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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1951. |
`HNO_3`(aq) is titrated with NaOH(aq) condutomatrically, graphical representation of the titration is :A. B. C. D. |
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Answer» Correct Answer - A |
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| 1952. |
Pick out the incorrect statementA. Equivalent conductance inc rease with dilutionB. Molar conductance increase with dilutionC. Specific conductance increase with dilutionD. Specific resistance increase with dilution |
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Answer» Correct Answer - C Equivalent and molar conductance increases with dilution. Specific conductance decreases with dilution. Reciprocal of specific conductance i.e., specific resistance increases with dilution. |
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| 1953. |
Which of the following is arranged in increasing order of ionic mobility?A. `I^(-)ltBr^(-)ltCl^(-)ltF^(-)`B. `F^(-)ltCl^(-)ltBr^(-)ltI^(-)`C. `F^(-)ltI^(-)ltCl^(-)ltBr^(-)`D. `F^(-)ltCl^(-)ltI^(-)ltBr^(-)` |
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Answer» Correct Answer - B |
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| 1954. |
A graph was plotted between molar conductivity of various electrolytes (NaCl, HCl and `NH_4OH`) and `sqrtC` (in mol`L^(-1)`) . Correct set is : A. I(NaCl),II(HCl) , III(`NH_4OH`)B. I(HCl),II(NaCl) , III(`NH_4OH`)C. I(`NH_4OH`), II(NaCl), III(HCl)D. I(`NH_4OH`), II(HCl) , III(NaCl) |
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Answer» Correct Answer - B |
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| 1955. |
A graph was plotted between the molar conductance of various electrolytes `(HCI, KCI` and `CH_(3)COOH)` and root of their concentrations in molar per litre. Which of the following is correct match?A. `I (CH_(3)COOH), II (KCI), II(HCI)`B. `I(HCI), II(KCI), II (CH_(3)COOH)`C. `I(CH_(3)COOH), II (HCI), II (KCI)`D. `I(KCI), II(CH_(3)COOH), II (HCI)` |
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Answer» Correct Answer - B Molar conductance of `HCI` will be maximum and `CH_(3)COOH` will be minimum because (i) `[H^(+)]` have expectional ionic mobility (ii) `CH_(3)COOH` is a weak electrolyte |
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| 1956. |
Compute the standard free energy chagne `(DeltaG^(@))` for the process `Zn^(2+)(aq)+2e^(-)rarrZn(s)`: `E^(@)Zn^(2+)=-0.76V`A. 146.68kJB. 73.34kJC. 220.2kJD. 1100kJ. |
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Answer» Correct Answer - A `E_("Cell")^(@)=E_("red")^(@)E_("red")^(@)` `=E_(Ag)^(@)+//Ag-E^(@)Cu^(2+)//Cu` `0.8-(+0.34)=+0.46V` |
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| 1957. |
Calculate the degree of dissociation of 0.02 M acetic acid at 298K, given that `lambda_(CH_(3)COOH)=11.7ohm^(-1)cm^(2)"mol"^(-1)` `lambda_((CH_(3)COO^(-)))=40.9ohm^(-1)cm^(2)"mol"^(-1)` `lambda(H^(+))^(@)=349.1ohm^(-1)cm^(2)"mol"^(-1)`A. 0.06B. 0.015C. 0.03D. 0.09. |
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Answer» Correct Answer - C `DeltaG^(@)=-nFE^(@)=-2xx96500xx(-0.76)` `=146680J=146.68kJ`. |
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| 1958. |
Calculate the degree of dissociation `(alpha) of CH_(3)COOH at 298K.` ltbr Given that `^^_(CH_(3)COOH)^(oo)=11.75cm^(2)mol^(-1)` `^^_(CM_(3)COO^(-))^(oo)=40.65cm^(2) mol ^(-1)` `^^_(H^(+))^(0)=349.15 cm^(2)mol ^(-1)` |
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Answer» Given that `^^_(CH_(3)COOH)^(oo) =11.75 cm ^(2) mol ^(-1)` `^^_(CH_(3)COO^(-))^(oo) =40.95 cm^(2) mol ^(-1)` `^^_(H)^(oo) =349.15 cm^(2)mol ^(-1)` `^^_(CH_(3)COOH)^(oo) =^^_(CH_(3) COO^(-))^(oo) +^^_(H^(+))^(oo)=40.95+349.15=390.1` Degree of dissociation `(alpha) =(^^(alpha))/(lamda^(oo))=(11.75)/(390.1)=0.30=3xx10^(-2)` `therefore alpha=3xx10^(-2)` |
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| 1959. |
Compouter the m olar conducitivity of a solution of `MgCl_(2)` at infinite dilution. Given that `lambda_(Mg)^(2+)=106.12ohm^(-1)cm^(2)"mol"^(-1)` `lambda _(Mg)^(2+)=106.12ohm^(-1)cm^(2)"mol"^(-1)`A. 182.46`ohm^(-1)cm^(2)"mol"^(-1)`B. `258.8ohm^(-1)cm^(2)"mol"^(-1)`C. `212.24ohm^(-1)cm^(2)"mol"^(-1)`D. `152.68ohm^(-1)cm^(2)"mol"^(-1)` |
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Answer» Correct Answer - B `426.1=lambda_((H^(+))^(0))+lambda_((CH_(3)COO^(-)))^(0)` .......(ii) `126.5 =lambda_((Na^(+)))^(0)+lambda_((Cl^(-))^(0)` Adding eqn. (i) and (ii) and sub tracting eqn. iii we get `lambda_((H^(+))^(0))+lambda_((CH_(3)COO^(-)))^(0)=426.1+91.0-126.5` `=390.6ohm^(-1)cm^(2)"mol"^(-1)` |
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| 1960. |
At infiinite dilution, the aqueous solution of `BaCl_(2)`. M olar conductivity of `Ba^(2+)` and `Cl^(-)` ions are =127.32 S `cm^(2)` /mol and 76.34S `cm^(2)`/mol respectively what is `^^_(m)^(@) ` for `BaCl_(2)` at same dilution?A. `280Scm^(2)"mol"^(-1)`B. `330.98S cm^(2)"mol"^(-1)`C. `90.98S cm^(2)"mol"^(-1)`D. `203.6S cm^(2) "mol"^(-1)` |
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Answer» Correct Answer - A `^^_(m)^(oo)=^^_(Ba^(2+))^(oo)+2lambda_(Cl^(-))^(oo)` `=127.32+2xx76.34` `=280S cm^(2) "mol"^(-1)` |
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| 1961. |
Calculate degree of dissociation of 0.02 M acetic acid at 298 K given that `mho_(m)(CH_(3)COOH)=17.37 cm^(2) mol^(-1),lambda_(m)^(@)(H+)=345.8 S cm^(2) mol^(-1), lambda_(m)^(@)(CH_(3)COO^(-))=40.2 Scm^(2) mol^(-1)` |
| Answer» Correct Answer - 0.045 | |
| 1962. |
Which of the following is anodic reaction.A. `SO_(4)^(2-)+H_(2)O rarr H_(2)SO_(4)+1//2O_(2)+2e^(-)`B. `H^(o+)+e^(-)rarr 1//2OH_(2)`C. `Ag^(o+)+e^(-) rarr Ag`D. None of these |
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Answer» Correct Answer - a At anode, oxidation occurs. So oxidatin reaction is only is `(a)` |
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| 1963. |
During electrolysis of fused calcium hydride, the hydrogen is produced atA. CathodeB. AnodeC. Hydrogen is not liberated at allD. `H_(2)` produced reacts with oxygen to form water. |
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Answer» Correct Answer - b `CaH_(2)overset(El ectrolysis)rarrCa^(2+)+2H^(c-)` At cathode `(` reduction `):Ca^(2+)+2e^(-) rarr Ca` At anode `(` oxidation`):2H^(c-) rarr H_(2)+2e^(-)` |
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| 1964. |
During the electrolysis of an aqueous salt solution, the `pH` in the space near one of the electrode was increased and the other one was decreased. The salt solution was:A. `Cu(NO_(3))_(2)`B. `ZnCl_(2)`C. `NaCl_(conc.)`D. `NaCl` very dilute |
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Answer» Correct Answer - D Anode: `2H^(+) + 2e rarr H_(2) pH` increases Cathode: `2OH^(-) rarr H_(2)O + 1//2O_(2) + 2e pH` decreases. |
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| 1965. |
`0.04 N` solution of a weak acid has specific conductance `4.23 xx 10^(-4) "mho cm"^(-1)`. If the degree of dissociation of acid at this dilution is `0.0612`, then equivalent conductivity at infinite dilution is :A. `150.8 "mho cm"^(2) eq^(-1)`B. `172.8 "mho cm"^(2) eq^(-1)`C. `180.6 "mho cm"^(2) eq^(-1)`D. `160.9 "mho cm"^(2) eq^(-1)` |
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Answer» Correct Answer - B `Lambda_(v) = kappa xx (1000)/(N)` `= 4.23 xx 10^(-4) xx (1000)/(0.04) = 10.575` Now `alpha = (Lambda_(v))/(Lambda_(oo))` `:. (Lambda_(oo)) = (10.575)/(0.0612) = 172.79` |
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| 1966. |
If the molar conductance values of `Ca^(2+)` and `Cl^(-)` at infinite dilution are respectively `118.88xx10^(-4) m^(2)` mho `mol^(-1)` and `77.33xx10^(-4) m^(2)` mho `mol^(-1)` then that of `CaCl_(2)` is : (in `m^(2)` mho `mol^(-1)`)A. `118.88xx10^(-4)`B. `154.66xx10^(-4)`C. `273.54xx10^(-4)`D. `196.21xx10^(-4)` |
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Answer» Correct Answer - C `Lambda_(m)^(@) CaCl_(2) =lambda^(@)Ca^(2+)+2lambda^(@)Cl^(-)` `=(118.88xx10^(-4))+2(77.33xx10^(-4))` `=273.54xx10^(4) m^(2)` mho `mol^(-1)` |
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| 1967. |
Two solutions of X and Y electrolytes are taken in two beakers and diluted by adding 500mL of water. `Lambda_m` of X increases by 1.5 times while that of Y increases by 20 times, what could be the eletrolytes X and Y?A. X `to` NaCl , Y `to` KClB. X `to` NaCl , Y `to CH_3COOH`C. X `to` KOH, Y `to` NaOHD. X `to CH_3COOH` , Y `to` NaCl |
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Answer» Correct Answer - B Electrolyte X is strong electrolyte as on dilution the number of ions remain same , only interionic attraction decreases and hence not much increase in `Lambda_m` is seen . While `Lambda_m` for a week electrolyte increases significantly |
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| 1968. |
Limiting molar conductivity for some ions is given below (in S `cm^(2) mol^(-1)` ) : `Na^+ -50.1 , Cl^(-) - 76.3 , H^(+) - 349.6 , CH_3COO^(-) -40.9 , Ca^(2+) -119.0` What will be the limiting molar conductivities `(Lambda_m^@)` of `CaCl_2, CH_3COONa ` and NaCl respectively ?A. 97.65 , 111.0 and 242.8 S `"cm"^(2) "mol"^(-1)`B. 195.3, 182.0 and 26.2 S `"cm"^(2) "mol"^(-1)`C. 271.6, 91.0 and 126.4 S `"cm"^(2) "mol"^(-1)`D. 119.0, 1024.5 and 9.2 S `"cm"^(2) "mol"^(-1)` |
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Answer» Correct Answer - C `Lambda_(m" " CaCl_2)^@=lambda_(Ca^(2+))^@ + 2lambda_(Cl^-)^@` =119.0 + 2 x 76.3 = 271.6 S `cm^2 mol^(-1)` `Lambda_(m " " CH_3COONa)^@=Lambda_(CH_3COO^-)^@+lambda_(Na^+)^@` =40.9+50.1=91 S `cm^2 mol^(-1)` `Lambda_(m " " NaCl)^@=lambda_(Na^+)^@ + lambda_(Cl^-)^@` = 50.1 + 76.3 = 126.4 S `cm^2 mol^(-1)` |
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| 1969. |
At `18^(@)C`, the conductance of `H^(+)` ions and `CH_(3)COO^(-)` ions at infinite dilution are 315 and 35 mho `cm^(2) eq^(-1)` respectively. The equivalent conductance of `CH_(3)COOH` at infinite dilution is (`mho cm^(2) " equiv"^(-1)`)A. 280B. 350C. 30D. j315 |
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Answer» Correct Answer - B (b) `Lambda_(CH_(3)COOH)^(oo)=lambda_(CH_(3)COO^(-))^(oo)+lambda_(H^(+))^(oo)=35+315` `=350" mho "cm^(2)"equiv"^(-1)`. |
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| 1970. |
What will be the molar conductivity of Al 3+ ions at infinite dilution if molar conductivity of `Al^(2) (SO_(4))_(3)`is 858 S `cm^(2)` `"mol" ^(-1)` and ionic conductance of `SO_(4)^(2-)` is 160 S `cm^(2) "mol" ^(-1)` at infinite dilution ?A. 189 S `"cm"^2 "mol"^(-1)`B. 698 S `"cm"^2 "mol"^(-1)`C. 1018 S `"cm"^2 "mol"^(-1)`D. 429 S `"cm"^2 "mol"^(-1)` |
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Answer» Correct Answer - A `Lambda_(Al_2(SO_4)_3)^@=2Lambda_(Al^(3+))^@+3Lambda_(SO_4^(2-))^@` `Lambda_(Al^(3+))^@=(Lambda_(Al_2(SO_4)_3)^@-3Lambda_(SO_4^(2-))^@)/2` `=(858-(3xx160))/2=189 S cm^2 mol^(-1)` |
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| 1971. |
Resistance of a decimolar solution between two electrodes 0.02 meter apart and 0.0004 `m^2` in area was found to be 50 ohm. Specific conductance (k0 is :A. `0.1Sm^(-1)`B. `1Sm^(-)`C. `10Sm^(-1)`D. `4xx10^(-4)Sm^(-1)` |
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Answer» Correct Answer - B |
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| 1972. |
Conducitivity `kappa`, is equal to____A. `(1)/(R)(l)/(A)`B. `(G^(**))/(R)`C. `wedge_(m)`D. `(l)/(A)` |
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Answer» Correct Answer - A::B `R=rho.(1)/(A)` or `R=(1)/(kappa)(1)/(A)` or `kappa`(conductivity)`=(1)/(R)(1)/(A)`, i.e., (a) As `(1)/(A)=G^(@)` (cell constant), `kappa(1)/(R)G^(@)=(G^(**))/(R)` , i.e., (b)`. |
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| 1973. |
`wedge_(m)^(0)H_(2)O` is equal to____A. `wedge_(m(HCl))^(0)+wedge_(m(NaOH))^(0)-wedge_(m(NaCl))^(0)`B. `wedge_(m(HNO_(3)))^(0)+wedge_(m(NaNO_(3)))^(0)-wedge_(m(NaOH))^(0)`C. `wedge_((HNO_(3)))^(0)+wedge_(m(NaOH))^(0)-wedge_(m(NaOH_(3)))^(0)`D. `wedge_(m(NH_(4)OH))^(0)+wedge_(m(HCl))^(0)-wedge_(m(NH_(4)Cl))^(0)` |
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Answer» Correct Answer - A::D `wedge_(m(H_(2)O))^(@)=lamda_(H^(+))^(@)+lamda_(OH^(-))^(@)`. We get this expressions from (a) and (d). |
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| 1974. |
Which one is the correct equation that represents the first law of electrolysis ?A. `mZ = ct`B. `m = cZt`C. `mc = Zt`D. `c = mZt` |
| Answer» Correct Answer - B | |
| 1975. |
Which of the following conditions is correct for operation of electrolytic cell ?A. `Delta G = 0, E = 0`B. `Delta G lt 0, E gt 0`C. `Delta G gt 0, E lt 0`D. `Delta G gt 0, E gt 0` |
| Answer» Correct Answer - C | |
| 1976. |
If a strip of Cu metal is placed in a solution of ferrous sulphateA. Copper will precipitate outB. Iron will precipitate outC. Copper will dissolveD. No reaction will take place |
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Answer» Correct Answer - D `Cu+FeSO_(4)to`No reaction because Cu has `E^(@)=0.34` volt and Fe has `E^(@)=-0.44`volt. |
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| 1977. |
The process in which chemical change occurs on passing electricity is termedA. ionisationB. neutralisationC. electrolysisD. hydrolysis |
| Answer» Correct Answer - C | |
| 1978. |
The theory of ionisation was presented by:A. FaradayB. ArrheniusC. OstwaldD. Rutherford |
| Answer» Correct Answer - B | |
| 1979. |
When a strip of copper is dipped in a solution of ferrous sulphate.A. Iron is dpeosited on the copper stripB. Copper is precipitatedC. Copper dissolvesD. no reaction occurs. |
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Answer» Correct Answer - D Copper is less reactive than iron, hence no reaction. |
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| 1980. |
Substance which give good conducting aqueous solution are called :A. weak electolytesB. strong electrolytesC. non-electrolytesD. catalysts |
| Answer» Correct Answer - B | |
| 1981. |
The number of ions given by one molecule of `K_(4)Fe(CN)_(6)` after complete dissociation is:A. 5B. 11C. 2D. 10 |
| Answer» Correct Answer - A | |
| 1982. |
Dissociation of an electrolyte in water into negative and positive ions is calledA. ionisationB. electrolysisC. decompositionD. hydrolysis |
| Answer» Correct Answer - A | |
| 1983. |
Degree of ionisation is equal to:A. total number of moles of the electrolyte present in solutionB. total number of moles of the electrolyte dissociated into ionC. number of moles dissociated/total number of moles dissolvedD. total number of moles dissolved/number of moles dissociated |
| Answer» Correct Answer - C | |
| 1984. |
How many moles of mercury will be produced by electrolysing 1.0 M `Mg(NO_(3))_(2)` solution with a current of 2.00 A for 3 hours? `[Hg(NO_(3))_(2)=200.6" g "mol^(-1)]` |
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Answer» Correct Answer - 0.112 mole `Hg^(2+)+2e^(-)toHg,Q=2Axx(3xx3600s)=21600C` `2F=2xx96500C` produce `Hg=1` mole `therefore21600`C will produce Hg `=(1)/(2xx96500)xx21600`mole=0.112 mole |
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| 1985. |
How many moles of mercury will be produced by electrolysing 1.0 M `Hg(NO_(3))_(2)` solution by a current of 2.0 A when passed for 3 hours ? |
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Answer» Quantity of charge passed (Q) `= Ixxt=(2.0 A)xx(3xx60xx60 s)` `=(2.0 Axx10800 s)`=21600 AS=21600 C `Hg(NO_(3))_(2)overset((aq))rarrHg^(2+)(aq)+2 NO_(3)^(-)(aq)` `{:(Hg^(+)(aq)+2e^(-)rarrHg(s)),(" "2xx96500 C" "1 mol):}` When `2xx96500` C is passed, mercury deposited =1 mol When 21600 C is passed, mercury deposited `=(1 mol)xx((21600 C))/((2xx96500 C))=0.112 mol` |
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| 1986. |
For which of the following electrolyte the value of `.^ _(m)` and `^^^_(eq)` are same?A. `Na_(2)SO_(4)`B. `BaCl_(2)`C. `KCl`D. `Al_(2)(SO_(4))_(3)`. |
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Answer» Correct Answer - C KCl solution h as same value of normality and molarity. |
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| 1987. |
The molar conductance of HCl, NaCl and `CH_(3)COONa` are 426,12 6 and 91 `Omega^(-1)cm^(2)"mol"^(-1)` respectively. The molar c onductance for `CH_(3)COOH` isA. `561Omega^(-1) "mol"^(-1)`B. `391Omega^(-1)cm^(2)"mol"^(-1)`C. `261Omega^(-1)cm^(2)"mol"^(-1)`D. `612Omega^(-1)cm^(2)"mol"^(-1)` |
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Answer» Correct Answer - B `^^_(CH_(3)COOH)^(oo)=+^^_( HCl)^(oo)+^^_(CH_(3)COONa^(-))^(oo)-^^_(NaCl)^(oo)` `=426+91-126=391Omega^(-1)cm^(2)"mol"^(-1)` |
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| 1988. |
The values of `^^_(m)^(oo)` for KCl and `KNO_(3)` are 149-86 and `154.96Omega^(-1)cm^(2)"mol"^(-1)`. Also `lambda_(Cl)^(oo)` is 71.44 `ohm^(-1)cm^(2) "mol"^(-1)` . The value of `lambda_(NO_(3)^(-))^(oo)` isA. 76.54 `ohm^(-1) cm^(2)"mol"^(-1)`B. `133-08 ohm^(-1)cm^(2)"mol"^(-1)`C. 37.7 `ohm^(-1)cm^(2)"mol"^(-1)`D. unpredictable. |
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Answer» Correct Answer - A `^^_(m)^(oo)(NO_(3)^(-))=^^_(m)^(oo)(KNO_(3))-^^_(m)^(oo)+^^_(KCl)+^^_(m)^(oo)(Cl^(-))` ` =(154-96-149.86+71.44)Omega^(-1)cm^(2)"mol"^(-1)` `(226.40-149.86)Omega^(-2)cm^(2)"mol"^(-1)` `=76.54Omega^(2)cm^(2)"mol"^(-1)` |
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| 1989. |
Which expressions can be used to calculate degree of ionisation of weak electrolyte of type `A^(-)B^(-)` ?A. `sqrt(K//C)`B. ` ^^ _(m)//^^_(m)^(oo)`C. Both A and BD. Neither A nor B. |
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Answer» Correct Answer - C See comprehensiv e review for details. |
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| 1990. |
The degree of dissociation of a weak electrolyte is given by :(a) α = \(\frac{∧_0}{∧_m}\)(b) α = ∧m x ∧0(c) α = \(\frac{∧_m}{∧_0}\)(d) α = ∧0 - ∧m |
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Answer» Option : (c) α = \(\frac{∧_m}{∧_0}\) |
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| 1991. |
Statement-I: In electrolysis the quantity needed for depositing 1 mole of silver is different from that required for 1 mole of copper. Because Statement-II: The molecular weights of silver and copper are different. |
| Answer» Correct Answer - B | |
| 1992. |
Statement-I: In electrolysis the quantity needed for depositing 1 mole of silver is different from that required for 1 mole of copper. Because Statement-II: The molecular weights of silver and copper are different.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If assertion and reason both are incorrect. |
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Answer» Correct Answer - C (c ) Correct reason : One mole of silver contains one gram equivalent while one mole of copper contians two gram equivalents. |
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| 1993. |
The reaction `(1)/(2)H_(2)(g)+AgCl(s)hArrH^(+)(aq)+Cl^(-)(aq)+Ag(s)` occurs in the galvanic cellA. `Ag|AgCl(s)|KCl("soln")|AgNO_(3)|Ag`B. `Pt|H_(2)(g)|HCl|("soln")|AgNO_(3)("soln")|Ag`C. `Pt|H_(2)(g)HCL("soln")|AgCl(s)|Ag`D. `Pt|H_(2)(g)|KCl("soln")|AgCl(s)|Ag` |
| Answer» Correct Answer - c | |
| 1994. |
The electrotrode potentials for `Cu^(2+)(aq.)+2e^(-) rarr Cu^(+)(aq.)` and `Cu^(+)(aq.)+e^(-) rarr Cu(s)` are `+0.15 V` respectively. The value of `E_(Cu^(2+))^(ɵ)//Cu` will be:A. `0.150 V`B. `0.500 V`C. `0.325 V`D. `0.650 V` |
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Answer» Correct Answer - C Many elements exist in a number of different oxidation states. The emf does not list the value of all the standard potentials of every possible half-reaction for these elements. But we can obtain these values, as long as each oxidation state of interest appers in a half-reaction in the series. To obtain this kind of information, we must combine half-reactions in the series to make a new half-reaction that is not in the series. The standard potential of the new half-reaction cannot bc found simply by a combination of the potentials of the original half-reactions. We could conver the `E^(o+)` of each half-reaction to the corresponding `DeltaG^(@)`, combine the `DeltaG^(@)` value to find the `DeltaG^(@)` back to `E^(@)`. But we shall use a different method. The `E^(@)` of each half-reaction is multiplied by the number of electrons in that half-reaction is multiplied by the number of electrons in that half-reaction. The resulting values arc combined. Finally, we divide the resulting potential by the number of electrons in the new half-reaction to get its potential. The procedure is sysbolized by `E_(T)^(@) = (n_(1)E_(1)^(@)+n_(2)E_(2)^(@)+.....)/(n_(T))` where `n_(1)` is the number of electrons in the first half-reaction and `E_(1)^(@)` is its standard potential, `n_(2)` is the number of electrons in the second half-reaction and `E_(2)^(@)` is its standard potential (and so on for as many half-reactions as are baing combined),and `n_(T)` is the number of electrons that appear in the resulting half-reaction. In using this relationship, be careful to make the sign of each `E^(@)` consistant with the way in which its half-reaction is combined to amke the new half-reaction. we being by writing the desired half-reaction. `Cu^(2+)(aq.)+2e^(-) rarr Cu(s)` Now we write half reactions that include these oxidation states of copper. These arc `Cu^(2+)(aq.)+e^(-) rarr Cu^(+)(aq.)," "E_(1)^(@) = +0.15 V` `Cu^(2+)(aq.)+e^(-) rarr Cu(s)," "E_(2)^(@) = +0.50 V` We see that the two equations should be added to produce the desired half-reaction. Thus `E_(T)^(@)=((1)(+0.15 V)+(1)(+.50 V))/(2)` `= 0.325 V` |
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| 1995. |
Assertion `(A):` The mobility of `Na^(o+)` is lower than that of `K^(o+)` ion. Reason `(R): ` The ionic mobility depends upon the effective radius of the ion.A. `S` is correct but `E` is wrongB. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Explanation is correct reason for statement. |
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| 1996. |
Assertion `(A:` Cell constant is the `EMF` of a cell. Reason `(R):` Cell constant is determined by using saturated `KCl` solution.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, `(R)` is correct. |
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Answer» Correct Answer - D `(A)` is incorrect. Cell constant `(G^(**))=(1)/(a)cm^(-1)` `(R)` is correct . `G^(**)` is determined by using `KCl` solution since its conductivity is known accurately at various concentrations and at differne temperature. |
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| 1997. |
The ionic equivalent conductivities of `C_(2)O_(4)^(2-),K^(o+),` and `Na^(o+)` ions are `x,y,` and `z S cm^(2) Eq^(-1)` respectively. Calculate `wedge_(eq)^(@)` of `(NaOOC-COOK).` |
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Answer» Total charge`=2` Number of equivalent of ion `=("Charge on the ions")/("Total charge")` `:. ` Equivalent of `C_(2)O_(4)^(2-)=(2)/(2)=1` Equivalent of `Na^(o+)=(1)/(2), Eq of K^(o+)=(1)/(2)` `:. wedge _(eq(NaOOC-COOK))^(@)=lambda_(eq(C_(2)O_(4)^(2-)))^(@)+(1)/(2)lambda_(eq(Na^(o+)))^(@)+(1)/(2)lambda_(eq(K^(o+)))^(@)` `=(x+(y)/(2)+(z)/(2))` |
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| 1998. |
Statement : The cathode of electrolytic cell during electrolysis of `NaCl_((aq.))` on additon of little litmus shows a blue colour. Explanation : At cathode : `2H^(+) + 2e rarr H_(2)`. The reaction at cathode give rise to an increase in `pH` ranging in alkaline medium and litmus shows blue colour.A. `S` is correct but `E` is wrongB. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Explanation is correct reason for statement. |
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| 1999. |
Assertion `(A):` Ionic conductivities increase with increase of temperature and pressure. Reason `(R): ` Viscosity of water decreases with increasew of temperature and increases with the increases of pressure.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, `(R)` is correct. |
| Answer» Correct Answer - D | |
| 2000. |
Assertion `(A): wedge_(m(H^(o+)))` and `wedge_(m(overset(c)(O)H))` ions are very much higher than those of other ions. Reason `(R): `It is due to proton jump from one water molecule to another resulting in a more rapid transfer of positive charge form one region to another.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, `(R)` is correct. |
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Answer» Correct Answer - A Refer Section |
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