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92801.

18 g glucose (molar mass "180 g mol"^(-1)) is present in 500cm^(3) of its aqueous solution. What is the molarity of the solution ? What additional data is required if the molality of the soluiton is also required to be calculated?

Answer»

SOLUTION :0.2 M, DENSITYOF solution is REQUIRED
Discussion
92802.

18 g glucose, C_(6)H_(12)O_(6), is dissolved in 1 kg of water in saucepan. At what temperasture will water boil at 1.013 bar ? K_(b) for water is 0.52 K kg mol^(-1).

Answer»

Solution :Moles of GLUCOSE `= 18g//180 g mol^(-1)=0.1` mol
Number of kilograms of SOLVENT = 1 kg
Thus MOLALITY of glucose solution `= 0.1 mol kg^(-1)`
For water, change in BOILING point
`Delta T_(b)=K_(b)xx m`
`=0.52 "K kg mol"^(-1)xx 0.1 mol kg^(-1) = 0.052` K
Since water boils at 373.15 K at 1.013bar pressure.Therefore, the boiling point of solution will be `373.15+0.052=373.202 K`.
Discussion
92803.

18 g glucose (C_(6)H_(12)O_(6)) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is :

Answer»

`759.0`
`7.6`
`76.0`
`752.4`

Solution :From Roult.s law
`(P_(O)-P_(S))/(P_(O))=X_("solute")`
`therefore (P_(O)-P_(S))/(P_(O))=(n_("solute"))/(n_("solute")+N_("SOLVENT"))`
`RARR (P_(O)-P_(S))/(P_(O))=(0.1)/((178.2)/(18)+0.1)`
`because P_(o)=760` torr `(100^(@)C)`
`therefore` Vapour pressure of `H_(2)O` in solution = 752.3 torr
Discussion
92804.

18 g glucose (C_(6)H_(12)O_(6)) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is

Answer»

7.6
76
752.4
`759.0`

Solution :`"18 g glucose "(C_(6)H_(12)O_(6))=(18)/(180)" MOLE"="0.1 mole"`
`"178.2 g water "(H_(2)O)=(178.2)/(18)" mole"="9.9 moles"`
`THEREFORE"Mole fraction of water in the solution"`
`=(9.9)/(0.1+9.9)=(9.9)/(10)=0.99`
Vapour pressure of water in aqueous solution
= Molefraction of water in the solution `xx` Vapour pressure of pure water
`=0.99xx"760 torr= 725.4 torr"`
Discussion
92805.

18-crown -6 ether is able to trap

Answer»

`K^+`
`NA^+`
`LI^+`
all of these

ANSWER :A
Discussion
92806.

18 g glucose and 6 g urea are dissolved in 1L of solution at 27^@C , the osmotic pressure in atm of the solution will be

Answer»

1.9
2.9
3.9
4.9

Answer :C
Discussion
92807.

1.7920 gm K_(2)SO_(4) is present in 1000 mL of solution. If such solution have osmotic pressure of 0.680 bar at 26^(@)C then find out van't Hoff factor value.

Answer»


ANSWER :`2.66`
Discussion
92808.

17.4% K_(2)SO_(4) solution at 27^(@)C is isotonic with 4% NaOH solution at the same temperature. If NaOH is 100% ionized, what is the degree of ionization of K_(2)SO_(4) in aqueous solution?

Answer»

Solution :As the solution are isotonic, the molar concentration of particles in the two solution must be same.
Molar concentration of `K_(2)SO_(4)=(17.4)/(174)molxx(1)/("100 mL")xx"1000 mL L"^(-1)=1.0M`
`""(because "molar mass of "K_(2)SO_(4)="174 g mol"^(-1))`
Molar concentration of NaOH `=(4)/(40)molxx(1)/(100mL)xx1000"mL L"^(-1)=1.0M`
`""(because" molar mass of NaOH = 40 g mol"^(-1))`
As NaOH completely ionizes as `NaOH rarr Na^(+) +OH^(-)`, molar COCENTRATION of particles = 2.0 M
If `alpha` is the degree of ionization of `K_(2)SO_(4)`, then
`{:(,K_(2)SO_(4)rarr2K^(+)+SO_(4)^(-)),("INITIAL conc.","1 M"),("Conc. after ionization",""1-alpha""2alpha""alpha):}`
Thus, the total concentration of the particles after ionization `=(1-alpha)+2alpha+alpha =1+2alpha`
As two solution are isotonic, `1+2alpha=2 or alpha=0.5`
Thus, degree of ionization of `K_(2)SO_(4)=0.5`, i.e., it is `50%` IONIZED.
Discussion
92809.

1.763g of hydrated BaCl_(2) was heated to dryness. The anhydrous salt remained was 1.505g. Wat is the formula of the hydrate?

Answer»

SOLUTION :`(BaCl_(2).2H_(2)O)`
Discussion
92810.

17.4% (mass/Vol) K_(2)SO_(4) solution at 27^(@)C is isotonic to 5.85% (mass/Vol) NaCl solution at 27^(@)C . If NaCl is 100% ionised, the % ionisation of K_(2)SO_(4) in aqueous solution is x xx10 then the value of x is

Answer»


ANSWER :5
Discussion
92811.

1.78 g of an optically active L-amino acid (A) is treated with NaNO_2//HCl at 0^(@) C. 448 cm^(3) of nitrogenwas at STP is evolved. A sample of protein has 0.25% of this amino acid by mass. The molar mass of the protein is

Answer»

`36,500 G "mol"^(-1)`
`34,500 g " mol"^(-1)`
`35,400 g " mol"^(-1)`
`35,600 g "mol"^(-1)`

Solution :`1.78g - 448 CM^(3)`
`?-22400 cm^(3)`
Mol mass of L - amino acid `=1.78 XX (22400)/(448) = 89`
`:. ` Mol mass of protein is
100-0.25
`?-89`
`= (100 xx 89)/(0.25 )` = 35,600 g / mol
Discussion
92812.

17.4% K_(2)SO_(4) solition at 27^(@)C is isotonic with 4% NaOHsolution at the same temperature. If NaOH is 100% ionised, What is the degree of innisation of K_(2)SO_(4) in aqueous solution ?

Answer»

Solution :Calculation of molar concerntration of both the solution.
`"Molar concentration of NaOH solution"("Mass of NaOH / Molar mass")/("Volume of solution solution in LITRES")`
`=((4g)//(40G mol^(-1)))/((0.1 L))=1.0 mol^(-1)=1.0 M`
molar concentration of `K_(2)SO_(4)` solution = `(Mass of K_(2)SO_(4)//Molar mass)/("Colume of solution in liters")`
`=((17.4g)//(174h mol^(-1)))/((0.1L))=10 molL^(-1)=1.0 M`
Calculation of Van't Hoff factor (i) for `K_(2)SO_(4)`
Since the two SOLUTIONS are isotonic, they have same osmotic pressure.
`alpha=1, n=2`
`alpha=(i-1)/(n-1)or1=(i-1)/(2-1)ori=2,c=1 M`
`pi_(NaOH)=iCRT=2xx(1M)xxRxxT`
C=1 M, i=? ( to be calcutaed)
`pi_(K_(2)SO_(4))=oCRT=i(1)xxRxxT`
As the twpsp,itopm are oisptpmec. `pi_(NaOH) or 2xxRxxT=ixxRxxT or o=2`
Calculation of dgree of ionisation of `K_(2)SO_(4)`
`K_(2)SO_(4): alpha=(i-1)/(n-1)=(2-1)/(3-1)=1/2=0.5`
The degree of ionisation of `K_(2)SO_(4)=0.5` or its percentage ionisation=o.5xx100=50%
Discussion
92813.

1.68 mg of an organic compound (A) with molecule formula (C_(9)H_(12)O_(3)) on Zeisel estimation produces an yellow precipitate of wt 4.7 mg the compound (A) is

Answer»




ANSWER :C
Discussion
92814.

1.615 g of anhydrous ZnSO_(4) was placed in moist air. After a few days, its mass was found to be 2.875 g. What is the molecular formula of the hydrated salt? (At. Mass : Zn = 65.5, S = 32, O = 16)

Answer»

Solution :Molecular MASS of ANHYDROUS `ZnSO_(4)=65.5+32+64=161.5`
1.615 G of anhydrous `ZnSO_(4)` combine with `H_(2)O=2.875-1.615g=1.260g`
`therefore"1.615 g of anhydrous "ZnSO_(4)" will combine with "H_(2)O=(1.260)/(1.615)xx161.5g`
`=126g=(126)/(18)"moles = 7 moles"`
Hence, formula of the hydrated salt = `ZnSO_(4).7H_(2)O.`
Discussion
92815.

1.60g of an oxide of iron, on heating in a stream of hydrogen gas, completey converted to 1.12g of iron. Find the empirical formula of the oxide

Answer»

SOLUTION :`(Fe_(2)O_(3))`
Discussion
92816.

160 ml of a gas are collected over water at 25^(@)C and 768.8mm Hg. If aqueous tension at 25^(@)C is 23.8 mm Hg, then pressure of dry gas at 25^(@)C is 1:29 PM

Answer»

76.8mm Hg
760 mm Hg
721.2mm Hg
600 mm Hg

Solution :PRESSURE of DRY GAS
= Pressure of gas collected over WATER -Aqueous TENSION
`= 745-23.8 = 721.2 mm`
Discussion
92817.

1.60 g ofa metal A and 0.96 g of a metal B when treated with excess of dilute acid, separately, produced the same amount of hydrogen. Calculate the equivalent weight of A if the equivalent weight of B is 12.

Answer»


ANSWER :20
Discussion
92818.

16 moles of hydrogen and 4 moles of nitrogen are sealed in a one litre vessel. The vessel is heated at a constant temperature until equilibrium is established, when it is found that the pressure in the vessel has fallen to 9/10 of its original value. Calculate K_C for the reaction N_2 + 3H_2hArr 2NH_3.

Answer»

SOLUTION :`6XX10^(-4) L^2 MOL^(-2)`
Discussion
92819.

1.6 mol of PCl_(5(g)) is placed in 4 dm^(3) closed vessel. When the temperaturee is raised to 500 K, it decomposes and at equilibrium, 1.2 mole of PCl_(5(g)) remains. What is the K_(c) vlaue for the decomposition of PCl_(5(g)) and Cl_(2(g))at 500 K

Answer»

`0.013`
`0.050`
`0.033`
`0.067`

Solution :`1.6` mol of `PCl_(5)` is placed in 4 `dm^(3)` CLOSED vessel. `PCl_(5(G))hArrPCl_(3(g))+Cl_(2(g))`
`{:(1.6mol,0,0,("initially")),((1.6-x)mol,xmol, xmol, ("At equlibrium")):}`
GIVEN that `1.6-x=1.2`
`thereforex=0.4` mol
Therefore, `[PCl_(5)]=(1.2)/(4)=0.3,[PCl_(3)]=(0.4)/(4)=0.1`
`&[Cl_(2)]=(0.4)/(4)=0.1`
`thereforeK_(C)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(0.1xx0.1)/(0.3)=0.033.`
Discussion
92820.

1.6 mol of PCl_(5) is placed in a 4 litre vessel. When the temperature is increased to 500 K, the PCl_(5) decomposes asPCl_(5)hArr PCl_(3)+Cl_(2)At equilibrium 1.20 mol of PCl_(5) remains' K_(c ) for the reaction is :

Answer»

`0.013`
`0.050`
`0.067`
`0.033`

ANSWER :A
Discussion
92821.

16 mL of hydrocarbon gas was exploded with excess of oxygen on cooling the volume of resulting gaseous mixture was reduced by 48 mL When KOH added there was further decreases of 48 mL in volume. Find the molecules formula.

Answer»

SOLUTION :`C_(3)H_(6)`
Discussion
92822.

16 grams of copper sulphate is dissolved in one litre of water . It is electrolysed using a current strength of 10 amp for a period of 965 sec. What is the concentration of copper sulphate after electrolysis ? Take atomic weight of copper as 64 and assume there is no loss in water during electrolysis.

Answer»


ANSWER :0.05M
Discussion
92823.

16kg oxygen gas expands at STP to occupy double of its original volume. The work done during the process is:

Answer»

260 kcal
187.6 kcal
130 kcal
272.8 kcal

Answer :D
Discussion
92824.

16 g off oxygen gas expands isothermally and reversibly at 300 K from 10 dm^(3) to 100 dm^(3). The work done is (in J)

Answer»

ZERO
`-2875J`
`+2875J`
infinite

Solution :`W=-2.303" nRT LOG"(V_(2))/(V_(1))`
`=-2.303xx(16)/(32)xx8.314xx300xx"log"(100)/(10)`
`=-2872J=-2875`J
Discussion
92825.

1.6 g of pyrolusite ore was treated with 50cm^(3) of 1.0 N oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250cm^(3) in a flask. 25cm^(3) of this solution when titrated with "0.1 N KMnO"_(4) required 32cm^(3) of the solution. Calculate the percentage of pure MnO_(2) in the ore.

Answer»

SOLUTION :`25xm^(3)" of OXALIC ACID left required "KMnO_(4)=32cm^(3)" of0.1 M"`
`therefore 250 cm^(3)" of the oxalic acid required "KMnO_(4) = 320 cm^(3)" of 0.1 N = 32 cm"^(3) " of 1 N"`
`therefore"Oxalic acid used up PYROLUSITE "=(50-32)cm^(3)" of 1 N "=18cm^(3)" of 1 N"`
`18cm^(3)" fo 1 N oxalic acid = 18 millieq. of oxalic acid"`
It must have reacted with equivalent amount of `MnO_(2)" of pyrolusite"`
`"Eq. wt. of "MnO_(2)=(55+32)/(2)=(87)/(2)=43.5`
`therefore"18 millieq. Of oxalic acid = 18 millieq. of "MnO_(2)=(18)/(1000)xx43.5g=0.783g`
`therefore""%" of pure "MnO_(2)" in pyrolusite "=(0.783)/(1.6)xx100=48.9%`
Discussion
92826.

1.6 g of an organic compound containing sulphur, when series of reagents ,produces H_(2)SO_(4) which on reaction with BaCl_(2) produces 0.233 g of BaSO_(4). Calculate % by mass Of S in the orgnic compound . (Given : Atomic weight of Ba = 137 )

Answer»


ANSWER :2
Discussion
92827.

1.5g of an organic compound in a quantitative determination of phosphorus gave 2.5090g of Mg_(2)P_(2)O_(7). Calculate the percentage of phosphorus in the compound

Answer»


ANSWER :0.4671
Discussion
92828.

1.575 gm of organic acid was dissolved in 250ml of water 20ml of this solution required 16 ml of N//8 alkali solution for completely neutralization. (M. wt of acid is 126) the basicity of acid is

Answer»


Solution :16ml `(N)/(8)` alkali solution = 20ml of acid solution
2ml 1 N alkali solution = 20ml of acid solution
? = 250 ML of acid solution
`implies` % 25ml IN of alkali solution obtained
25ml 1N of alkali solution .......... 1.575grams of acid
1000ml 1 N alkali solution ........?
`implies (1.575)/(25) xx 1000` GRAMS of acid
`implies` equalant Acid = 63
Molecular WEIGHT of acid = `"equivalent acid" xx "VALANCY"`
`implies` valancy = 2 126 = 63X valancy
Discussion
92829.

1.5276g of CdCl_(2) was found to contain 0.9367g of cadmium. Calculate the atomic mass of cadmium.

Answer»


Solution :EQUIVALENT mass of CADMIUM `=("Mass of CD")/("Mass of CL")xx35.5`
`=(0.9367)/(0.5909)xx35.5=56.27`
Discussion
92830.

1.525g of an organic compound was Kjeldahlised and the ammonia so produced was passed into 30mL of N HCl solution. The remaining HCl was further neutralised by 120mL of (N)/(10) NaOH solution. Calculate the percentage of nitrogen in the compound

Answer»

Solution :m.e. of `NaOH = (1)/(10) XX 120= 12`
m.e of remaining HCl= 12
m.e. of HCl (total) `=1 xx 30= 30`
m.e. of HCl neutralised by `NH_(3)= 30-12=18`
`therefore` m.e. of `NH_(3)= 18`
Eq of `NH_(3)= (18)/(1000)= 0.018`
`therefore` mole of `NH_(3)= 0.018`
Now mole of N in `NH_(3) = 1 xx` mole of `NH_(3)`
`=1 xx 0.018 = 0.018`
Weight of nitrogen (N) = `0.018 xx 14= 0.252g`
PERCENTAGE of nitrogen `= (0.252)/(1.525)xx 100= 16.52%`
Discussion
92831.

152.06xx0.24 = 36.4944. The correct answer to this problem is proper number of significant digit is

Answer»

`36.4944`
`36.494`
`37`
`36.5`

Solution :`underset("five S.F.")(152.06)xxunderset("TWO S.F.")(0.24) = underset("Two S.F.")(36.4944)`
Here ANSWER should have two SIGNIFICANT figures. Therefore rounding off the answer to two significant DIGIT we GET 37 as the answer.
Discussion
92832.

1.500g of hydroxide of a metal gave 1.000g of its oxide on heating. What is the equivalent mass of the metal ?

Answer»


Solution :By the law of EQUIVALENTS: `("E + Eq. Mass of" OH^(-))/("E + Eq. Mass of O") =("mass of metal HYDROXIDE")/("mass of metal oxide")`
or `(E+17)/(E+8) = 1.5/1 RARR E=10`
Discussion
92833.

1.502 g sample of steel yields 0.259 g of nickel dimethylglyoximate NiC_(8)H_(14)N_(4)O_(4)(molar mass 289 g mol^(-1)). What is the percentage of nickel in the steel? (Atomic mass of Ni=59)

Answer»

Solution :`underset("59 g")(Ni^(2+))+underset("Dimethylglyoxime")(2C_(4)H_(8)N_(2)O_(2))rarrunderset("289 g")(NiC_(8)H_(14)N_(4)O_(4))+2H^(+)`
Thus, 289 g of glyoximate is obtained from Ni = 59 g
`THEREFORE` 0.259 g of glyoximate is obtained from Ni `= (59)/(289)xx0.259 g`
This amount of Ni is PRESENT in 1.502 g of the sample of steel.
`therefore % " of Ni in steel "= (59)/(289)XX(0.259)/(1.502)xx100=3.52`
Discussion
92834.

150 mL of C_(2)H_(5)OH(density = 0.78 g mL^(-1)is diluted to one litre by adding water, molality of the solution is

Answer»

2.54
11.7
2.99
29.9

Solution : `m=(W_(2) xx 1000)/(M_(2) xx W_(1)(G))`
150 mL `C_(2)H_(5) OH=150 xx 0.78 g =117.0 g=W_(2)`
Water =850 g = `W_(2)`=46 g `"mol"^(-1)
`"Molality"= `(117 xx 1000)/ (46 xx 850)=2.99`
Discussion
92835.

15% of oxygen is converted to ozone. The mass of ozone that can be prepared from 67.2 L of oxygen at N.T.P. will be

Answer»

14.4 g
28.8 g
52.0 g
64 g

Solution :`underset("3 vol")(3O_2) to underset("2vol")(2O_3)`
67.2 L of `O_2` produces OZONE =` (2 XX 67.2)/(3) = 44.8 L`
sicne of efficiency is only 15%
`O_3` produced ` = (44.8 xx 15)/(100) = 6.72 L`
mass of `O_3` produced ` = (6.72 xx 48)/(22.4)= 14.4 g `
Discussion
92836.

1.5 % Urea solution and 5.25% unknown substance's solutions are isotonic. The density of both the solutions are 1 gm//cm^(3) then what is the molecular weight of unknown substance ?

Answer»

90.0 gram/mole
115.0 gram/mole
1050.0 gram/mole
210.0 gram/mole

SOLUTION :For isotonic solution `(W_("unknown"))/(M_("unknown"))=(W_("UREA"))/(M_("urea"))`
`M_("unknown")=(5.25xx60)/(1.5)=210` gram/mole
Discussion
92837.

1.5 moles of O_(2) combines with Mg to form MgO. The mass of Mg that has combined is (Mg= 24)

Answer»

48g
24g
72g
36g

Solution :`2MG + O_(2) rarr 2MgO`
48gm 1mole
1 mole of `O_(2)` COMBINES with = 48g Mg
`rArr 1.5` mole `O_(2)` combines with = 72 GM Mg.
Discussion
92838.

15 moles of H_(2) and 5.2 moles of I_(2) are mixed and allowed to attain equlibrium at 500^(@)C. At equlibrium, the concentration of HI is found to be 10 moles. The equlibrium constant for the formation of HI is

Answer»

50
15
100
25

Solution :`{:(H_(2)" "+" "I_(2)" "HARR" "2HI),(15""5.2""0),((15-5)" "(5.2-5)""10):}`
`K_(C)=([HI]^(2))/([H_(2)][I_(2)])=(10xx10)/(10xx0.2)=50`
Discussion
92839.

15 moles of H_(2) and 5.2 moles of I_(2) are mixed and allowed to attain equilibrium at 500^(@)C. At equilibrium, the concentration of HI is found to be 10 moles. The equilibrium constant for the formation of HI is

Answer»

50
15
100
25

Solution :`UNDERSET(15-5)underset(15)(H_(2))+underset(5.2-5)underset(5.2)(I_(2))hArrunderset(10)underset(0)(2HI)`
`K_(C)=[HI]^(2)/[[I_(2)])=(10xx10)/(10xx0.2)=50`
Discussion
92840.

15 ml of N/10 NaOH solution completely neutralises 12 ml of H_(2)SO_(4) solution. The normality of H_(2)SO_(4) solution will be

Answer»

`N//5`
`N//10`
`N//8`
`N`

SOLUTION :`N_(1)V_(1)=N_(2)V_(2),(1)/(10)xx15=N_(2)xx12impliesN_(2)=(1)/(8)N`
Discussion
92841.

15 mL of a gaseous hydrocarbon required 45 mL oxygen for complete combustion .30 mL of CO_2 is formed. The formula of the hydrocarbon is :

Answer»

`C_2H_6`
`C_2H_4`
`C_3H_6`
`C_2H_2`

ANSWER :B
Discussion
92842.

15 ml of 0.2 N alkali is required to complete neutralization of 30 ml acid solution. Concentration of the acid solution is

Answer»

0.1 N
0.3 N
0.15 N
0.4 N

Solution :`N_(1)V_(1)=N_(2)V_(2),15xx0.2=30xx X``thereforex=0.1N`
Discussion
92843.

15mL gaseoushydrocarbon(A)was requiredforcompletecombustion 357 ml of air(21% oxygen byvolume ) andgaseous produects occupied 327ml (all volume geingmesuredat STP ). (A) underset( ("monochlorination"))underset( Cl_(2).hv)to (B) ( number of isomers including stereiosmers )(A) and (B) underset( "(dichlorination )")overset( Cl_(2)hv)to( C)( numberof isomers incuding stereoisomrs ) whichisomer of ( C)on reaction withdiethylmalonate (DEM) will givecyclobutane derivative ?

Answer»




ANSWER :A
Discussion
92844.

15mL gaseoushydrocarbon(A)was requiredforcompletecombustion 357 ml of air(21% oxygen byvolume ) andgaseous produects occupied 327ml (all volume geingmesuredat STP ). (A) underset( ("monochlorination"))underset( Cl_(2).hv)to (B) ( number of isomers including stereiosmers )(A) and (B) underset( "(dichlorination )")overset( Cl_(2)hv)to( C)( numberof isomers incuding stereoisomrs ) whichof( C)on reactionwithNa or mg will givecyclopropane ?

Answer»




ANSWER :A
Discussion
92845.

15mL gaseoushydrocarbon(A)was requiredforcompletecombustion 357 ml of air(21% oxygen byvolume ) andgaseous produects occupied 327ml (all volume geingmesuredat STP ). (A) underset( ("monochlorination"))underset( Cl_(2).hv)to (B) ( number of isomers including stereiosmers )(A) and (B) underset( "(dichlorination )")overset( Cl_(2)hv)to( C)( numberof isomers incuding stereoisomrs ) the numberfo isomers of ( c)is "

Answer»

2
3
4
5

Answer :D
Discussion
92846.

15mL gaseoushydrocarbon(A)was requiredforcompletecombustion 357 ml of air(21% oxygen byvolume ) andgaseous produects occupied 327ml (all volume geingmesuredat STP ). (A) underset( ("monochlorination"))underset( Cl_(2).hv)to (B) ( number of isomers including stereiosmers )(A) and (B) underset( "(dichlorination )")overset( Cl_(2)hv)to( C)( numberof isomers incuding stereoisomrs ) themolecularformulaof thehydrocarbon (A) is :

Answer»

`C_(2)H_(6)`
`C_(2)H_(4)`
`C_(3)H_(6)`
`C_(3)H_(8)`

Answer :D
Discussion
92847.

15mL gaseoushydrocarbon(A)was requiredforcompletecombustion 357 ml of air(21% oxygen byvolume ) andgaseous produects occupied 327ml (all volume geingmesuredat STP ). (A) underset( ("monochlorination"))underset( Cl_(2).hv)to (B) ( number of isomers including stereiosmers )(A) and (B) underset( "(dichlorination )")overset( Cl_(2)hv)to( C)( numberof isomers incuding stereoisomrs ) thenumber of isomers of (b) is :

Answer»

2
3
4
5

Answer :A
Discussion
92848.

1.5 gm sample of bleaching power was suspended in water. If was treated with CH_(3)COOH followed by the addition of excess of Kl. The liberated iodine required 150 mL of (M)/(10) hypo solution for complete titration. The percentage of available chlorine in the sample is

Answer»


ANSWER :35.5
Discussion
92849.

15 gm of a substance dissolved in 450 gms of water produces a depression of -0.34^@ C in the freezing point . Calulation the Mol. Wt. of the solid ( K_f for water is 1.86 K kgmol ^(-1))

Answer»

Solution :`DELTA T_f =K_f XX w/M xx 100/W` < br> `implies12 =18.5xx15/Mxx100/ 150`
`IMPLIES M = (18.5xx15xx100)/(1.2xx150)`
`= 154.16 gm MOL^(-1)`
Discussion
92850.

15 gm of a substance dissolved in 150 gms of water produces a depression of -1.2^@ C in the freezing point . Calulation the Mol. Wt. of the solid ( K_f for water is 1.86 K kg mol ^(-1) )

Answer»

Solution :`Delta T_f =K_f XX w/M xx 100/W` < br> `implies12 =18.5xx15/Mxx100/ 150`
`implies M = (18.5xx15xx100)/(1.2xx150)`
`= 154.16 gm mol^(-1)`
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