "molL"^(-1)`
Discussion
92921.

12 grams of silver was extracted from a sample of an ore from whidi the only source of silver was Ag_(2)S . How many grams of Ag_(2)Swere in the original sample?

Answer»


Solution :`(2AG^(+)+ S^(2-) to Ag_(2)S)`
MASS of `Ag_(2)S` required
`=("Molar mass of" Ag_(2) S xx 12)/(2 xx 108) = 13.8 G`
Discussion
92922.

12 gm of an impure sample oxide (As_(2)O_(3)which is acidic in nature ) was dissolved in water containing sodium bicarbonate ( which is basic in nature ) and the resulting solution was diluted to 250 ml. 25 ml of this solution was completely oxidized by22.4 ml of a solution 24.8 gm of hydrated sodium thiosulphate (Na_(2)S_(2)O_(3).5H_(2)O) in one litre . Calculate the percentage of arsenious oxide in the sample.

Answer»

SOLUTION :In this reaction , `As_(2)O_(3)` acts as acidic oxide and `NaHCO_(3)` as a base, givingacidbase neutralization reaction, which is non-redoxprocess. Here , `As_(2)O_(3)` does not act as basic oxide. HENCE , it will form `As_(2)(CO_(3))_(3)`, which does not exist as non-metals do not form carbonates.
n-factor of `As_(2)O_(3)` is 6 and that of `NaHCO_(3)` is 1. After the reaction, `As^(3+)` is oxidized by `l_(2)" to " As^(+5)` while `l_(2)` is reduced to `L^(-)` .
Normality of `Na_(2)S_(2)O_(3).5H_(2)O = (24.8)/(248) = 0.1`
Normality of `l_(2)` = Normality of `Na_(2)S_(2)O_(3). 5H_(2)O = (24.8).248`
` :. ` Equivalents of `l_(2) = 0.1 xx 22.4 xx 10^(-3)` = Equivalent of `As^(3+)` reacted in 25 ML` = 2.24 xx 10^(-3)`
`:. ` Equivalents of `As^(3+) ` reacted in 250 ml= `2.24 xx 10^(-2)`
Moles of `As^(3+)` in 250 ml ` = (2.24 xx 10^(-2))/2 = 1.12 xx 10^(-2)`
Moles of `As_(2)O_(3)` reacted ` = (1.12 xx 10^(-2))/2 = 5.6 xx 10^(-3) `
Percentage of `As_(2)O_(3) = (5.6 xx 10^(-3) xx 198 xx 100)/12 = 9.24 %`
Discussion
92923.

12 gm of urea is dissolved in 2 liter solution a 300 K temperature. How many gram of NaCl should be dissolved in 10 liter solution so that it becomes iso - osmotic with urea solution ? [At. Wt. of Na = 23, Cl = 35.5 gm/mole]

Answer»

29.25 gm
7.31 gm
5.85 gm
19.5 gm

Solution :`("MOLE of urea")/("volume of urea")=("mole of NACL")/("volume of NaCl")`
`therefore (0.2)/(2)=("mole of NaCl")/(10)`
`therefore` mole of NaCl = 1 mole
But total No. of particles of NaCl = 2
so `("1 mole")/(2)=0.5`mole
WEIGHT of NaCl = mole `xx` weight
`= 0.5xx58.5=29.25` GRAM .
Discussion
92924.

12 g of urea is dissolved in 1 litre os water and 68.4 g of sucrese in also dissolven in 1 lite of water. The lowering in the vapour pressure of first case is :

Answer»

equal to secound
greater than second
less than secound
double than second.

Solution :Both have same lowering in V.P. as they have same no. of moles.
`"UREA=((12G))/((60g MOL^(-1)))=0.2 mol`
`"Sucrose"=(68.4 G)/((342g mol^(-1)))=0.2 mol`
Discussion
92925.

1.2g of an organic compound on Kjeldahlization liberates ammonia which consumes 30cm^(3) of 1N HCl. The percentage of nitrogen in the organic compound is

Answer»

30
35
46.67
20.8

Solution :`1.4xx"NORMALITY of acid"XX"volume"`
% of nitrogen`("Used with NH_(3)")/("Wieight of ORGANIC compound")`
`=(1.4xx1xx30)/(1.2)=35`
Discussion
92926.

12 gof Mgwill react completely with an acidto give

Answer»

1 mole of `O_(2)`
`(1)/(2)` mole of `H_(2)`
1 mole of `H_(2)`
2 MOLES of `H_(2)`

Answer :B
Discussion
92927.

12 g of Mg was burnt in a closed vessel containing 32 g oxygen , which of the following is/are correct.

Answer»

2 gm of MG will be lift unburnt.
0.75 gm-molecules of `O_(2)` will be left unreacted .
20 gm of MgO will be formed.
The mixture at the end will WEIGHT 44 g.

Answer :B::C::D
Discussion
92928.

12 g of magnesium (atomic mass 24) on reacting completely with acid gives hydrogen gas, the volume of which at N.T.P. would be

Answer»

22.4 L
11.2 L
44.8 L
6.1 L

Solution :`underset(12g)(Mg) + 2HCl to MgCl_2 + underset(22.4L)(H_2)`
24 G of Mg given `H_2` at N.T.P. = 22.4 L
12 g of Mg GIVE `H_2` at N.T.P. ` = (22.4)/(24) xx 12 = 11.2 L`
Discussion
92929.

12 g of H_(2)SO_(4) are dissolved in water to make 1200 ml of solution. The normality of the solution is

Answer»

1
0.2
0.1
2

Answer :B
Discussion
92930.

12 g of an alkyl earth metal gave 14.8 g of its nitride .atomic weight of that metal is ,

Answer»

20
12
40
14.8

Answer :C
Discussion
92931.

1.2 g mixture of Na_(2)CO_(3) and K_(2)CO_(3) was dissolved in water to form 100cm^(3) of a solution. 20cm^(3) of this solution required 40cm^(3) of 0.1 N HCl for neutralisaiton. Calculate the weight of Na_(2)CO_(3) and K_(2)CO_(3) in the mixture.

Answer»

Solution :Suppose weight of `Na_(2)CO_(3)` in the mixture = xg `therefore` Weight of `K_(2)CO_(3)` in the mixture `=(1.2-x)g`
`"Eq. wt of "Na_(2)CO_(3)=(46+12+148)/(2)=53,`
`"Eq. wt. of "K_(2)CO_(3)=(78+12+48)/(2)=69`
`"No. of g eq. of "N_(2)CO_(3) and K_(2)CO_(3)" in the mixture "=(x)/(53)+(1.2-x)/(69)`
`"40 cc of 0.1 N HCl contain g eq. of HCl "=(0.1)/(1000)xx40=4xx10^(-3)`
`"Thus,20 cc of the mixture sol NEUTRALIZE HCl "=4xx10^(-3)"g eq."`
`therefore"100 cc of the mixture sol. will neutralise HCl"=(4xx10^(-3))/(20)xx100=2xx10^(-2)"g eq=0.02 g eq."`
As substance react in equivalent amounts ,`(x)/(53)+(1.20-x)/(69)=0.02`
or `69x+63.6-53x=0.02xx53xx69=73.14 or 16x=9.54 or x=0.596g`
Thus, `Na_(2)CO_(3)=0.596g and K_(2)CO_(3)=1.2-0.596=0.604 g`
Discussion
92932.

1,2-Dimethy1cyclohexene reacts with HBr in C CI_(4) to form mainly

Answer»

`E-1-`bromo`-1,2-`dimethy`1`cyclohexene
meso`-1-`bromo`-1,2-`dimethy`1`cyclohexane
`(+-)Z-1-`bromo`-1,2-`dimethy`1`cyclohexane
`(+-)E-1-`bromo`-1,2-`dimethy`1`cyclohexane

Solution :Addition of hydrogen halide to an ALKENE in nonpolar solvents such as `C CI_(4)` is predominantly trans or anti on account of the formation of a bridged carbocation.

This stereospecific reaction proceeds through a bridged cation like the bromonium ion wher `H` replaces `Br`. The intermediate may actually be a `pi` complex. In the absence of a solvent that can stabilize a free `R^(+),Br^(-)` attacks the protonated complex from the opposite face RESULTING in trans (anti) addition.
Since water is a very good ion-solvator, THR protonated complex COLLAPSES to the free `R^(+)` thet can now recat FORM either face giving both cis and trans addition.
Note: Both cis and trans addition occur with conc. aq. acid. With dilute aq. acid, addition of `H_(2)O` also occurs.
Discussion
92933.

1,2-dichloroethane is known as

Answer»

GEMINAL dihalide
vicinal dihalide
haloform
all of the above

Solution :`CH_(2)ClCH_(2)Cl` is 1,2 dischloro ETHANE is a vecinal dihalid.
Discussion
92934.

1.2-dibromopropane on treatment with X moles of NaNH_2 followed by treatment with ethyl bromide gave a pentyne, the value of X is

Answer»

one
two
three
four

Solution :
THUS, X is three.
HENCE, (C) is the correct ANSWER.
Discussion
92935.

1,2-dibromoethane when heated with alcoholic potash gives

Answer»

Ethane
Acetylene
Ethylene
Methane

Solution :`{:(CH_2Br,),("|",+"2KOH(ALCOHOLIC)"oversetDeltatoCH-=CH+2KBr),(CH_2Br,):}`
Discussion
92936.

1,2-Dibromoethane when heated with alcoholic potash followed by NaNH_2 in liq. NH_3 gives

Answer»

ethane<BR>acetylene
ethylene
methane

Solution :`underset(1 , 2 "Dibromo ethane")(BrCH_2) - CH_2 Br overset(KOH(alc))underset(HBR)(to) CH_2 = CHBR overset(NaNH_2//"liq." NH_(3))(to) Hunderset("Acetylene")(C-= CH)`
Discussion
92937.

1,2- dibromoethane is added to prevent deposition of lead metal in :

Answer»

WATER pipes
petrol engines
Electric heaters
metal WORKING LATHE machines

Answer :B
Discussion
92938.

1,2 di bromo cyclohexane on dehydrohalogenation gives

Answer»




NONE of these

Solution :
Discussion
92939.

1^@.2^@. And 3^@ amines can be best distinguished by :

Answer»

`HNO_2` TREATMENT `
Exthaustive alkylation
MUSTARD oil REACTION
CARBYLAMINE reaction

Answer :A
Discussion
92940.

""_(11)Na^(24) is radioacitive and it decays to

Answer»

`""_(9)F^(20)` and `alpha`-PARTICLES
`""_(13)AI^(24)` and positron
`""_(11)Na^(23)` and neutron
`""_(12)Mg^(24)` and `BETA`-particles.

Solution :`""_(11)^(24)Na` is `beta`- radioactive.
`""_(11)^(24)Na to ""_(12)^(24)Mg+ underset(beta-"particle")(""_(-1)^(0)e)`
Discussion
92941.

._(11)Na^(24) half-life is 15 hours. On heating it will

Answer»

Reduce
Remain unchanged
Depend on temperature
Become double

Solution :`t_(1//2)` is INDEPENDENT of all external factors and is CONSTANT for a GIVEN SPECIES
Discussion
92942.

_(1)^(1)H is a stable isotope . _(1)^(3)H is expected to disintegrated by :

Answer»

`ALPHA`-EMISSION
`BETA`-emission
Positron emission
protonemission

Answer :B
Discussion
92943.

1.1g of CoCl_(3).6NH_(3) (mol.wt. =267) was dissolved in 100g of H_(2)O. The freezing point of the solution was -0.29^(@)C. How many moles of solute particles exist in solution for each mole of solute introduced ? K_(f) for H_(2)O=1.86^(@)C.m^(-1)

Answer»

Solution :Molality (experimental) `=(DeltaT_(f))/(K_(f))=(0.29)/(1.86)`
`=0.156` mole/1000g
Molality (THEORETICAL) `=("moles of solute")/("wt. of solvent in G")xx1000`
`=(1.1)/(267)xx(1000)/(100)=0.0412` mole/1000g
Thus, number of moles of solute PARTICLES produced by 1 mole of solute
`=(0.156)/(0.0412)=4`...........(Eqn. 10)
Discussion
92944.

116mg of a compound on vaporisation in Victor Meyer's apparatus displaces 44.8mL of air measured at STP. The molecular mass of the compound is.

Answer»

116
232
58
44.8

Solution :MOLAR MASS of compound
=Mass of 22400 ML vapour at STP
`=(0.116xx22400)/(44.8)=58`
Discussion
92945.

116 mg of a compound onvaporisation is used for a Victor Mayer's apparatus displaces 44.8 mlof air measured at S.T.P. The molecularweight ofthe compound is

Answer»

116
232
58
44.8

Solution :`116 mg` compound means `116 XX 10^(-3) gm` compound since `1 mg` CONTAIN`10^(-3)` gm
Mol. WT. of compound
`= (" mass of the substance")/(" volume of the vapour at S.T.P") xx 22400`
`= (116 xx 10^(-1))/(44.8) xx 224000 = 58`.
Discussion
92946.

116 mg of a compound on vaporisation in a Victor Meyer's apparatus displaces 44.8 ml of air measured at STP. The molecular weight of the compound is

Answer»

116
232
58
44.8

Solution :22400 ML of AIR at STP will be DISPLACED by
`=(0.116)/(44.8)xx22400g=58g.`
Discussion
92947.

116 g of A_(3)B_(4) has 1.5 moles of A. Molecular weight of A_(3)B_(4) is

Answer»

164
232
77
cannot be CALCULATED

ANSWER :B
Discussion
92948.

1.16 g CH_(3)(CH_(2))_(n)COOH was burnt in excess air and the resultant gases (CO_(2) and H_(2)O) were passed through excess NaOH solution. The resulting solution was divided in two equal parts. One part requires 50 mL of 1 N HCl for neutralization using phenolphthalein as indicator. Another part required 80 mL of 1 N HCl for neutralization using methyl orange as indicator. Amount of excess NaOH solution taken initially

Answer»

3.2 gm
6.4 gm
1.2 gm
None of these

ANSWER :B
Discussion
92949.

1.16 g CH_(3)(CH_(2))_(n)COOH was burnt in excess air and the resultant gases (CO_(2) and H_(2)O) were passed through excess NaOH solution. The resulting solution was divided in two equal parts. One part requires 50 mL of 1 N HCl for neutralization using phenolphthalein as indicator. Another part required 80 mL of 1 N HCl for neutralization using methyl orange as indicator. What is the value of n

Answer»

4
3
2
1

Answer :A
Discussion
92950.

1.16 g CH_(3)(CH_(2))_(n)COOH was burnt in excess air and the resultant gases (CO_(2) and H_(2)O) were passed through excess NaOH solution. The resulting solution was divided in two equal parts. One part requires 50 mL of 1 N HCl for neutralization using phenolphthalein as indicator. Another part required 80 mL of 1 N HCl for neutralization using methyl orange as indicator. Produced mole of the CO_(2)

Answer»

0.1
0.01
0.06
None of these

ANSWER :C
Discussion