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90251.

A coordination compound with molecular formula CrCl_(3).4H_(2)O precipitates one mole of AgCl with AgNO_(3) solution. Its molar conductivity is found to be equivalent to two ions. What is the structural formula and name of the compound ?

Answer»

Solution :The structural formula of the coordination compound is :
`[Cr(H_(2)O)_(4)Cl_(2)]Cl`
The NAME of the compound is :
Tetraaquadichloridochromium(III) chloride
Discussion
90252.

A coordination compound of cobalt has the molecular formula cotaining five ammonia molecules, one intro group and two chlorine atoms for one cobalt atom. One mole of this compoundproduces three moles of ions in an aqueous solution. The aqueous solution on treatment with an excess of AgNO_(3) gives two moles of AgCl as a precipitate. The formula of this complex would be

Answer»

`[CO(NH_(3))_(4)(NO_(2))Cl][(NH_(3))_(4)Cl]`
`[Co(NH_(3))_(5)Cl][ClNO_(2)]`
`[Co(NH_(3))_(5)(NO_(2))]Cl_(2)`
`[Co(NH_(3))_(5)][(NO)_(2))_(2)Cl_(2)]`

Solution :( C) `[Co(NH_(3))_(5)(NO_(2))]Cl_(2) OVERSET(aq)IFF[Co(NH_(3))_(5)(NO_(2))]^(2+)(aq)+2Cl^(-)(aq)`(no. of ions=3)
`2Ag^(+)+2Cl^(-) to 2AgCl darr `(white).
Discussion
90253.

A coordination compound has the formula CoCl_(3).4NH_(3). It does not liberate ammonia but forms a precipitate with AgNO_(3). Write the structure and IUPAC name of the complex compound.

Answer»

Solution :`[Co(NH_(3))_(4)Cl_(2)]Cl`. Its IUPAC name is tetraamminedchlorocobalt(lll) chloride.
It FORMS a PRECIPITATE with `AgNO_(3)` because of the PRESENCE of ionisable CHLORINE.
Discussion
90254.

A coordination compound has the formula, CoCl_(3).4NH_(3). It does not liberate ammonia but precipitates chloride ions as silver chloride. Give the IUPAC name of the complex and write its structural formula.

Answer»

Solution :REMEMBERING that coordination number of Co is 6, the FORMULA of the complex will be `[Co(NH_(3))_(4)Cl_(2)]Cl`.The name will be tetraamminedichloridocobalt (III) chloride.
Discussion
90255.

A coordination compound CrCl_(3).4H_(2)O precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write structural formula of the compound and name it.

Answer»

Solution :`[Co(H_(2)O)_(4)Cl_(2)]CL` [Tetraaquadichloridocobalt(III) chloride).
Discussion
90256.

A coordination compound CrCl_(3).4 H_(2)O precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write structural formula of the compound and name it.

Answer»

SOLUTION :Total 2 ions means one ionizable ion and one complex ion.
`therefore` FORMULA is `[Co(H_(2)O)_(4)Cl_(2)]Cl` (tetraaquadichloridocobalt (III) chloride).
Discussion
90257.

A coordination complex of type MX_(2)Y_(2)[M=metal ion, X,Y=monodentata ligands ], can have either a tetrahedral of a square planar geometry. The maximum number of possible isomers in these two cases are respectively

Answer»

0 and 2
2 and 1
1 and 3
3 and 2

Answer :A
Discussion
90258.

A coordination complex of type MX_(2)Y_(2) (M-metal ion: X, Y-monodentate lingads), can have either a tetrahedral or a square planar geometry. The maximum number of posible isomers in these two cases are respectively:

Answer»

1 and 2
2 and 1
1 and 3
3 and 2

Solution :In tetrahedral all the posiiton are adjacent to ONE another HENCE only one isomer is possible
Discussion
90259.

A Coordination complex has the formula PtCl_(4).2KCl. Electrical conductance measurements indicate the presence of three ion in one formula unit. Treatment with AgNO_(3) produces no precipitate of AgCl. What is the coordination no. of Pt in this complex

Answer»

5
6
4
3

Solution :`PtCl_(4).2KCL to K_(2)[PtCl_(6)]`
C.N. = 6
Discussion
90260.

A coordiantion compound has the formula CoCl_(3).4NH_(3). It does not liberate NH_(3) but forms a precipitate with AgNO_(3). Write the structure and IUPAC name of the complex compound. Does it show geometrical isomerism?

Answer»

Solution :Formula : `[Co(NH_(3))_(4)Cl_(2)]Cl`
Name : Tetraaminedichloridocobalt(III) chloride
Yes, it show geometrical isomerism.
Discussion
90261.

A coordiantion complex compound of cobalt has the molecular formula containing four ammonia molecules, one nitro group and two chlorine atoms. A solution of this complex was prepared by dissolving 2.44 g of it in water and making the volume to 200 mL. Excess of AgNO_(3) solution was then added to it and the precipitate of AgCl formed was filtered and dried. The weight of AgCl thus obtained was found to be 1.435 g (Atomic mass of Co=59, Ag=108) The type of isomerism shown by the complex will be

Answer»

linkage isomerism
Geometrical isomerism
Ionization isomerism
All the three

Solution :DUE to `NO_(2)` GROUP, it can show linkage isomerism. It can have the ionisation ISOMER `[Co(NH_(3))Cl_(2)]NO_(2)`. It can exist in cis and trans FORM (i.e., geometrical isomers)
Discussion
90262.

A coordination complex compound of cobalt has molecular formula containing five ammonia molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three mole ions in an aqueous solution. On reacting this solution with excess of silver nitrate solution, two moles of AgCl get precipitated. The formula of this compound would be

Answer»

`[CO(NH_(3))_(4)(NO_(2))CL][NH_(3)Cl]`
`[Co(NH_(3))_(5)Cl][Cl(NO_(2))]`
`[Co(NH_(3))_(5)(NO_(2))]Cl_(2)`
`[Co(NH_(3))_(5)][(NO_(2))Cl_(2)]`

Solution :`[Co(NH_(3))_(5)Cl]Cl_(2) hArr [Co(NH_(3))_(5)Cl]^(2+)+2Cl^(-)` THREE ions. With `AgNO_(3)`, it gives 2 moles of AgCl.
Discussion
90263.

A coordiantion complex compound of cobalt has the molecular formula containing four ammonia molecules, one nitro group and two chlorine atoms. A solution of this complex was prepared by dissolving 2.44 g of it in water and making the volume to 200 mL. Excess of AgNO_(3) solution was then added to it and the precipitate of AgCl formed was filtered and dried. The weight of AgCl thus obtained was found to be 1.435 g (Atomic mass of Co=59, Ag=108) The name of the complex would be

Answer»

tetraamminedichloridocobalt (III) nitrite
tetraamminenitrocobalt (III) CHLORIDE
tetraamminechloridonitrocobalt (III) chloride
tetraamminedichloridonitrocobalt (III)

Solution :The NAME of the complex is tetraammine chloridonitrocobalt (III) chloride
Discussion
90264.

A coordiantion complex compound of cobalt has the molecular formula containing four ammonia molecules, one nitro group and two chlorine atoms. A solution of this complex was prepared by dissolving 2.44 g of it in water and making the volume to 200 mL. Excess of AgNO_(3) solution was then added to it and the precipitate of AgCl formed was filtered and dried. The weight of AgCl thus obtained was found to be 1.435 g (Atomic mass of Co=59, Ag=108) The formula of the complex would be

Answer»

`[CO(NH_(3))_(4)(NO_(2))]Cl_(2)`
`[Co(NH_(3))_(4)Cl(NO_(2))]Cl`
`[Co(NH_(3))_(4)Cl_(2)]NO_(2)`
`[Co(NH_(3))_(4)Cl_(2)(NO_(2))]`

Solution :Molar mass of `Co(NH_(3))_(4)(NO_(2))Cl_(2)=59+68+46+71=244 g mol^(-1)`
`therefore` 2.44 g of the complex `=(2.44)/(244)=0.01` mol.
Molar mass of AGCL = 108+35.5=143.5 g
`therefore 1.435 g AgCl = (1.435)/(143.5)=0.01` mol
This shows that one mole of the complex gives 1 mole of `Cl^(-)` ions in the solution. Hence, the FORMULA of the complex must be `[Co(NH_(3))_(4)Cl(NO_(2))]Cl`.
Discussion
90265.

a. Convert cyclohexyl amine into cyclopentyl amine . b. Convert cyclohexene oxide into aminocyclohexamol.

Answer»

SOLUTION :` (##KSV_CHM_ORG_P2_C15_E01_013_S01.png" WIDTH="80%">
Discussion
90266.

(a) Convert benzene into (i) Acetophenone (ii) Benzaldehyde (iii) Benzophenone. (b) Write a note on aldol condensation.

Answer»

Solution :
(b) Aldol condensation : Aldehydes and KETONES having `alpha` - H atom when treated with dil. NaOH solution UNDERGO condensation.
e.g. `underset("Acetaldehyde")(CH_(3)CHO+CH_(3)CHO) overset(Dil. NaOH)to underset("Acetaldol")(CH_(3)-underset(H)underset(|)overset(OH)overset(|)C-CH_(2)CHO)`
Discussion
90267.

(A) Conversion of fresh precipitate to colloidal stateis called peptization (R ) It is caused by the addition of common ions

Answer»

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 :B
Discussion
90268.

(A) Conversin of PbS to PbSO _(4) consumes four moles of ozone (R ) O_(3) to O _(2) + (O)

Answer»

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
90269.

A contributions of both heat (enthalpy) and randomness(entropy) shall be considered to the overall spontaneity of process.When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change (DeltaG). DeltaG=DeltaH-TDeltaS where,DeltaH=Enthalpy change, DeltaS=Entropychange , T=Temperature in kelvin. If DeltaGlt0, Process is spontaneous , DeltaG=0, Process is at equilibrium , DeltaGgt0, Process is non-spontaneous. Quick lime (CaO) is produced by heating limestone (CaCO_3) to drive off CO_2 gas. CaCO_3(s)toCaO(s)+CO_2(g),DeltaH^@=180 KJ,DeltaS^@=150 J//K Assuming that variation of enthalpy change and entropy change with temperature to be negligible, choose the correct option :

Answer»

Decomposition of `CaCO_s (s)` is NEVER sppotaneous
Decomposition of `CaCO_s (s)` becomes spontaneous when temperature is less than `900^@C`
Decomposition of `CaCO_3 (s)` becomes non-spontaneous when temperature is greater than `1000^@`
Decomposition of `CaCO_3 (s)` becomes spontaneous when temperature is greater than `927^@C`

Solution :`T=(DeltaH)/(DELTAS)=(180xx10^3)/150=1200 K`
Temperature =1200-273=`927^@C`
For `DeltaG lt 0,T gt 927^@C`
Discussion
90270.

A convenient reagent to distinguish ethyl alcohol from n-propyl alcohol is

Answer»

Lucas REAGENT
Tollen's reagent
Schiff's reagent
Iodine with AQ. NAOH solution

Answer :D
Discussion
90271.

A contributions of both heat (enthalpy) and randomness(entropy) shall be considered to the overall spontaneity of process.When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change (DeltaG). DeltaG=DeltaH-TDeltaS where,DeltaH=Enthalpy change, DeltaS=Entropychange , T=Temperature in kelvin. If DeltaGlt0, Process is spontaneous , DeltaG=0, Process is at equilibrium , DeltaGgt0, Process is non-spontaneous. 5 mol of liquid water is compressed from 1 bar to 10 bar at constant temperature.Change is Gibb's energy (DeltaG) in Joule is:[Density of water =1000 kg//m^3]

Answer»

18
225
450
900

Solution :dG=VdP-SdT
at CONSTANT T
dG=VdP
`because` for `H_2O` volume will be almost constant on INCREASING PRESSURE.
`:. DeltaG=V(DeltaP)`
`DeltaG=V(P_2-P_1)=5xx18xx10^(-6)(101-1)xx10^5` Joule `=5xx18xx10^(-1)xx10^(2)=900` Joule
Discussion
90272.

A container whose volume is V contains an equilibrium mixture that consits of 2mol each of PCl_(5) , PCl_(3) and Cl_(2) ( all as gases ). The pressure is 30.3975 kPa and temperature is T. A certain amount of Cl_(2) ( g) is now introduced keeping the pressure and temperature constant until the equilibrium volume is 2V. Calculate the amount of Cl_(2) that was added and the value of K_(p).

Answer»

Solution :`PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))`
At equilibrium 2mol2 mol2 mol
Total pressure = 30.3975 Kpa = 3 atm = P ( say )
`K _(p) = ( P_(PCl_(3)) xx P_(Cl_(2)))/(P_(Cl_(5))) = (( P)/( 3) xx ( P)/( 3))/((P)/( 3))`
`K_(p ) = ( P)/( 3) = 1`.....(1)
When CHLORINE is added to the system. The system will behave to nullify the effect and hence formation of `PCl_(5)` will be preferred.
Since P and T are CONSTANT are
`( V_(1))/( V_(2)) = ( n _(1))/( n _(2)) = n _(1) = 6" " V_(1) = V " " V_(2) = 2V `
`:. n _(2) = 12 ` moles

Say a moles of `Cl_(2)` were added
`PCl_(5) hArr PCl_(3) = Cl_(2)`
Initial 222
FInal2`+x`2-x`2+a-x`
`n_(T )- 12 = 6+a -x `
`K_(p) = ( (( 2+ax) - x)/( 12) xx ( 2-x)/( 12))/((2+x)/(12)) xxP=((2+a-x)(2-x))/(4( 2+x)) =1 `.....(2)
Solving equation ( 1) and ( 2) we get
`a = ( 20)/(3)` moles
Hence `( 20)/( 3)` moles of `Cl_(2)` was added
Discussion
90273.

A container with a volume of 20.0 L holds N_(2(g)) and H_(2)O_((l)) at 300 K and 1.0 atm. The liquid water is then decomposed completely into H_(2(g)) and O_(2(g)) by any means, at constant temperature, if the final pressure becomes 1.86 atm, what was the mass of water (jn gm) present initially. Neglect the initial volume of water: [Given : vapour pressure of water at 300 K=0.04 atm L-atm / K -mol]

Answer»

<P>

SOLUTION :`P_(gas)=1-0.04 =0.96 gm`
`{:(H_(2)O_((L))HARR,H_(2(g)),+,1/2 O_(2(g))),("x=0.5 mole",x,,x/2""=3/2 x =1.86 -0.96 =0.9),(=9gm,,,),(0.9xx20=3/2x xx24,,,x=0.6),(0.5=12/24 =x,,,):}`
Discussion
90274.

A contributions of both heat (enthalpy) and randomness(entropy) shall be considered to the overall spontaneity of process.When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change (DeltaG). DeltaG=DeltaH-TDeltaS where,DeltaH=Enthalpy change, DeltaS=Entropychange , T=Temperature in kelvin. If DeltaGlt0, Process is spontaneous , DeltaG=0, Process is at equilibrium , DeltaGgt0, Process is non-spontaneous. For the change H_2O(s),(273 K,2 atm)to H_2O(l),(273 K, 2 atm), choose the correct option.

Answer»

<P>`DeltaG=0`
`DeltaGlt0`
`DeltaGgt0`
None

Solution :`H_2O(s)toH_2O(L)`
At P =1 ATM, `DeltaG=0`, there is EQUILIBRIUM at 293 K
Hence, as `P uarr`, more `H_2O(l)` is formed.
Discussion
90275.

A container whose volume is V contains an equilibrium mixture that consists of 2 mol each or PCI_5, PCI_3 and CI_2 (all as gases) . The pressure is 30.3975 kpa and temperature is T. A certain amount of CI_2 (g) is now introduced keeping introduced keeping the pressure and temperature constant until the equilibrium volume is 2V. Calculate the amount of CI_2 that was added and the value of k_p.

Answer»


ANSWER :`(20)/3` MOLE
Discussion
90276.

A container has hydrogen and oxygen mixture in ratio of 4 : 1 by weight, then

Answer»

INTERNAL ENERGY of the mixture decreases
Internal energy of the mixture increases
Entropy of the mixture increases
Entropy of the mixture decreases

Solution :Entropy of the mixture increases DUE to increase in IMPURITY.
Discussion
90277.

A container has 3.2 g of a certain gas at NTP. What would be the mass of the same gas contained in the same vessel at 200^(@)C and 16 atm pressure.

Answer»


Solution :Mol. WT. `(M) = (3.2)/(V) xx 22.4`, V = vol. in li. Apply `pV = (w)/(M)RT` ,
w = wt. of the GAS in 473 K and 16 atm.
Discussion
90278.

A container contains 1 mole of a gas at 1 atm pressure and 27^@ C and its volume is 24.6 liters. If pressure is 10 atm and temperature 327^@ C then the new volume is approximately :

Answer»

2 liters
48 liters
10 liters
4.92 liters

Answer :D
Discussion
90279.

A container contains a certain gas of mass m at high pressure. A little amount of the gas has been allowed to escape from the container and after some time, the pressure of the gas becomes half and its absolute temperature two-third. The mass of the gas escaped is

Answer»

`(2M)/3`
`m/2`
`m/4`
`m/6`

ANSWER :C
Discussion
90280.

A constant current was flowing for 2 hours through a KI solution oxidizing iodide ions to iodine (2I^(-)toI_(2)+2e^(-)). At the end of the experiment, liberated iodine consumed 21.75 mL of 0.0831 M sodium thisulphate solution following the redox change I_(2)+2S_(2)O_(3)^(2-)to2I^(-)+S_(4)O_(6)^(2-). calculate the average rate of current flow.

Answer»

SOLUTION :21.75 mL of 0.831 M `Na_(2)S_(2)O_(3)=(0.0831)/(1000)xx21.75`mole
as 2 moles of `Na_(2)S_(2)O_(3)` react with 1 mole of `I_(2)`
`thereforeI_(2)` liberated`=(1)/(2)xx(0.0831)/(1000)xx21.75`mole
1 mole of `I_(2)` is liberated by 2F or `2xx96500`COULOMBS
`therefore(0.0831xx21.75)/(2000)` mole will be liberated by CHARGE
`=2xx96500xx(0.0831)/(2000)`coulombs=174.4 coulombs
`Q=Ixxt` or `I=(Q)/(t)=(174.4)/(2xx60xx60)=0.0242A`
Discussion
90281.

A constant current of 30 amperes ispassed through an aqueous solution of NaCl for 1 hour. How many grams of NaOH will be formed in the reaction ? Also find out the volume of Cl_(2) evolved at S.T.P.

Answer»

Solution :The redox reaction TAKING place on electrolysis is an follows :
`2Cl^(-)(aq) to Cl_(2)(g)+2e^(-)`
`(2H_(2)O(L)+2e^(-) to H_(2)(g)+2OH^(-)(aq))/(2Cl^(-)(aq)+2H_(2)O(l) to H_(2)(g)+Cl_(2)(g)+2OH^(-)(aq)`
Step I. Calculation of the mass of NaOH formed
Quantity of charge passed=(30 amp)xx(60xx60 s)=108000 Coulombs
Now, 96500 C of charge from NaOH=40 g
108000 C of charge will form NaOH `=(40g)xx((108000C))/((96500C))=44.77 g=1.12 mol`.
Step II. Calculation of VOLUME of `Cl_(2)` evolved at S.T.P.
According to the equation,
No. of moles of `Cl_(2)` evolved`=1//2xx"No"`. of mole of NaOH formed `=1//2xx1.12=0.56 mol`
1 mole of `Cl_(2)` at S.T.P. will correspond to volume =22.4L
0.56 mole of `Cl_(2)` at S.T.P. will correspond to valume`=((22.4L))/((1 mol))xx(0.56 mol)=12.54 L`
Discussion
90282.

A constant current of 1.50 amp is passed through an electrolytic cell containing 0.10 N solution of AgNO_(3) and a silver anode and a platinum cathode are used. After some tine, the concentration of the AgNO_(3) solution may be

Answer»

EQUAL to 0.10 M
less than 0.10 M
greater than 0.10 M
none of these

Answer :A
Discussion
90283.

A constant current flowed for 30 min throiugh a solution of Kl oxidising the iodide ion to ioidine At the end of experiment the iodine eas titrated with 10ml 0.075 M Na_(2)S_(2)O_(3) solution calculate the strength of current

Answer»

0.082 A
0.632 A
0.999 A
0.0402 A

Solution :0.0402 A
Discussion
90284.

A constant current flowed for 2 hours through a potassium iodide solution oxidising the iodide ion to iodine (2I^(-) rarr I_(2) + 2e^(-)). At the end of the experiment, the iodine was titrated with 21.75 mL of 0.0831 M soldium thiosulphate solution. (I_(2) + 2S_(2)O_(3)^(2-) rarr 2I^(-) +S_(4) O_(6)^(2-)) What was the average rate of current flow in apmeres ?

Answer»


ANSWER :0.0242 AMP
Discussion
90285.

A considerable part of the harmful ultraviolet radiation of the sun does not reach the surface of the earth. This is because high above the earth.s atmosphere there is a layer of:

Answer»

CARBON dioxide
Hydrogen
Ozone
Amonia

Answer :C
Discussion
90286.

A conjugated diene wil have two double bonds in :

Answer»

isolatedpositions
adjacentposition
alternatepositions
Noneof the above

ANSWER :C
Discussion
90287.

(A) Conductivity of silicon increases by doping it with group-15 elements. (R) Doping means introduction of small amount of impurities like P, As or Bi into the pure crystal.

Answer»

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 :B
Discussion
90288.

A conductivity cells has been calibrated with a 0.01 M 1:1 electrolyte solution (specific conductance,) K= (1.25 xx 10^(-3) S cm^(-1)) in the cell and the measured resistance was 800 ohms at 25^(@)C. The cell constant will be

Answer»

`1.02 cm^(-1)`
`0.102 cm^(-1)`
`1.00 cm^(-1)`
`0.5 cm^(-1)`

Solution :`K= 1.25 xx 10^(-3) S cm^(-1) : rho = 1/K = (1)/(1.25 xx 10^(-3))`
`R = rho (1)/(A), therefore 800= (1)/(1.25 xx 10^(-3)) xx (l/A)`,
Where `l/A =` Cell CONSTANT
Discussion
90289.

A conductivity cell having cell constant 8.76 cm^(-1) placed in 0.01 M solution of an electrolyte offered a resistance of 1000 ohms . What is the conductivity of the electrolyte ?

Answer»

`8.76 XX 10^(-4) ohm^(-1) cm^(-1)`
`8.76 xx 10^(-3) ohm^(-1) cm^(-1)`
`8.76 xx 10^(-2) ohm^(-1) cm^(-1)`
`8.76 xx 10^(-1) ohm^(-1) cm^(-1)`

Solution :`k = (1)/(R) xx` CELL const = `(1)/(1000) xx 8.76`
`= 8.76 xx 10^(-3) OMEGA^(-1) cm^(-1)`
Discussion
90290.

A conductivity cell is filled with a 0.02 M KCl solution which has a specific conductance of 2.768xx10^(-3)"ohm"^(-1) cm^(-1). If its resistance is 82.4 ohm at 25^(@), the cell constant is :

Answer»

`0.6182cm^(-1)`
`0.2281cm^(-1)`
`0.4821cm^(-1)`
`0.5381cm^(-1)`

ANSWER :B
Discussion
90291.

A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the cross sectional area of each electrode is 4.5 sq. cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15 Omega. Find the specific conductance of the solution.

Answer»

Solution :`k = 1/R(L/A)`
`k = 1/(15 omega) xx (1.5 xx 10^(-2) m)/(4.5 xx 10^(-4)m^2) = 2.22 SM^(-1)`
`l = 1.5 CM = 1.5 xx 10^2 m`
`A = 4.5 cm^2 xx (10^(-4)) m^2`
`R = 15 Omega` .
Discussion
90292.

A conductivity cell has been callibrated with a 0.01 M 1 : 1 electropyte solution (specific conductance, k = 1.25 xx 10^(-3) S cm^(-1) ) in the cell and the measured resistance was 800 ohms at 25^@C. The cell constant will be :

Answer»

`1.02 CM^(-1)`
`0.102 cm^(-1)`
`1.00 cm^(-1)`
`0.5 cm^(-1)`

ANSWER :C
Discussion
90293.

A conductivity cell has been calibrited with 0.01 M electrolyte solution (k=1.25xx10^(-3)" S "cm^(-1)) in the cell and the measured resistance is 800 ohms at 25^(@)C. The cell constant will be :

Answer»

`1.02 CM^(-1)`
`0.102 cm^(-1)`
`1.00 cm^(-1)`
`0.5 cm^(-1)`

Solution :(c ) Specific CONDUCTANCE (k) `=(1)/(R )xx(1)/(a)`
`(L)/(a)=kxxR`
`=(1.25xx10^(-3)" OHM"^(-1)cm^(-1))xx(800" ohm")`
`=1.0 cm^(-1)`
Discussion
90294.

A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (kappa=1.25 times 10^(-3)" "S" "cm^(-1)) in the cell and the measured resistance was 800Omega at 25^(@)C. The cell constant is,

Answer»

`10^(-1)CM^(-1)`
`10^(1)cm^(-1)`
`1 cm^(-1)`
`5.7 TIMES 10^(-12)`

Answer :C
Discussion
90295.

A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (k = 1.25 xx 10^(-3)S cm^(-1)) in the cell and the measured resistance was 800 Omega at 25^@C. The cell constant is,

Answer»

`10^(-1) c m^(-1)`
`10^1 cm^(-1)`
`1 cm^(-1)`
`5.7 XX 10^(-12)`

SOLUTION :`R = rho.l/A`
Cell constant = `R/rho = k.R(1/(rho) = k) = 1.25 xx 10^(-3) mu^(-1)cm^(-1) xx 800 Omega = 1 cm^(-1)`.
Discussion
90296.

A conductivity cell has been calibrated with a 0.01 M 1:1 electrolyte solution (specific conductance, k=1.25xx10^(-3)S" "cm^(-1)) in the cell and the measured resistance was 800 ohms at 25^(@)C. The constant will be

Answer»

1.02 cm
0.102 `cm^(-1)`
1.00 `cm^(-1)`
0.5 `cm^(-1)`

Solution :`K=1.25xx10^(-3)S" "cm^(-1):rho=(1)/(K)=(1)/(1.25xx10^(-3))`
`R=rho(l)/(A)`
`therefore800=(1)/(1.25xx10^(-3))XX((l)/(A))`, where `(l)/(A)=`cell CONSTANT
`(l)/(A)=800xx1.25xx10^(-3)=1`.
Discussion
90297.

A conductivitycell findwith 0.1 M KCIgivesat 25^(@) C aresistance of 85.5 other . Theconductivityof 0.1 M KCIat 25^(@) C is0.01286 "ohm"^(-1) cm^(-1) . Thesamecell filledwith 0.005 M HCIgivesa resistanceof 529ohms.What isthe molarconductivityof HCIsolutionat 25^(@)C?

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Solution :Given :Resistance Of KCIsolution= `R_(KCI) = 85.5 Omega`
Conductivityof KCIsolution`= k_(KCI) = 0.01286 ohm^(-1)CM^(-4)`
Concentration = C=0.005 M HCI
Resistanceof HCIsolution = `R _(soln) =529OHMS `
Molarconductivity of HCI`= ^^_(m(HCI)) = ? `
CONDUCTIVITY `= (" Cell constant ")/("Resistance ")`
`k_(KCI) = ( b) /(R_(KCI))`
`:. b =(k_(KCI))xx R_(KCI) = 0.01286 xx 85.5 = 1.1 cm^(-1)`
Conductivityof HCIsolution
`k_("soln")= ( b )/(R_("soln"))= (1.1)/(529)= 2.08 xx 10^(-3)ohm^(-1) cm^(-1)`
Molarconductivity`^^_(m_(HCI)) = .(k_("soln")xx 1000)/(C )`
`=(2.08 xx 10^(-3) xx 1000)/(0.005)`
`= 416 " ohm"^(-1) cm^(2)mol^(-1)`
Discussion
90298.

A conductivitycell filledwith 0.02 M AgNO_(3) Givesat 25^(@) Cresistance of 947ohms . Ifthe cellconstantis 2.3 cm^(-1)whatis the molarconductivityof 0.02 MAgNO_(4)at 25^(@) C ?

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SOLUTION :Molar CONDUCTIVITY`= ^^_(m) = 121 .5 Omega^(-1) CM^(2)mol^ (-1)`
Discussion
90299.

A conductivitycell filled with0.02 M AgNO_(3) gives at 25^(@) Cresistance of 947 ohms. If thecellconstant is 2.3 cm^(-1) . Whatis themolarconductivity of0.02M AgNO_(2)" at"25^(@)C

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Solution :Given :Concetration=C =0.02 M `AgNO_(2)`
Resistanceof solution= `R_("SOLN") = 947 Omega`
Cellconstant`= b =2.3 cm^(-1)`
MOLARCONDUCTIVITY= `^^_(m) = ?`
Conductivityof soln `= (" cellconstant ")/(R_("soln "))= ( b)/(R_("soln ")) = (23)/(947)`
`= 0.002429 Omega^(-1)cm^(-1)`
`"Molarconductivity" ^^_(m) = (KXX 1000)/( C) = (0.002429 xx 1000)/(0.02)`
`= 121. 5 Omega^(-1)cm^(2) mol^(-1)`
Discussion
90300.

A conductivitycell filledwith0.01 M KC1givesat 25^(@) Ca resistance of484ohms.The conductivityof 0.01 M KC1at 25^(@) Cis 0.00141 Omega^(-1)cm^(-1) 0.001 Mof NaC1 solutionwhen filledin thesame cellgivesa resistance of5490 ohmsat 25^(@) C Calculatethe molarconductivityof NaC1solution .

Answer»

SOLUTION :`^^_(m) =124 .3 OMEGA^(-1)cm^(2)MOL^(-1)`
Discussion