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90051.

A decimolar solution of potassium ferrocyanide is 50% dissociated at 300K. It's Osmotic pressure is 7.482 x10^(x) Nm^(-2) . Find x ?

Answer»


ANSWER :5
Discussion
90052.

A decimolarsolution of potassium ferrocyanide is 50% dissociatd at 300 K . Calculate the osmotic pressureof the solution . Given solution constant (R )= 8.314"JK"^(-1) mol^(-1)

Answer»

Solution :Molar mass of SUCROSE `(C_(12) H_(22) O_(11)) = 342 " gmol"^(-1)`
Molarof mass of urea `(NH_(2)CONH_(2)) = 60 g " mol"^(-1)`, Molar mass of glucose`(C_(6) H_(12)O_(6)) = 180g " mol"^(-1) `
Molarmassof NaCl = 58.5 g`" mol"^(-1)`,
Molar conc. of sucrose = `(34.2)/(342) = 0.1 M ,""`Molar conc of urea `= 60/60 = 1 M ,`
Discussion
90053.

A decapiptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed product. The number of glycine units present in the decapeptide is

Answer»

6
5
4
3

Answer :A
Discussion
90054.

A decimolar solution of K_4[Fe(CN)_6] at 300K is 50% dissociated, then, osmotic pressure of the solution is

Answer»

`3.61 ` ATM
7.38 atm
12.32 atm
21.34 atm

ANSWER :B
Discussion
90055.

A decapeptide (Mol .Wt. 796) on complete hydrolysis gives glycine (Mol.Wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptides is

Answer»


Solution :Mol. Wt. of decapeptide `=796`.
No. of `H_2O` molecules used for hydrolysis of decapeptide `=9`.
i.e., Decapeptide `+9H_2Oto` GLYCINE + Alanine +Phenylanine
Total wt. of amino acids obtained after addition of 9 molecules of WATER `=796+9xx18=958`
Total wt. of glycine in the hydrolysed products `=(958xx47)/(100)450.26`
But mol. wt. of glucine `=75`
`THEREFORE ` No. of glycine UNITS in the decapeptide `=(450.26)/(75)=6`
Discussion
90056.

(A) De-emulsification is a reverse process of emulsion formation (R ) Duringde-emulsification , the emulsion breaks to give oil and water.

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Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A)
Both (A)and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is FALSE
Both (A) and (R ) are false

ANSWER :A
Discussion
90057.

A dark voilent solid X reacts with NH_(3) to form a mild explosive Y which decomposes to give a voilet coloured gas. X also reacts with H_(2) to give an acid Z. Z can also be prepared by heating its salt with H_(3) PO_(4) The dark violent solid X is

Answer»

`Cl_(2)`
`SO_(2)`
`Br_(2)`
`I_(2)`

Solution :`2NH_(3) + underset("(X)")(3I_(2)) RARR NI_(3). underset("(Y)")NH_(3)+3HI , underset("(X)")(I_(2)) + H_(2)rarr underset("(Z)")(2HI)`
Discussion
90058.

A dark voilent solid X reacts with NH_(3) to form a mild explosive Y which decomposes to give a voilet coloured gas. X also reacts with H_(2) to give an acid Z. Z can also be prepared by heating its salt with H_(3) PO_(4) The mid Explosive Y is

Answer»

`NI_(3)`
`NH_(4)I`
`HI`
`NI_(3). NH_(3)`

SOLUTION :`2NH_(3) + underset("(X)")(3I_(2)) rarr NI_(3). underset("(Y)")NH_(3)+3HI , underset("(X)")(I_(2)) + H_(2)rarr underset("(Z)")(2HI)`
Discussion
90059.

A dark green bead in the borax bead test indicates the presence of

Answer»

`CR^(3+)`
`Mn^(2+)`
`Co^(2+)`
`Ni^(2+)`

SOLUTION :`Cr^(3+)` forms dark GREEN chromium metaborate `Cr(BO_(2))_(3)`
Discussion
90060.

A Daniel cell produces a current of 3.6mA at 1.1 V for 15 minutes. What is the maximum possible work you can get from it?

Answer»


ANSWER :`1.1 XX 3.5 xx 10^(-3) xx 900`
Discussion
90061.

A daniell cell is set up by dipping a zinc rod weighing 100 g in 1 litre of 1.0 M CuSO_(4) solution. How long would the cell run if it delivers a steady current of 1.0 ampere? (Atomic masses Cu=63.5,Zn=65)

Answer»

82.47 hrs
53.61 hrs
41.23 hrs
26.80 hrs

Solution :1L of 1.0 M `CuSO_(4)` solution CONTAINS `CU^(2+)=1` mole
`100" g Zn"=(100)/(65)` mole
THUS, in the cell reaction `Zn+Cu^(2+)toZn^(2+)+Cu,Cu^(2+)` ions are the limiting reagent. The cell will run till all the `Cu^(2+)` ions are deposited, i.e., 1 mole of Cu is deposited. quantity of electricity requried for DEPOSITION of 1 mole of Cu=2F
Time, `t=(2xx96500C)/(1A)=2xx96500s`
`=(2xx96500)/(3600)hr=53.61hr`
Discussion
90062.

(A) d block elements are transitioned between s-block metals and p-block metals. (R) d block elements are more-electropositive than s-block elements.

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Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is FALSE
Both (A) and (R) are false

Solution : (A) is true but (R) is false
Discussion
90063.

A d-block element forms octahedral complex but its magnetic moment remains same either in strong field or in weak field ligand. Which of the following is/are correct ?

Answer»

Element always forms colourless compound
Number of electrons in `t_(2_(G))` orbitals are higher than in `e_(g)` orbitals
It can have either `d^(3)` or `d^(8)` configuration
It can have either `d^(7)` or `d^(8)` configuration

Solution :* For `d^(3)` configuration :
In strong ligand field : `t_(2)^(3)EG^(0)` : In WEAK ligand field `:t_(2g)^(3)eg^(0)`
* For `d^(8)` configuration :
In strong ligand field : `t_(2g)^(6)eg^(2)` , In weak ligand field : `t_(2g)^(6)eg^(2)`
Discussion
90064.

A d-block element forms octahedral complex but its magnetic moment remains same either in strong field or in weak field ligand. Which of the following is/are correct?

Answer»

Element ALWAYS forms colourless compound
Number of electron in `t_(2g)` ORBITALS are higher than in `e_(g)` orbitals
It can have either `d^(3) or d^(8)` configuration
It can have either `d^(7) or d^(8)` configuration

Solution :Irrespective nature of ligand field, magnetic moment of OCTAHEDRAL complexes having metal ions in `d^(1),d^(2),d^(3),d^(8),d^(9)` configuration, remains same.
Element may or may not form colourless complex.
Discussion
90065.

A cylindrical piece of Mg is 10 cm long and has a diameter of 8 cm. The density of Magnesium is 0.7 gm//cm^(3). How many atoms does piece of Mg contain. [N_(A) = 6 xx 10^(23)] [Divide your answer by 22 xx 10^(23) ]

Answer»


Solution :Volume of cylindrical piece `= pi r^(2)L`
`= pi xx ((8)/(2))^(2) xx 10`
`= (22)/(7) xx 4 xx 4 xx 10`
`= ((22)/(7) xx 160) cm^(3)`
Mass of piece `= (0.7 xx (22)/(7) xx 160) cm^(3)`
`= (22 xx 16) gm`
Atoms of `Mg = ((22 xx 16)/(24)) N_(A) RARR (44)/(3) N_(A)`
`= (44)/(3) xx 6 xx 10^(23)`
`= (88 xx 10^(23))/(22 xx 10^(23)) = 4`
Discussion
90066.

A cylindrical tube of length 30 cm is partitioned by a tight-fitting separator. Theseparator is very thin, very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperature in the parts A and B are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position at which the length of the tube at part A is 20 cm. Find the final equilibrium position of the separator, reached after a long time.

Answer»

Solution : APPLY `(p_1V_1)/(T_1) = (p_2V_2)/(T_2)` for PART A under the given two CONDITIONS and again forpart B. Remember that at both the equilibrium POSITIONS, both sides will have the same pressure.
10 cm ALONG part A
Discussion
90067.

A cylinder with movable piston is fitted with H_(2) gas at 27^(@)C that occupies 250ml. If the maximum capacity of the cylinder is 1L, the highest temperature to which the cylinder can be heated at constant pressure without having the piston to come out is :

Answer»

`1200^(@)C`
`977^(@)C`
`927^(@)C`
`1023K`

Solution :`T_(1) = 273 + 27 = 300K`,
`V_(1) = 250 ML`
`V_(2) = `1000ML, `T_(2) = ?`
At constant pressure,
`(V_(1))/( T_(1)) = ( V_(2))/( T_(2))`
or `T_(2) = ( V_(2) T_(1))/( V_(1))`
or ` = ( 1000xx300)/( 250) = 1200K`
`T_(2) = 1200 - 273 = 927^(@)C`
Discussion
90068.

A cylinder of V litre capacity containing NH_(3) gas is inverted over another vessel of V litre capacity containing HCl gas at same temperature and pressure . After some time the pressure in cylinder will :

Answer»

1. become double
2. REMAIN same
3. drop considerably
4. become `3//2` of ORIGINAL pressure

Solution :Because of the formation of solid `NH_(4)Cl`.
Discussion
90069.

A cylinder of gas supplied by Bharat Petroleum is assumed to contain 14 kg of butane. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder lasts (Delta_(C ) H^(@) of C_(4) H_(10) = - 2658 kJ mol^(-1))

Answer»

15 DAYS
20 days
50 days
32 days

Solution :Calorific value of butan = `(DeltaH_(c))/("MOL. wt.")=(2658)/(58)=45.8 KJ//gm`
Cylinder consist 14 Kg of butane means 14000 gm of butane
`because` 1 gm gives 45.8 kJ
`therefore` 14000 gm gives`14000xx45.8 = 641200 kJ`
So gas full fill the requirement for `(641200)/(20,000)=32.06` days.
Discussion
90070.

A cylinder of gas supplied by Bharat Petroleum is assumed to contain 14 kg of butane. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder last …….. days (Delta H_(C) of C_(4)H_(10) = - 2658 kJ per mole)

Answer»

15 days
20 days
50 days
32 days

Solution :MOLAR mass of butane, `C_(4)H_(10) = 12 xx 4 + 1 xx 10 = 58 g mol^(-1)`
58 g of butane gives 2658 KJ of HEAT energy.
14 kg of butane will give heat energy
`= (2658)/(58) xx 14 xx 10^(3)`
`= 641.5862 xx 10^(3) kJ`
Daily energy requirement for cocking = 20.000 kJ
`= 2 xx 10^(4) kJ`
No. of days CYLINDER will last `= (641.5862 xx 10^(3) kJ)/(2 xx 10^(4) kJ "day"^(-1))`
= 32.08 days.
Discussion
90071.

A cylinder of gas is assumed to contain 11.6 kg of butane (C_4H_10). If a normal family needs 20000kJ ofenergy per day, then the cylinder will last for : (Given that ΔH for combustion of butane is – 2658kJ)

Answer»

20 DAYS
18 days
26.6 days
32 days

SOLUTION :`DeltaH_"released"=(11.6xx10^3)/58xx2658` KJ
No of DAY =`(11.6xx10^3xx2658)/(58xx20000)`
=26.58
`approx 26.6`
Discussion
90072.

A cylinder of gas is assumed to contain 11.2 kg of butane (C_(4)H_(10)). If a normal family needs 20000 kJ of energy per day. The cylinder will last: (Given that DeltaH for combustion of butane is -2658 kJ)

Answer»

20 DAYS
25 days
26 days
24 days

Answer :C
Discussion
90073.

A cylinder of gas is assumed to contain 11.2 kg of butane. If a normal family needs 20,000 kJ of energy per day for cooking, how long will the cylinder last if the enthalpy of combustion, Delta H = - 2658 kJfor butane?

Answer»

30.5 days
25.66 days
40.6 days
10.66 days

Solution :Given `C_4 H_10(g) + 13/2 O_2(g)to 4CO_2 (g) + 5H_2O (g) , Delta H = - 2658 kJ`
MOLECULAR weight of `C_4H_10 = 58 `g/mol
58 g of BUTANE on COMBUSTION PRODUCES = 2658 kJ heat
` therefore ` 11.2 kg of butane will produce
`= (2658)/(58) xx 11.2 xx 10^3 kJ` heat
= 513268.97kJ
The family needs 20,000 kJ of energy per day
` therefore ` Number of days ` = (513268.97)/(20000) kJ = 25.66 days`
Discussion
90074.

A cylinder of gas is assumed to contain 11.2 kg of butane (C_(4)H_(10)). If a normal family needs 20000 kJ of energy per day, the cylinder will last in (given that DeltaH for combustion of butane is -2658 kJ)

Answer»

20 days
25 days
26 days
24 days

Solution :Cylinder contains 11.2 KG or 193.10 mole of butane.
[`because` moleclar mass of butane=58]
`because`Energy released by 1 mole of butanne=-2658 KJ
`THEREFORE` Energy released by 193.10 mole of butane
`=-2658xx193.10`
`=5.13xx10^(5)kJ`
`therefore`Cylinder will LAST in `(5.13xx10^(5))/(20000)=25.66` or 26 days.
Discussion
90075.

A cylinder of compressed gas contains nitrogen and oxygen in the ratio 3:1 by mole. If the cylinder is known to contain 2.5xx10^(4)g of oxygen, what is the total mass of the gas mixture?

Answer»

Solution :NUMBER of moles of oxygen in the CYLINDER
`=("Mass in gram")/("Molecular mass in gram")=(2.5xx10^(4))/(32)`
=781.25
`therefore` Number of moles of `N_(2)=3xx781.25=2343.75`
Mass of nitrogen in the cylinder `=2343.75xx28`
=65625g
`=6.5625xx10^(4)G`
Total mass of the gas in the cylinder
`=2.5xx10^(4)+6.5625xx10^(4)=9.0625xx10^(4)g`
Discussion
90076.

A cylinder fitted with a movable piston contains liquid water in equiliberium with water vapour at 25^o C .Which operation result in a decrease in the equiliberium vapour pressure ?

Answer»

MOVING the oiston downward a short DISTANCE
removing a small AMOUNT of vapour
Removing a small amount of the LIQUID water
Dissolving salt in the water

Answer :D
Discussion
90077.

A cylindercontainer of volume 4.48 liters is containing equal no .Ofmoles of a monoatomic gasin two section A and B separatedby an adiabatic frictionless piston as show in figure . The initial temperatureandpressue of gases was 273 K and 1 atm.Now gas section'A' is slowlyheated till the volumeof sectionB becomes 1//2 sqrt(2)of initinalvolume. Findtotal changein DeltaH forsectionA and B . ( in cal/mole ) "" [C_(v) "ofmonoatomic gas" =3//2 R, gamma = 5//3] [Use R = 2 cal //mol &sqrt(2)=1.4]

Answer»

Solution :Initial volume ofsec. A = sec. B= 2.24 LITRES
Find volume ofsexc.`B = [(2.24)/(2sqrt(2))]` litres
Thegasin sec . 'B' compressed reversiblyand adiabatically
`rArr T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1) rArr T_(2) = T_(1)((V_(1))/(V_(2)))^(r-1) rArr T_(2)= T_(1) (2sqrt(2)) ^(2//3) = 2T_(1)`
{ Formonoatom gas`gamma = (5)/(3)`}
The final pressure in sec. 'B'
`P_(f)= (P_(1)V_(1))/(T_(1)),(T_(2))/(V_(2)) = P_(1) XX ((T_(2))/(T_(1)))((V_(1))/(V_(2))) P_(1) xx 2xx 2sqrt(2)"" = 4 sqrt(2) "atm`
`rArr` Pressure in sec. 'A' = `4sqrt(2)"atm"``rArr` Final temperature in sec. `AT_(2) = ((P_(2))/(P_(1)))((V_(2))/(V_(1)))T_(1)`
`T_(2) = ((P_(2))/(P_(1))){V_(1) + V_(1)-(V)/(2sqrt(2))} T_(1)`
`4sqrt(2) (2 -(1)/(2sqrt(2)))T_(1) ""=[((4sqrt(2)-1)(4sqrt(2)))/(2sqrt2)]T_(1)`
`(4 sqrt(2)-1)2 T_(1)`
`[8sqrt(2)T_(1)- 2T_(1)- T_(1)]={4sqrt(2)T_(1) - 3T_(1)}`
`DeltaH_(A) = (0.1) (5)/(2) R{8sqrt(2)-3} rArr DeltaH_(B) = (0.1) (5)/(2)R{273}`
`""2(4sqrt(2)-1)`
`DeltaH_(T) = (0.1)(5)/(2)R(273){2} (4sqrt(2)-1)""=1255.8 or 5 xx 1255.8= 6279`
Discussion
90078.

(A) Cyclohexanone exhibits keto-enol tautomerism. (R) In cyclohexanone, one form contains the keto group (C=O) while the other contains enolic group (-C=C-OH).

Answer»


ANSWER :A
Discussion
90079.

A cyclotron is used to

Answer»

ACCELERATE neutrons
Accelerate electrons
Accelerate protons
Accelerate `ALPHA`-particles

Answer :C
Discussion
90080.

(A) Cyclobutane is less stable than cyclopentane. (R) Presence of bent bonds causes loss of orbital overlap.

Answer»


ANSWER :A
Discussion
90081.

A cycloalkane having molecular mass 84 and four secondary carbon atoms will forms monochloro structure isomers on chlorination. Identify the structure of cycloalkane.

Answer»


SOLUTION :N//A
Discussion
90082.

A cyclindricalcontainer ofvolume44.8 litresis containing equalno.of moles (in integerno. ) of an idealmonoatomicgasin twosections A and B separeted byanadiabaticfrictionless pistonas show in figure The initinaltemperatureand pressureof gasin bothsectionB becomes(1//8)^(th) of initial volume . Given : R = 2 cal //"mole -K", C_(v,m)of monoatmicgas =(3)/(2)R, At1 atm & 0^(@) idealgas occupy22.4 liter.Changeinenthalpyfor sectionA in K cal.

Answer»

<P>48.3
80.53
4.83
8.05

Solution :`DELTAH = nC_(p)DeltaT`
In A no. ofmole =`(PV)/(RT)= 10`lt brgt`DeltaH= 10 xx (5)/(2) xx 2[1638-27.3]`
=`80535 CAL = 80.535K cal`
Discussion
90083.

A cyclindricalcontainer ofvolume44.8 litresis containing equalno.of moles (in integerno. ) of an idealmonoatomicgasin twosections A and B separeted byanadiabaticfrictionless pistonas show in figure The initinaltemperatureand pressureof gasin bothsectionB becomes(1//8)^(th) of initial volume . Given : R = 2 cal //"mole -K", C_(v,m)of monoatmicgas =(3)/(2)R, At1 atm & 0^(@) idealgas occupy22.4 liter.Findtemperaturein containerA will be

Answer»

1638 K
6988 k
3274K
51 K

Solution :Afterheating P inbothcontainer is EQUAL
`(P_(1)V_(1))/(T_(1))=(P_(1)V_(1))/(T_(2))`
`V_(2)` of container `A = 22.4 +22.4 xx(7)/(8)`
`= 42 L`
`(1 xx22.4)/(27.3)=(32 xx 42)/(T_(2))`
`T_(2)=1638 K`
Discussion
90084.

A cyclindricalcontainer ofvolume44.8 litresis containing equalno.of moles (in integerno. ) of an idealmonoatomicgasin twosections A and B separeted byanadiabaticfrictionless pistonas show in figure The initinaltemperatureand pressureof gasin bothsectionB becomes(1//8)^(th) of initial volume . Given : R = 2 cal //"mole" -K, C_(v,m)of monoatmicgas =(3)/(2)R, At1 atm & 0^(@) idealgas occupy22.4 liter.What will be thefinalpressure in contianerB:

Answer»

2 ATM
8 atm
16 atm
32 atm

Solution :`P_(1)V_(1)^(gamma)= P_(2)V_(2)^(gamma)`
`V_(2)=(1)/(8) V_(1)"" gamma=(5)/(3)`
`1 XX 22.4^(gamma) =P_(2)(2.8)^(gamma)`
Discussion
90085.

A cyclic steroisomers having the molecular formula C_(4)H_(7)Cl are classified and tabulated. Find out the correct set of numbers.

Answer»

`{:("Geometrical","OPTICAL"),("6","2"):}`
`{:("Geometrical","Optical"),("4","2"):}`
`{:("Geometrical","Optical"),("6","0"):}`
`{:("Geometrical","Optical"),("4","0"):}`

ANSWER :A
Discussion
90086.

A cyclic process ABCD is shown in VT diagram for an ideal gas. Which of the following diagram represents the same process ?

Answer»




ANSWER :D
Discussion
90087.

A cyclic process ABCD is show in PV diagram for a ideal gas . Whichof thefollowingdigram reppresents the sameprocess ?

Answer»




Solution :In CD volume constant
In case of ADand BCtemp of BCgreater than AD
`A to B` Isobaric
`B to C` Isothermal
`D to A ` Isothermal `T_(Bto C)GT T_(D to A)`
`C to D `Isochoric
`BECOME T `const ``THEREFORE DeltaU = 0`
Discussion
90088.

A cyanohydrin of a compound X on hydrolysis gives lactic acid. The X is

Answer»

HCHO
`CH_(3)CHO`
`(CH_(3))_(2)CO`
`C_(6)H_(5)CH_(2)CHO`

ANSWER :B
Discussion
90089.

A current strength of 9.65 amperes is passed through excess fused AlCl_(3) for 5 hours . How many litres of chlorine will be liberated at STP ? (F = 96500 C)

Answer»

`2.016`
`1.008`
`11.2`
`20.16`

Solution :Quantity of electricity = `I xx t`
`= 96.5 xx 5 xx 60 xx 60`
`= 173,700 C `
Faraday of electricity = `(173,700)/(96500)= 1.8 F `
`2 Cl^(-) - 2e^(-) to Cl_(2)`
1 F of electricity liberates 35.5 G or 11.2 L of `Cl_2` at S.T.P.
`therefore` Volume of `Cl_2` LIBERATED by 1.8 F electricity .
`= 1.8 xx 11.2 = 20.16 L`
Discussion
90090.

A current strength of 3.86 A was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40g/mol and 1F = 96500 C).

Answer»

4
2
8
6

Solution :`m = ZIT ""41 "min" 40 "SEC" = 2500 "seconds"`
`= (40 xx 3.86 xx 2500)/(2 xx 96500) "" Z = (m)/(n xx 96500) = 40/(2 xx 96500)`
` = 2g`.
Discussion
90091.

A current strength of 3.86 A was passed through molten Calcium oxide for 41 minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40 g / mol and 1F = 96500C).

Answer»

4
2
8
6

Answer :B
Discussion
90092.

A current of strength 2.5 amp was passed through CuSO_(4) solution for 6 minutes 26 seconds. The amount of copper deposited is:- (Atomic weight of Cu=63.5) (1 faraday=96500 coulombs)

Answer»

0.3175 g
3.175 g
0.635 g
6.35 g

Solution :`Q=2.5xx386=96500C`
`2F(2xx96500C)` deposited `C u=63.5g`
`therefore`HENCE 965 C will deposited, Cu=0.3175gm.
Discussion
90093.

A current of I ampere was passed for t second through three cell P,Q and R connectedin series. These contains respectively silver nitrate. mercuric nitrate and mercurous nitrate. At the cathode of the cell P, 0.216 g of Ag was deposited. The weights of mercury deposited in the cathode of Q and R respectively are :

Answer»

`0.4012 and 0.8024g`
`0.4012 and 0.2006 G`
`0.2006 and 0.4012 g`
`0.1003 and 0.2006 g`

ANSWER :C
Discussion
90094.

A current of dry at was passed through a solution of 2.5 g of a non-volatile substance 'X' in 100 g of w ater and then through water along. The loss of weight of the former was 1.25 g and that of the latter was 0.05 g. Calculate (i) mole fraction of the solute in the solution (ii) molecular weight of the solute.

Answer»

SOLUTION :Loss in weight of solution `prop p_(s) prop 1.25g`, Loss in weight of solvent (WATER) `prop p^(@)-p_(s) prop 0.05g`
`p^(@)=(p_(0)-p_(s))+p_(s) therefore""p^(@) prop 0.05 +1.25, i.e., p^(@) prop 1.30g`
Hence, `""(p^(@)-p_(s))/(p^(@))=(0.05)/(1.30)=0.0385, i.e, x_(2)=0.0385`
`"Now,"(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(1)//M_(2))/(w_(1)//M_(1)) ,`
`0.0385=(2.5//M_(2))/(100//18) or (2.5)/(M_(2))=(100)/(18)xx0.0385 or M_(2)=11.7`
Discussion
90095.

A current of dry air was passed through a solution containing 2.0 g of a solute A_(2)B in 98 g of water and then through pure water. The loss in weight of the solution was 0.98 g and the loss in weight of pure water is 0.01 g. If the molar mass of A_(2)B is 90 g//mol and it dissociates into A^(+) and B^(-2) ions, then its percentage dissociations in water is.

Answer»


ANSWER :75
Discussion
90096.

A current of dry air was passed first through a series of bulbs containing a solution of C_(6)H_(5) - NO_(20 in ethanol of molality 0.725 and then through a series of bulbs containing pure ethanol. (T = 284 K) loss in weight of the solvent bulbs was 0.0685 g. Calculate the loss in weight of the solution bulbs.

Answer»

4.60 g
5.20 g
2.50 g
2.05 g

Answer :D
Discussion
90097.

A Current of dry air was first passed through the bulb containing solution of 'A' in water and then through the bulb containing pure water. The loss in mass of a solution bulb is 1.92g gm. Where as that in pure water bulb is 0.08g, then mole fraction of 'A' is.

Answer»

0.86
0.2
0.96
0.04

Answer :D
Discussion
90098.

A current of dry air was bubbled through a bulb containing 26.67g of an organic compound in 200g of water , then through a bbulb at the same temperature, containing water and finally through a tube containing anhydrous CaCl_(2) the loss in mass of bulb containing water was 0.087g and gain in mass of CaCl_(2) tube was 2g, then molecular mass of organic compound is

Answer»

180
53.75
280.7
140.7

Answer :B
Discussion
90099.

A current of dry air was bubbled through a bulb containing 26.66 g of an organicsubstance in 200 g of water, then through a bulb at the same temperature containing pure water, and finally through a tube containing fused calcium chloride. The loss in weight of water bulb was 0.087 g and the gain in weight of the calcium chloride tube was 2.036 g. Determine the molecular weight of the organic substance.

Answer»


ANSWER :56.1
Discussion
90100.

A current of 9.65 amperes is passed through excess of fused AlCl_(3) for 5 hours. How many litres of chlorine will be liberated at S.T.P. ? (1F=96500C)

Answer»

`2.016" L "`
1.008L
11.2 L
20.16 L

Solution :(d) The anodic reaction is :
`2Cl^(-) to underset(22.4L)(Cl_(2))+underset(2F)(2e^(-))`
The QUANTITY of charge (Q) passed`=ixxt`
No. of FARADAYS of `=(9.65xx5xx60(C ))/(96500(C ))`
charge passed =1.8 F
1F of charge evolves `Cl_(2)=11.2 L`
1.8 F of charge evolves `Cl_(2)=(11.2L)/((1F))xx(1.8F)=20.16" L"`
Discussion