Explore topic-wise InterviewSolutions in Current Affairs.

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1.

The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below : {:("Concentration/M", 0.001, 0.010, 0.020, 0.050, 0.100),(10^(2) xx k//S m^(-1), 1.237, 11.85, 23.15, 55.53, 106.74):} Calculate Lambda_(m) for all concentration and draw a plot between Lambda_(m) and C^(1//2). Find the values of Lambda_(m)^(@).

Answer»

Solution :`1 S cm^(-1) = 100 S m^(-1)`.

The graph is EXTRAPOLATED to MEET Y-axis. It meets the Y-axis at 124.05.
`Lambda_(m)^(@)`= Intercept on Y-axis = 124.05 S `cm^(2) mol^(-1)`
Discussion
2.

The conductivity of sodium chloride at 298K has been determined at different concentrations and the results are given below: Calculate Lamda_(m) for all concentration and draw a plot between Lamda_(m) and c^((1)/(2)). Find the value of Lamda_(m)^(@).

Answer»

Solution :* Here, the value of given concentration and concentration and conductivity and the values of `c^((1)/(2))` and `Lamda_(m)` re given in the following table.

* Plotting the graph, from the calculation of above table `c^((1)/(2))` is on X-axis and its respective value of `Lamda_(m)` on the Y-axis, we will get following graph.

* This graph having negative slopping linear line.
Value of `Lamda_(m)^(0)" S "m^(2)MOL^(-1)`:
Value of `c^((1)/(2))=0` which is the value of `Lamda_(m)^(0)`
`k=1,237xx10^(-1)S " "m^(-1),c=0.001M`
Then, `k=1.237xx10^(-4)" S "cm^(-1),c^((1)/(2))=0.0316M^((1)/(2))`
`therefore Lamda_(m)=(k)/(c)=(1.237xx10^(-2)" S "cm^(-1))/(0.001" mol "L^(-1))XX(1000cm^(3))/(L)`
`=123.7" S "cm^(2)mol^(-1)` Given,
`k=11.85xx10^(-2)" S "m^(-1),c=0.010M`
Then, `k=11.85xx10^(-4)" S "cm^(-1),c^((1)/(2))=0.01M^((1)/(2))`
`=118.5" S "cm^(2)mol^(-1)` Given,
`k=23.15xx10^(-2)" S "m^(-1),c=0.020M`
Then, `k=23.15xx10^(-4)" S "cm^(-1),c^((1)/(2))=0.1414M^((1)/(2))`
`therefore Lamda_(m)=(k)/(c)=(23.15xx10^(-4)" S "cm^(-1))/(0.020" mol "L^(-1))xx(1000cm^(3))/(L)`
`=115.8" S "cm^(2)mol^(-1)` Given,
`k=55.53xx10^(-2)" S "m^(-1),c=0.050M`
Then, `k=55.53xx10^(-4)" S "cm^(-1),c^((1)/(2))=0.2236M^((1)/(2))`
`therefore k=(k)/(c)=(55.53xx10^(-4)" S "cm^(-1))/(0.050" mol "L^(-1))xx(1000cm^(3))/(L)`
`=111.1" S "cm^(2)mol^(-1)` Given,
`k=106.74xx10^(-4)" S "m^(-1),c=0.3162M`
`therefore Lamda_(m)=(k)/(c)=(106.74xx10^(-4)" S "cm^(-1))/(0.100" mol "L^(-1))xx(1000cm^(3))/(L)`
`=106.74" S "cm^(2)mol^(-1)`
Now, we have the following data:

Since the line interrupts `Lamda_(m)` at `124.0" S "cm^(2)`
`mol^(-2),Lamda_(m)^(@)=124.5" S "cm^(2)mol^(-1)`.
Discussion
3.

The conductivity of NaCl at 298K has been determined at different concentrations and the results are given below: {:("Concentration/M",0.001,0.010,0.020,0.050,0.100),(10^(2)xxkappa//S" "m^(-1),1.237,11.85,23.15,55.53,106.74):} Calculate wedge for all concentrations and draw a plot between wedge and c^(1//2). find the value of wedge^(8).

Answer»

Solution :1 S `cm^(-1)=100" S "m^(-1)`
`(1" S "cm^(-1))/(100" S "cm^(-1))=1` (UNIT conversion FACTOR)


`(AA)^(@)=`intercept on the `(AA)_(m)` axis=124.0 S `cm^(2)mol^(-1)` (on extrapotation to zero CONCENTRATION).
Discussion
4.

The conductivity of pure water in a conductivity cell with electrodes of cross sectional area 4 cm^2 and 2 cm apart is 8xx10^(-7)S cm^(-1) What current would flow through the cell under an applied potential difference of 1 volt ?

Answer»


ANSWER :A
Discussion
5.

The conductivity of pure water in a conductivity cell with electrodes of cross sectional area 4 cm^2 and 2 cm apart is 8xx10^(-7)S cm^(-1) What is resistance of conductivity cell ?

Answer»


ANSWER :A::B
Discussion
6.

The conductivity of saturated solution of AgCl is found to be 1.86 xx 10^(-6) ohm^(-1)cm^(-1) and that of water is 6xx 10^(-8) ohm^(-1)cm^(-1). If lamda^0 AgCl is

Answer»

`1.3 xx 10^(-5)`
`1.69 xx 10^(-10)`
`2 xx 10^(-10)`
`2.7 xx 10^(-10)`

ANSWER :B
Discussion
7.

The conductivity of ____________ semiconductor is increased by adding an appropriate amount of suitable ____________________.

Answer»

SOLUTION :IMPURITY
Discussion
8.

The conductivity of metals increases with:

Answer»

INCREASE in temperature
decrease in temperature
Nochange ovserved
Increase then decrease

Answer :B
Discussion
9.

The conductivity of intrinsic semiconductors can be increased by adding a suitable impurity. This process is called __(P)__. This can be done with an impurity which is __(Q)_ rich or deficient as compared to the semiconductor. Such impurities introduce __(R)__ defects in them. Electron rich impurities result in __(S)__ type semiconductors while electron deficit impurities result in __(T)___ type semiconductors .

Answer»

<P>P-doping , Q-proton , R-point , S- p , T-n
P-doping , Q-electron , R-non-stoichiometric, S- p , T-n
P-energy GAP , Q- CHARGED , R-impurity , S-n , T-p
P-doping , Q-electron , R-electronic , S-n , T-p

ANSWER :D
Discussion
10.

The conductivity of a solution containing 1.0 g of anhydrous BaCl, in 200 cm of the solution has been found to 0.0058 S cm^(-1). Calculate the molar conductivity and equivalent conductivity of the solution.

Answer»


SOLUTION :Molarity of solution = `(1//208 xx 1000)/(200) = 0.0240 M`
`Lambda_(m) = (0.0058 xx 1000)/(0.0240) = 241.67 OHM^(-1) cm^(2) MOL^(-1)`
NORMALITY of solution = `(1//104 xx 1000)/(200) = 0.0480 N`
`Lambda = (0.0058 xx 1000)/(0.0480) = 120.83 ohm^(-1) cm^(2) "EQUIV"^(-1)`
Discussion
11.

The conductivity of a solution containing 1 gram of anhydrous BaCl_(2) in 200 cm^(3) of water has been found to be 0.0058 S cm^(-1). What are the molar conductivity and equivalent conductivity of the solution? (At. Wt. of Ba=137 and Cl=35.5)

Answer»

Solution :We are given: Conductivity `(kappa)=0.0058" S "cm^(-1)`
Molar MASS of `BaCl_(2)=137+2xx35.5=208" G "mol^(-1)`
As 1 gram of `BaCl_(2)` is present in 200 `cm^(3)` of the solution, therefore, molar concentration (C)
`=(1)/(208)xx(1)/(200)xx1000" mol "L^(-1)=0.0240" mol "L^(-1)`
`therefore`Molar conductivity, `wedge_(m)=(kappaxx1000)/(c_(m))=(0.0058" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.0240" mol "L^(-1))=241.67" S "cm^(2)mol^(-1)`
Further in case of `BaCl_(2)`, equivalent weight`=("Mol. wt")/(2)=(208)/(2)=104`
`therefore` Concentration of the solution in gram equivalent per litre `(c_(EQ))=(1)/(104)xx(1)/(200)xx1000=0.0480`
`therefore`Equivalent conductivity, `wedge_(eq)=(kappaxx1000)/(c_(eq))=(0.0058" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.0480" g "eqL^(-1))=120.83" S "cm^(2)eq^(-1)`.
Discussion
12.

The conductivity of a saturated solution of BaSO_(4) at 298 K , is found to be 3.758 xx 10^(-6) ohm^(-1) cm^(-1) and that of water used is 1.36 xx 10^(-6) ohm^(-1) cm^(-1) . Molar ionic conduntances at infinite dilution for Ba^(2+) and SO_(4)^(2-) ions are 110 and 136.6 ohm^(-1) cm^(2) mol^-1 respectively . Calculate the solubility of BaSO_(4) at 295 K (Atomic masses : Ba = 137 , S = 32 , O = 16)

Answer»


SOLUTION :`kappa (BaSO_(4)) = kappa ` (solution ) - `kappa` (water)
`= 3.758 xx 10^(-6) - 1.36 xx 10^(-6)`
`= 2.398 xx 10^(-6) ohm^(-1) cm^(-1)`
`Lambda_(m)^(@) = 110 + 136.6 = 246.6 ohm^(-1) cm^(2) mol^(-1)`
Molar MASS = `137 + 32 + 4 xx 16 = 233`
Solubility = `(kappa xx 1000)/(Lambda_(m))`
`(2.398 xx 10^(-6) xx 1000)/(246.6)`
`= 9.724 xx 10^(-6) mol L^(-1)`
or `= 9.724 xx 10^(-6) xx 233 = 2.266 xx 10^(-3) G L^(-1)`
Discussion
13.

The conductivity of a saturated solution of AgCl at 288 K is found to be 1.382xx10^(-6)" S "cm^(-1). Find its solubility. Given ionic conductances of Ag^(+) and Cl^(-) at infinite dilution are 61.9 S cm^(2)mol^(-1) and 76.3"S "cm^(2)mol^(-1). respectively

Answer»

SOLUTION :`wedge_(m)^(@)(AgCl)=lamda_(ag^(+))^(@)+lamda_(Cl^(-))^(@)=61.9+76.3=138.2" S "cm^(2)mol^(-1)`
Solubility `=(kappaxx1000)/(wedge_(m)^(@))=(1.382xx10^(-6)xx1000)/(138.2)=10^(-5)" mol "L^(-1)=10^(-5)xx143.5" G "L^(-1)=1.435xx10^(-3)g" "L^(-1)`.
Discussion
14.

The conductivity of a saturated solutior of a sparingly soluble salt MX_2 is found to be 4 xx 10^(-5) Omega^(-1) cm^(-1)." If "lambda_(m)^(oo) (1/2 M^(2+))=50 Omega^(-1) cm^(2) mol^(-1) and lambda^(oo) (X^(-))=50 Omega^(-1) cm^(2) mole^(-1), the solubility product of the salt is about

Answer»

`2 xx 10^(-10)M^(3)`
`2 xx 10^(-12)M`
`8 xx 10^(-12)M`
`8 xx 10^(-14)M`

ANSWER :C
Discussion
15.

The conductivity of a metal depends upon

Answer»

nature and structure of the metal.
number of VALENCE ELECTRONS per ATOM,
temperature.
all the above.

Answer :D
Discussion
16.

The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 xx 10^(-4) S cm^(-1). Calculate the molar conductivity (Lambda_(m)) of the solution.

Answer»


ANSWER :16.5 S `CM^(2) MOL^(-1)`
Discussion
17.

The conductivity of 0.20M solution of KCl at 298K is 0.0248 S cm^(-1) . Calculate its molar conductivity.

Answer»

SOLUTION :`"MOLAR conductivity"=(kxx1000)/(M)=("0.0248 S cm"^(-1)XX"1000 cm"^(3)L^(-1))/("0.2 mol L"^(-1))`
`="124.0 S cm"^(2)"mol"^(-1)`
Discussion
18.

The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 xx 10^(-4 )S cm^(-1).Calculate molar conductivity (Lambda_(m))of the solution.

Answer»

SOLUTION :Use the relation,
`Lambda_(m) = (1000 k)/C`, where C stands for CONCENTRATION
SUBSTITUTING the values, we get
`Lambda_(m) = (1000 xx 1.65 xx 10^(-4))/0.01 = 16.5 S cm^(2) MOL^(-1)`
Discussion
19.

The conductivity of a 0.01 M solution of a 1:1 weak electrolyte at 298 K is 1.5 times 10^(-4)" S " cm^(-1). i) molar conductivity of the solution ii) degree of dissociation and the dissociation constant of the weak electrolyte Given that lambda_("cation")^(@)=248.2" S "cm^(2)mol^(-1) lambda_("anion")^(@)=51.8" S "cm^(2)mol^(-1)

Answer»

SOLUTION :Given
C = 0.01 M `""lambda_("cation")^(@)=248.2" S "cm^(2)mol^(-1)`
`K=1.5 times 10^(-4)" S " cm^(-1) " "lambda_("anion")^(@)=51.8" S "cm^(2)mol^(-1)`
(i) Molar conductivity
`wedge_(m)^(@)=(""^(@)(SM^(-1)) times 10^(-3))/(C("in M"))mol^(-1)m^(3)`
`" "^(@)=1.5 times 10^(-4)" S " cm^(-1)`
`""=(1.5 times 10^(2) times 10^(-3))/(0.01)S" "mol^(-1)m^(2)`
`1cm^(-1)=10^(2)m^(-1)`
`""=1.5 times 10^(-3)" S "m^(2)mol^(-1)`
`""=1.5 times 10^(2)`
(ii) Degree of DISSOCIATION `alpha=WEDGE^(@)/wedge_(infty)^(@)`
`wedge_(infty)^(@)=lambda_("cation")^(@)+lambda_("anion")^(@)`
`" "=(248.2+51.8)" S "cm^(2)mol^(-1)`
`" "=300" S " cm^(2)mol^(-1)`
`" "=300 times 10^(-14)" S "m^(2)mol^(-1)`
`alpha=(1.5 times 10^(-3)" S"m^(2)mol^(-1))/(300 times 10^(-4)Sm^(2)mol^(-1))`
`alpha=0.05`
Discussion
20.

The conductivity of a 0.01 M solution of a 1:1 weak electrolyte at 298 K is 1.5xx10^(-4)Scm^(-1). (i) molar conductivity of the solution (ii) degree of dissociation and the dissociation constant of the weak electrolyte Given that lambda_("cation")^(@)=248.2" S cm"^(2)" mol"^(-1) lambda_("anion")^(@)="51.8 S cm"^(2)" mol"^(-1)

Answer»

Solution :(i) Molar conductivity
`C=0.01M`
`kappa=1.5xx10^(-4)"S cm"^(-1)(or)kappa=1.5xx10^(-2)"S m"^(-1)`
`Lambda_(m)=(kappaxx10^(-3))/(C )"Sm"^(-1)"mol"^(-1)m^(3)=(1.5xx10^(-2)xx10^(-3))/(0.01)"Sm"^(2)"mol"^(-1)`
`Lambda_(m)=1.5xx10^(-3)"Sm"^(2)"mol"^(-1)`
(ii) Degree of dissociation
`ALPHA=(Lambda_(m))/(Lambda_(oo))(or)alpha=(Lambda_(m))/(Lambda_(m)^(@))`
`Lambda_(oo)^(@)=lambda_("cation")^(@)+lambda_("anion")^(@)`
`=(248.2+51.8)"S cm"^(2)"mol"^(-1)`
`="300 S cm"^(2)"mol"^(-1)`
`=300xx10^(-4)"S m"^(2)"mol"^(-1)`
`alpha=(1.5xx10^(-3))/(300xx10^(-4))=0.05`
`K_(a)=(alpha^(2)C)/(1-alpha)`
`=((0.05)^(2)(0.01))/(1-0.05)`
`K_(a)=2.63xx10^(-5)`
Discussion
21.

The conductivity of 10^(-3) mol/L acetic acid at 25 °C is 4.1 xx 10^(-5) cm^(-1) . Calculate its degree of dissociation,Lambda_(m)^(@) if for acetic acid at 25 °C is 390.5 S cm^(2) mol^(-1).

Answer»

Solution :Apply the following relation:
`Lambda_(m) = (1000 k)/C`
Substituting the values in the above EQUATION, we have
`Lambda_(m) =(1000 xx 4.1 xx 10^(-5))/10^(-3) = 41 S cm^(2) mol^(-1)`
Degree of dissociation is given by the following equation:
`alpha = Lambda_(m)^(C )/Lambda_(m)^(@) = 41/(390.5)= 0.105` or `10.5%`
Discussion
22.

The conductivity of 0.20M solution of KCl of 298K is 0.0248 S cm^-1. Calculate its molar conductivity.

Answer»

SOLUTION :`A_m=(K times 1000)/(MOLARITY)`
`=(0.0248 S cm^-1 times1000cm^3 L^-1)/(0.20 mol L^-1)`
`=124 S cm^2 mol^-1`
Discussion
23.

The conductivity of 0.20M solution of KCL at 298K is 0.025 S cm^-1. Calculate its molar conductivity.

Answer»

SOLUTION :CONCENTRATION of solution =0.20 M
Conductivity =`0.025 S cm^-1`
Molar conductivity `A_m`
`=("Conductivity (K) TIMES 1000")/("Concentration of solution ")`
`=(0.025 times 1000)/0.20`
`=125 S cm^2 mol^-1`
Discussion
24.

The conductivity of 0.20 M solution of KCl at 298K is 0.0248 S cm^(-1). Calculate its molar conductivity.

Answer»

SOLUTION :`wedge_(m)=(kappaxx1000)/("MOLARITY")=(0.0248" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.20" mol "L^(-1))=124" S "cm^(2)mol^(-1)`
Discussion
25.

The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm^(-1)?. Calculate its molar conductivity.

Answer»

Solution :Given that k = `0.025 S cm^(-1)`
`C = 0.2 "moles" L^(-1) = 0.2` moles `(1000 cm^(3))^(-1) =(0.2 "moles")/(1000 cm^(3))`
`Lambda_(m) = k/C = (0.025 S cm^(-1))/(0.2 "moles") XX 1000 cm^(3) = 125 S cm^(2) "mol"^(-1)`
Discussion
26.

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^(-1). Calculate its molar conductivity.

Answer»

SOLUTION :* Molar conductivity `Lamda_(m)`:
Molar conductivity `(Lamda_(m))=(kxx1000)/(c)`
Where, conductivity `k=0.0248" S "cm^(-1)`
`therefore Lamda_(m)=(kxx1000)/("Molarity")`
`=((0.0248" S "cm^(-1))xx(1000cm^(3)DM^(-3)))/(0.20" mol "dm^(-3))(OR(m)L^(-1))/(OR" mol "L^(-1))`
`=124.0S" "cm^(2)mol^(-1)`
Discussion
27.

The conductivity of 0.1 M solution of AgNO_3 is 9.47 xx 10^(-3) S cm^(-1) at 291 K . The ionic conductivities of Ag^(+) and NO_(3)^(-) at the same temperature are 55.7 and 50.8 S cm^(2) "equiv"^(-1)respectively . Calculate the degree of dissociation of AgNO_3 in 0.1 M solution .

Answer»


Solution :`Lambda = (KAPPA xx 1000)/(C) = (9.47 xx 10^(-3) xx 1000)/(0.1)`
`= 94.7 S CM^(2) "equiv"^(-1)`
`Lambda^(oo) = lambda^(oo) (Ag^(+)) + lambda^(oo) (NO_(3)^(-)) = 106.5 S cm^(2) "equiv"^(-1)`
`ALPHA = (94.7)/(106.5) = 0.089 or 89 %`
Discussion
28.

The conductivity of 0.2 M solution of KCl at 298 K is 0.0248 S cm^(-1). Calculate its mlar conductivity.

Answer»


ANSWER :`124Omega^(-1)CM^(2)MOL^(-1)`
Discussion
29.

The conductivity of 0.20 M solution of KCI at 298 K is 0.0248 S cm^(-1).Calculate its molar conductivity.

Answer»

Solution :Applying the FOLLOWING relation and substituting the values, we get:
`Lambda_(m) = (k xx 1000)/("MOLARITY") = (0.0248 S CM^(-1) xx 1000 cm^(3) L^(-1))/(0.20 mol L^(-1)) = 124 S cm^(2) mol^(-1)`.
Discussion
30.

The conductivity of 0.1 M KCl solution is 1.29 S m^(-1). If the resistance of the cell filled with 0.1 M KCl is 100 Omega, Calculate the cell constant.

Answer»

Solution :The cell constant filled with electrolytic solution is GIVEN as ,
Cell constant = conductivity `xx` RESISTANCE.
Cell constant `=1.29m^(-1)XX100 Omega`
`=129m^(-1)=1.29cm^(-1)`.
Discussion
31.

The conductivity of 0.01 mol L^(-1)KCl solution is 1.41xx10^(-3)" S "cm^(-1). What is the molar conductivity (S cm^(2) mol^(-1)) ?

Answer»

14.1
1.41
1410
141

Solution :(d) MOLAR conductivity`(Lambda_(m))=(1000xxk)/(M)`
`=(1000xx(1.41xx10^(-3)" S " cm^(-1)))/((0.01" MOL "cm^(-3)))=141" S "cm^(-2)mol^(-1)`
Discussion
32.

The conductivity of 0.001028 M acetic acid is 4.95xx10^(-5)" S "cm^(-1). Calculate its dissociation constant if wedge^(@) for acetic acid is 390.5 S cm^(2)mol^(-1).

Answer»

Solution :For the given concentration of acetic acid solution,
`wedge_(m)=(kappaxx1000)/(c)=(4.95xx10^(-5)S" "CM^(-1)xx1000cm^(3)L^(-1))/(0.001028" mol "L^(-1))=48.15" S "cm^(2)mol^(-1)`
`alpha=(wedge_(m))/(wedge_(m)^(@))=(48.15" S "cm^(2)mol^(-1))/(390.5" S "cm^(2)mol^(-1))=0.1233`
`{:(,CH_(3)COOH,HARR,CH_(3)COO^(-),+,H^(+)),("INITIAL conc.",c,,0,,0),("Equilibrium conc.",c-calpha=c(1-alpha),,calpha,,calpha):}`
`K=(calpha.calpha)/(c(1-alpha))=(calpha^(2))/(1-alpha)=((0.001028" "molL^(-1))(0.1233)^(2))/(1-0.1233)=1.78xx10^(-5)mol" "L^(-1)`
Discussion
33.

The conductivity of 0.01 M NaCl solution is 0.00147" ohm"^(-1)cm^(-1). What happens to the conductivity if extra 100 mL is added to the above solution.

Answer»

REMAINS same
First increases and then decreases
Increases
Decreases.

Solution :(d) Conductivity or SPECIFIC CONDUCTANCE (k)decreases with dilution
Discussion
34.

The conductivity of 0.001028 mol L^(-1)acetic acid is 4.95 xx 10^(-5) S cm^(-1). Calculate its dissociation constant if Lambda_(m)^(0) for acetic is 390.5 S cm^(2) mol^(-1) ?

Answer»

14. 1
141
1410
1.41

Answer :B
Discussion
35.

The conductivity of 0.001M acetic acid is 4 times 10^-8 S//cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm^2//mol.

Answer»

SOLUTION :`A_m=(K TIMES 1000)/m`
`=(4 times 10^-5 times 1000)/0.001`
`alpha=A_m/A_m^0=40/390=0.102`
Discussion
36.

The conductivity measurement of a coordination compound of cobalt (III) shows that it dissociates into 3 ions in solution. The compound is

Answer»

hexaamminecobalt (III) chloride
pentaamminesulphateocobalt (III) chloride
pentaamminechloridocobalt(III) sulphate
pentaamminechloridocobalt (III) chloride

Solution :Pentaamminechloridocobalt (III) chloride i.e
`[Co(NH_(3))_(5)CL]Cl_(2)hArr ubrace([Co(NH_(3))_(5)Cl]^(2+)+2CL^(-))_(3"ions")`
Discussion
37.

The conduction of electricity occurs by a direct low of electrons under the influence of applied potential are known as

Answer»

ELECTRO conductor
electronic conductor
electrolytic conductor
electrochemical conductor

Answer :B
Discussion
38.

The conductivity, K, of an electrolytic solution depends upon the concentration of electrolyte, nature of the solvent and _________

Answer»

SOLUTION :TEMPERATURE
Discussion
39.

The conductivity of 0.001 M acetic acid is 5 xx 10^(-5) S cm^(-1) and ^^""^(0) is 390.5 S cm^(2) "mol" ^(-1) then the calculated value of dissociation constant of acetic acid would be

Answer»

`81.78 XX 10^(-4)`
`81.78 xx 10^(-5)`
`18.78 xx 10^(-6)`
`18.78 xx 10^(-5)`

ANSWER :C
Discussion
40.

The conductance of a 0.0015M aqueous solution of a weak monobasic acid was determined by using a conductivity acid was determined by using a cunductivity cell consisting of platinised pt electrodes. The distance between the electrodes is 120cm with an area of croos section of 1 cm^(2). The conductance of this solution was found to be 5xx10^(-7) S. the pH of the solution is 4. The value of limiting moalr conductivity (A_(m)^(@)) of this weak monobasic acid in aqueous solution is Zxx10^(2)S cm^(-1) "mol"^(-1). the value of Z is

Answer»


ANSWER :`6XX10^(2)S CM^(-1) "MOL"^(-1)`
Discussion
41.

The conductance of a salt solution (AB) measured by two parallel electrodes of area 100 cm^(2) separated by 10 cm was found to be 0.0001 Omega^(-1). If volume enclosed between two electrode contains 0.1 mole of salt, and the molar conductivity (S cm^(2) mol^(-1)) of salt at same concentration is 1.0 xx 10^(-x), x is

Answer»


Solution :`C = 10^(-4)Omega , a = 100 cm^(2) , L =10 cm , R= P. (1)/(a) implies (1)/(C) = (1)/(K) xx (1)/(a)implies K = C xx (1)/(a) = 10^(-4) xx (10)/(100)= 10^(-5)`
` M = (N)/(V) = (0.1)/( 100 xx 100) xx 1000 = 0.1M , ^^_M = (K xx 1000)/(M) = (10^(-5) xx 1000)/(0.1) = 0.1 = 1 xx 10^(-x) = 1 xx 10^(-1)`
x=1
Discussion
42.

The conductance of an emulsion increase on adding common slat, What type of emulsion is this ?

Answer»

SOLUTION :Oil-in-water TYPE.
Discussion
43.

The conductance of solution of an electrolyte is equal to that of its specific conductance. The cell constant of the conductivity cell is equal to

Answer»

Resistance
Faraday
Zero
Unity

Answer :D
Discussion
44.

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of Pt electrodes. The distance between the electrodes is 120 cm with an area of cross-section of 1 cm^(2). The conductance of this solution was found to be 5xx10^(-7)S. the pH of the solution is 4. the value of the limiting molar conductivity (wedge_(m)^(@)) of the weak monobasic acid is aqueous solution is Zxx10^(2).S" "cm^(2)mol^(-1). the value of Z is

Answer»

Solution :CONC `C=0.0015M`, conductance, `G=5xx10^(-7)S`
`l=120cm,a=1cm^(2)`
`wedge_(m)^(c)=(Kxx1000)/("Molarity")=(6xx10^(-5)xx1000)/(0.0015)`
`=40" S "cm^(2)mol^(-1)`
`{:(HA,HARR,H^(+),+,A^(-)),(C,,0,,0),(C-Calpha,,Calpha,,Calpha):}`
`[H^(+)]=Calpha=10^(-4)M""(becausepH=4)`
`theeforealpha=(10^(-4))/(C)=(10^(-4))/(0.0015)`
`alpha=(wedge_(m)^(c))/(wedge_(m)^(@)) therefore(10^(-4))/(0.0015)=(40)/(wedge_(m)^(@))`
or `wedge_(m)^(@)=(40xx0.0015)/(10^(-4))=6xx10^(2)" S "cm^(2)mol^(-1)`
Discussion
45.

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm^(2). The conductance of this solution was found to be 5 xx 10^(-7)S. The pH of the solution is 4. The value of limiting molar conductivity (Lambda^(@)) of this monobasic acid in aqueous solution is Z xx 10^(2)S cm^(-1) mol^(-1). The value of Z is ......... .

Answer»


ANSWER :6
Discussion
46.

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm^(2). The conductance of this solution was found to be 5xx10^(-7)S. the pH of the solution is 4. The value of limiting molar conductivity (wedge_(m)^(0)) of this weak monobasic acid is aqueous solution is Zxx10^(2)S" "cm^(-1)mol^(-1). the value of Z is:-

Answer»

Solution :`C=0.0015M""l=120cm`
`G=5xx10^(-7)s""a=1cm^(2)`
`G=KXX(a)/(l)`
`5xx10^(-7)=Kxx(1)/(120)`
`K=6xx10^(-5)S" "cm^-1`
`(^^)_(m)^(c)=(Kxx1000)/(M)=(6xx10^(-5)xx1000)/(0.0015)`
`pH=4implies[H^(+)]=10^(-4)=calpha=0.0015alpha`
`alpha=(10^(-4))/(0.0015)""alpha=((^^)_(m)^(c))/((^^)_(m)^(0))impleis(10^(-4))/(0.0015)=((6xx10^(-5)xx1000)/(0.0015))/((^^)_(m)^(0))`
`(^^)_(m)^(0)=6xx10^(2)s" "cm^(2)"mole"-1`
Discussion
47.

The conductance of a 0.0015 M aqueous solution of a week monobasic acid was determind by using a conductivity cell consisting of plantnized Pt electrodes. The distance between the electrods is 120 cm with an area of cross section was found to be5xx10^-7S. The pH of the solution is 4. The value of limiting molar conductivity(Lambda_m^0) of the weak monobasic acid is aqueous solution is Zxx10^2S cm^(-1). The value of Z is

Answer»


ANSWER :6
Discussion
48.

The conductance of 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cellconsisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm^(2). The conductance of this solution was found to be 5xx10^(-7)" S ". The pH of the solution is 4. The value of limiting molar conductivity Lambda_(m)^(0) ofweak monobasic acid in aqueous solution is Zxx10^(2)" S "cm^(-1). The value of Z is

Answer»

SOLUTION :z=6
For solution, CONSULT SOLVED EXAMPLE 3.68
Discussion
49.

The conditions that are necessary in the preparation of aryl halides from arenes

Answer»

LOW temperature
Absence of sunlight
Presence of HALOGEN carrier
all of the above

Answer :D
Discussion
50.

The condition to obtain standard emf is

Answer»

1M solution of REACTANTS and PRODUCTS
`25^(@)C`
both (a) and (b)
neither (a) and (b)

ANSWER :C
Discussion