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Conductivity and its types - class-XII

Description: conductivity and its types
Number of Questions: 47
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Tags: electrochemistry
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What do you mean by equivalent Conductivity?

  1. It is defined as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte in solution.

  2. It is defined as the conducting power of all the ions produced by dissolving ten gram equivalent of an electrolyte in solution.

  3. It is defined as the conducting power of all the ions produced by dissolving hundred gram equivalent of an electrolyte in solution.

  4. It is defined as the conducting power of all the ions produced by dissolving thousand gram equivalent of an electrolyte in solution.


Correct Option: A
Explanation:

Equivalent Conductivity is defined as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte in solution. It is expressed as and is related to specific conductance as. (M is Molarity of the solution)

The resistance of $1\ N$ solution of $CH _{3}COOH$ is $250\ ohm$ when measured in a cell of cell constant $1.15\ cm^{-1}$. The equivalent conductance will be:

  1. $4.6\ ohm^{-1} cm^{2} eq^{-1}$

  2. $9.2\ ohm^{-1} cm^{2} eq^{-1}$

  3. $18.4\ ohm^{-1} cm^{2} eq^{-1}$

  4. $0.023\ ohm^{-1} cm^{2} eq^{-1}$


Correct Option: A
Explanation:

$K = C \times \dfrac {l}{A} = \dfrac {1}{250}\times 1.15 = 4.6\times 10^{-3}\ ohm^{-1} cm^{-1}$
$\wedge _{e} = K\times \dfrac {1000}{N} = 4.6\times 10^{-3} \times \dfrac {1000}{1} = 4.6\ ohm^{-1} cm^{2} eq^{-1}$.

The specific conductance of a $0.01\ M$ solution of $KCl$ is $0.0014\ ohm^{-1} cm^{-1}$ at $25^{\circ}C$. Its equivalent conductance is____________.

  1. $14$

  2. $140$

  3. $1.4$

  4. $0.14$


Correct Option: B
Explanation:

$\kappa =0.0014\ S{ cm }^{ -1 }\ \Lambda _{ eq }=\cfrac { 1000\times \kappa  }{ C } \ C=0.01M\ { \Lambda  } _{ eq }=\cfrac{1000 \times 0.0014}{0.01}=140$

The resistance of a N/10 KCI solution is 245$\Omega $. Calculate the equivalent conductance of the solution if the electrodes in the cell are 4cm apart and each having an area of 7.0sq,cm.

  1. $23.32S{ cm }^{ 2 }{ eq }^{ -1 }$

  2. $23.23S{ cm }^{ 2 }{ eq }^{ -1 }$

  3. $2.332S{ cm }^{ 2 }{ eq }^{ -1 }$

  4. none of these


Correct Option: A
Explanation:

Given resistance = 245$\Omega $

Formula for the specific conductance (K) = $ \dfrac{1}{R} $ $\times \dfrac{l}{a}$
$ \dfrac{1}{245} $ $\times \dfrac{4}{7}$
 K = 2.33 $\times 10^-$$^3$ 
Formula for the euivalent conductance=  K $\times$ $\dfrac{1000}{C}$
= 2.33 $\times 10^-$$^3$ $\times 10000$
= 23.32S $cm^2$ eq $^-$$^1$

The equivalent conductance of a weak monobasic acid at infinite dilution is $100cm^3$ $eq^{-1}$ and that of its $0.01$M solution is $5cm^2$ $eq^{-1}$ at $25^o$C. The dissociation constant $K _a$ of the acid is:

  1. $2.5\times 10^{-4}$

  2. $5\times 10^{-4}$

  3. $1.25\times 10^{-5}$

  4. $2.5\times 10^{-5}$


Correct Option: D
Explanation:

The degree of dissociation, $\alpha=\cfrac {\text{Equivalent conductance at given dilution}}{\text{Equivalent conductance at infinite dilution}}$

                                                   $=\cfrac {5cm^2eq^{-1}}{100cm^2eq^{-1}}=\cfrac {1}{20}$
For weak monobasic acid,
$K _a=\cfrac {C \alpha^2}{1-\alpha}$
       $=C\alpha^2(\because 1>> \alpha)$
       $=0.01\times \left(\cfrac {1}{20}\right)^2= 2.5 \times 10^{-5}$

Equivalent conductance and molar conductance of $Fe _2(SO _4) _3$ are related?

  1. $\bigwedge _e=\bigwedge _m$

  2. $\bigwedge _{eq} = \dfrac{\bigwedge _m}{3}$

  3. $\bigwedge _{eq} = 3 \bigwedge _m$

  4. $\bigwedge _{eq} = \dfrac{\bigwedge _m}{6}$


Correct Option: D

Molar ionic conductance of ${Ca}^{+2}$ is $x$ $S{m}^{2}$ ${mole}^{-1}$. Equivalent conductance of calcium phosphate is ____ $S{m}^{2}g$ ${eq}^{-1}$

  1. $x+y$

  2. $(3x+2y)$

  3. $6(3x+2y)$

  4. $\cfrac{3x+2y}{6}$


Correct Option: A
Explanation:

Calcium phosphate $\rightarrow { Ca } _{ 3 }{ \left( { PO } _{ 4 } \right)  } _{ 2 }$

ionic conductance of ${ Ca }^{ 2+ }=x$ ${ sm }^{ 2 }{ mol }^{ -1 }$
ionic conductance of ${ PO } _{ 4 }^{ 3- }=Y$ ${ sm }^{ 2 }{ mol }^{ - }$
${ Ca } _{ 3 }{ \left( { PO } _{ 4 } \right)  } _{ 2 }\rightarrow 3{ Ca }^{ 2+ }+2{ PO } _{ 4 }^{ 3- }$
$\therefore $  equivalent conductivity $=x+y$

Which of the following has least conductivity in aqueous solution?

  1. $CO(NH _3) _4Cl _3$

  2. $Co(NH _3) _3Cl _3$

  3. $Co(NH _3) _5Cl _3$

  4. $Co(NH _3) _6Cl _3$


Correct Option: A
Explanation:

Conductivity $\alpha $ no. of ions dissociated in solution.

$Co{ \left( { NH } _{ 3 } \right)  } _{ 4 }{ Cl } _{ 3 }\rightarrow { \left[ Co{ \left( { NH } _{ 3 } \right)  } _{ 4 }{ Cl } _{ 2 } \right]  }^{ + }+{ Cl }^{ \left( - \right)  }$
                                                    2 ions
$Co{ \left( { NH } _{ 3 } \right)  }{ Cl } _{ 3 }\rightarrow \left[ Co\left( { NH } _{ 3 } \right) { \left( Cl \right)  } _{ 3 } \right] $
                                                          1 ion
$Co{ \left( { NH } _{ 3 } \right)  } _{ 5 }{ Cl } _{ 3 }\rightarrow { \left[ Co{ \left( { NH } _{ 3 } \right)  } _{ 5 }Cl \right]  }^{ 2+ }+2{ Cl }^{ \left( - \right)  }$
                                                          3 ions
$Co{ \left( { NH } _{ 3 } \right)  } _{ 6 }{ Cl } _{ 3 }\rightarrow { \left[ Co{ \left( { NH } _{ 3 } \right)  } _{ 6 } \right]  }^{ 3+ }+3{ Cl }^{ \left( - \right)  }$
                                                             4 ions
$\therefore  Co{ \left( { NH } _{ 3 } \right)  } _{ 6 }{ Cl } _{ 3 }$ has highest conductivity.

The equivalent conductance of $CH _3COONa, \ HCl$ and $NaCl$ at infinite dilution are $91, 426$ and $126 \ S \ cm^3 \ eq^{-1}$ respectively at $25^oC$. The equivalent conductance of $1 \ M \ CH _3COOH$ solution is $19.55 \ S \ cm^2 \ eq^{-1}$. The pH of the solution is:

  1. $5.3$

  2. $4.3$

  3. $2.3$

  4. $1.3$


Correct Option: D
Explanation:

$\lambda^o _{CH _3COONa}= \lambda^o _{CH _3COO^-}+ \lambda^o _{Na}=91 \ Scm^2eq^{-1}$.....(i)

$\lambda^o _{HCl}= \lambda^o _{H}+ \lambda^o _{Cl^-}= 426 \ Scm^2eq^{-1}$.....(ii)

$\lambda^o _{NaCl}= \lambda^o _{Na}+ \lambda^o _{Cl^-}=126 \ Scm^2eq^{-1}$.....(iii)

$\lambda^o _{CH _3COOH}= \lambda^o _{CH _3COO^-}+ \lambda^o _{H^+}= (i) + (ii) – (iii) = 391 \ Scm^2eq^{-1}$

$ \lambda _{CH _3COOH}= 19.5 \ Scm^2eq^{-1}$ (given)

Degree of dissociation $= \cfrac {\lambda _m}{\lambda^o _m}= \cfrac {19.55}{391}=0.05$

$CH _3COOH \longrightarrow CH _3COO^- + H^+$

$1(1-0.05)$                $0.05$             $0.05$

$[H^+]=0.05 \ M$

$pH= -log [H^+]=1.3$

The electrolytic conductance is a direct measure of:

  1. resistance

  2. potential

  3. concentration

  4. dissociation


Correct Option: A
Explanation:

By measuring conductance, we can measure resistance, by using formula.

$R=\cfrac{1}{C}$             C=conductance

Equivalent conductance of $BaCl _2, H _2SO _4$ and $HCl$ are $x _1, x _2$ and $x _3 S cm^2 equiv^{-1}$ at infinite dilution.If specific conductance of saturated $BaSO _4$ solution is of $y S cm^1$ then $K _{sp}$ of $BaSO _4$ is:

  1. $\frac{10^3y}{2(x _1 + x _2 - 2x _3)}$

  2. $\frac{10^6y^2}{4(x _1 + x _2 - 2x _3)^2}$

  3. $\frac{10^6y^2}{2(x _1 + x _2 - x _3)^2}$

  4. $\frac{x _1 + x _2 - 2x _3}{10^6y^2}$


Correct Option: A
Explanation:

${ \Lambda  } _{ eq }\quad (Ba{ SO } _{ 4 })={ \Lambda  } _{ eq }\left( Ba{ Cl } _{ 2 } \right) +{ \Lambda  } _{ eq }\left( { H } _{ 2 }{ SO } _{ 4 } \right) -2{ \Lambda  } _{ eq }\left( HCl \right) $

                            $=\left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right) S{ cm }^{ 2 }{ eq }^{ -1 }$
$\because { \Lambda  } _{ eq }=K\times \cfrac { 1000 }{ N } \quad \Rightarrow \quad N=\left( \cfrac { y\times { 10 }^{ 3 } }{ { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } }  \right) \quad \left( \because { n } _{ f }=2 \right) $
$\therefore \quad M=\left{ \cfrac { y\times { 10 }^{ 3 } }{ 2\left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right)  }  \right} $
$\therefore \quad { K } _{ sp }={ \left( M \right)  }^{ 2 }=\left{ \cfrac { { y }^{ 2 }\times { 10 }^{ 6 } }{ 4{ \left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right)  }^{ 2 } }  \right} $

For $HCl$ solution at ${25}^{o}C$ equivalent conductance at infinite dilution is $425 \ {ohm}^{-1}{cm}^{2}{equiv}^{-1}$. The specific conductance of a solution of $HCl$ is $3.825$ ${ohm}^{-1}{cm}^{-1}$. If the apparent degree of dissociation is $90$% the normality of the solution is :

  1. $0.90N$

  2. $1.0N$

  3. $10\ N$

  4. $1.2N$


Correct Option: C

The equivalent conductivity of $0.1 N \ CHNCH _{3}COOH$ at $25^{0}C$ is 80 and at infinite dilution it is 400, the degree of dissociation of $CH _{3}COOH$ is :

  1. 1

  2. 0.2

  3. 0.1

  4. 0.5


Correct Option: B
Explanation:

Given:-

$\wedge _{eq}(CH _3COOH)= 80 S cm^{-2} eq^{-1}$

$\wedge^{\infty} _{eq}(CH _3COOH)$ at infinite dilution= $400S cm^2 eq^{-1}$

$\alpha \longrightarrow$ Degree of dissociation

$\alpha= \cfrac {\wedge^m _{eq}}{\wedge^{\infty} _{eq}}=\cfrac {80}{400}= 0.2$

$\alpha= 0.2$

In infinite dilusions, the equivalent conductances of $Ba^{2+}$ and $Cl^{-}$ are $127$ and $76 ohm^{-1} \, cm^{-1} \, eqvt^{-1}$. The equivalent conductivity of $BaCl _2$ at indefinite dilution is?

  1. $101.5$

  2. $203.5$

  3. $139.5$

  4. $279.5$


Correct Option: C
Explanation:

The equivalent conductance of BaCl2 at infinite dilution, 


λ of BaCl2=1/2 λ of Ba2+ + λof Cl
       

 =127/2+76 

=139.

Option C is correct answer

${\text{N}}{{\text{a}} _{\text{3}}}{\text{Al}}{{\text{F}} _{\text{6}}}\,\,$ is added to $\,{\text{A}}{{\text{l}} _{\text{2}}}{{\text{O}} _{\text{3}}}$

  1. Improve the electrical conductivity of the cell

  2. Increases rate of production

  3. Increases the melting point

  4. Decrease the electrical conductivity


Correct Option: A

The equivalent conductance of $0.02$ M acctic acid  $1.62.*{10^{ - 3}}$.  Degree of ironisation $'a'$ of $C{H _3}COOH$ is:
$({x _H} = 349.83oh{m^{ - 1}}and\lambda C{H _3}CO{O^ - } = 40.89ohm{s^{ - 1}}$

  1. $0.01$

  2. $0.02$

  3. $0.03$

  4. $0.04$


Correct Option: B

Equivalent conductivity of $BaCl _2,H _2SO _4$ and HCI, are $x _1,x _2$ and $x _3scm^{-1}eq^{-1}$ at infinite dilution. If conductivity of saturated $BaSO _4$ solution is x $Scm^{-1}$, then $K _{sp}$ of $BaSO _4$ is

  1. $\dfrac {500x} {(x _1+x _2-2x _3)}$

  2. $\dfrac {10^6x^2} {(x _1+x _2-2x _3)^3}$

  3. $\dfrac {2.5\times10^5 x^2} {x _1-2x _2-x _3)^2}$

  4. $\dfrac {0.25 x^2} {x _1 + x _2-x _3)^2}$


Correct Option: A

Molar conductance of $C{a^{2 + }}$ and $C{l^ - }$ are  $120\,c{m^2}\,$ $mo{l^{ - 1}}$ and $77\,S\,c{m^2}\,mo{l^{ - 1}}$ respectively. What is the equivalent conductance of $CaC{l _2}$ ?

  1. $98.5\,Sc{m^2}e{q^{ - 1}}$

  2. $137\,Sc{m^2}e{q^{ - 1}}$

  3. $197\,Sc{m^2}e{q^{ - 1}}$

  4. $247\,Sc{m^2}e{q^{ - 1}}$


Correct Option: D

At infinite dilution equivalent conductance of ${B^{ + 2}}$ & CI ions are 127  &  76$oh{m^{ - 1}}$ $c{m^{ - 1}}$` $e{q^{ - 1}}$ respectively. Equivalent conductance  of $BaC{I _2}$ at infinite diluition is : 

  1. 139.5

  2. 101.5

  3. 203

  4. 279


Correct Option: C

Equivalent constant of standard $BaSO _{4}$ is $400ohm^{-1}\ cm^{2}$ equiv$^{-1}$ and specific conduction is $8\times 10^{-5}\ ohm^{-1}\ cn^{-1}$. Hence $K _{SP}$ of $BaSO _{4}$ is

  1. $4\times 10^{-8}M^{2}$

  2. $1\times 10^{-8}M^{2}$

  3. $2\times 10^{-4}M^{2}$

  4. $1\times 10^{-4}M^{2}$


Correct Option: B

Which of the following is correct regarding current carrying ions in the solution of $C _{2}H _{5}COOH$ upon dilution?

  1. The number of ions in $1 cm^{3}$, as well as in total volume increases.

  2. The number of ions in $1 cm^{3}$ decreases, whereas that in the total volume remains constant.

  3. The number of ions in $1 cm^{3}$ decreases, but that in the total volume increases.

  4. The number of ions in $1 cm^{3}$, as well as in total volume decreases.


Correct Option: C
Explanation:

For weak electrolytes, as the concentration decreases, the percentage dissociation increases and the total number of ions increases. But as the volume increases, number of ions per unit volume decreases.


Hence, option C is correct.

The resistance of $N/2$ solution of an electrolyte in a cell was found to be $45$ ohm. The equivalent conductivity of a solution, if the electrodes in the cell are $2.2$ cm apart and have an area of $3.8 cm^2$ will be:

  1. $52.72\ S cm^2 eq^{-1}$

  2. $22.57\ S cm^2 eq^{-1}$

  3. $27.52\ S cm^2 eq^{-1}$

  4. $25.72\ S cm^2 eq^{-1}$


Correct Option: D
Explanation:
Cell constant = $\dfrac{l}{a}=\dfrac{2.2}{3.8}=0.579 cm^{-1} $

Specific conductance = cell constant $\times$ conductance

Specific conductance = $\dfrac{0.579}{45} $

Equivalent conductance= $\dfrac{0.579}{45} \times \dfrac{1000}{0.5} $

Equivalent conductance = $25.73 Scm^{2} eq^{-1} $

Hence, option D is correct.

The resistance of 0.1 N solution of a salt is found to be $2.5\times10^{3}$. The equivalent conductance of the solution is: (cell constant=1.15 $cm^{-1}$)

  1. 3.6

  2. 4.6

  3. 5.6

  4. 6.6


Correct Option: B
Explanation:

The relationship between the specific conductance, resistance and the cell constant is $ \kappa =\cfrac { 1 }{ R } \times \cfrac { l }{ a }$.
Substitute $ R=2.5\times { 10 }^{ 3 }\quad ohm $ and $ \cfrac { l }{ a } =1.15\quad {cm }^{ -1 } $.
Hence $ \kappa =\cfrac { 1 }{ 2.5\times { 10 }^{ 3 }\quad ohm } \times1.15\quad { cm }^{ -1 }=\cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \quad { ohm }^{ -1 }\quad { cm }^{ -1 } $.
The relationship between the equivalent conductance and specifc conductance is $  { \Lambda  } _{ eq }=\cfrac { \kappa \times 1000 }{ M }  $.
Substitute $ M=0.1\quad N $ and $ \kappa=\cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \quad { ohm }^{ -1 }\quad { cm }^{ -1 } $.
Hence $ { \Lambda  } _{ eq }=\cfrac { \cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \times 1000 }{ 0.1 } \quad =\quad 4.6\quad \quad { ohm }^{ -1 }\quad { cm }^{ 2 }\quad { equiv }^{ -1 } $.

At 291 K, the equivalent conductivities at infinite dilution of $NH _4Cl$, $NaOH $ and $NaCl $ are $129.8$, $217.5$ and $108.9$ S $cm^2 eq^{-1}$ respectively. The equivalent conductivity at infinite dilution of $NH _4OH$ is:

  1. $208.4 S cm^2 eq^{-1}$

  2. $238.4 S cm^2 eq^{-1}$

  3. $283.4 S cm^2 eq^{-1}$

  4. None of these


Correct Option: B
Explanation:
Equivalent conductance of $NH _4OH $ = Equivalent conductance of $NH _4Cl $+$NaOH$-$NaCl$

                                                                $= 129.8+218.4-108.9$

Equivalent conductance of $NH _4OH = 239.3$

Therefore, option B is the correct answer.

The correct order of equivalent conductivity at infinite dilution of $LiCl,\ NaCl$ and $KCl$ is:

  1. $LiCl > NaCl > KCl$

  2. $KCl > NaCl > LiCl$

  3. $NaCl > KCl > LiCl$

  4. $LiCl > KCl > NaCl$


Correct Option: B
Explanation:

The correct order of equivalent conductivity at infinite dilution of $LiCl,\ NaCl$ and $KCl$ is $KCl > NaCl > LiCl$.

Ionic mobility depends upon size of the ion. The ionic size in case of hydrated cation, is $K^+(aq) < Na^+(aq) < Li^+(aq)$.

As the size of the hydrated ion increases, the equivalent conductivity at infinite dilution decreases.

$\lambda {eq}$ x Normality = _________

  1. K x 10$^3$

  2. K x 10$^4$

  3. K x 10$^5$

  4. K x 10$^6$


Correct Option: A
Explanation:

As we know,
                     $\lambda _{eq} = k\times1000/N$

              Then, $\lambda _{eq} \times normality$ = K x 10$^3$


   So. the correct option is $A$

Equivalent conductance of an electrolyte containing $NaF$ at infinite dilution is $90.1\ Ohm^{-1} cm^{2}$. If $NaF$ is replaced by $KF$ what is the value of equivalent conductance?

  1. $90.1\ Ohm^{-1} cm^{2}$

  2. $11.2\ Ohm^{-1} cm^{2}$

  3. $0$

  4. $222.4\ Ohm^{-1} cm^{2}$


Correct Option: A
Explanation:

At infinite dilution the equivalent conductance of strong electrolytes furnishing same number of ions is same.

For this case, Both NaF and KF are strong electrolyte and also furnish 2 ions in the solution.
Hence, Both will have same equivalent conductance.
So, option A is correct.

The resistance of $N/10$ solution is found to be $2.5\times 10^{3}ohm$. The equivalent conductance of the solution is (cell constant $= 1.25\ cm^{-1})$.

  1. $2.5\ ohm^{-1} cm^{2} equiv^{-1}$

  2. $5\ ohm^{-1} cm^{2} equiv^{-1}$

  3. $2.5\ ohm^{-1} cm^{-2} equiv^{-1}$

  4. $5\ ohm^{-1} cm^{-2} equiv^{-1}$


Correct Option: B
Explanation:

Given Data : 1)  Resistance = $R$ = $2.5 \times 10^3$

                     2) Cell constant = $k$ = $\cfrac{l}{a}$= $1.15$$cm^2$
                     3)  Normality =$N$= $0.1 N$
To Find : Equivalent conductance = $\Lambda$$ _e$$ _q$

The relation between $K$ ,$R$ and $ k$ is,
$K$ = $\cfrac{1}{R} \times \cfrac{l}{a}$
where $K$ is specific conductance.
$\therefore$ Substituting the given values we get,

$K$ = $\cfrac{1}{2.5×10^3}$ × $1.15cm^-$$^1$

     = $\cfrac{1.15}{2.5×10^3}$ $ohm$$^-$$^1$$cm$$^-$$^1$

The relation between $\Lambda$$ _e$$ _q$ and $K$ is,

$\Lambda$$ _e$$ _q$ = $\cfrac{K×1000}{N}$

Substituting the value of $M$ and $K$ we get,

$\Lambda _{eq}= \cfrac{1.15}{2.5\times10^3\times0.1} \times 1000$

       = $4.6$$ohm^-$$^1$$cm$$^2$$equi$$^{-1}$      [Note:$ \text {Normality= no. of equiv.} /cm^3$]

Here appproximation is taken,

      $\approx$ $5$ $ohm$$^-$$^1$$cm$$^2$$equi$$^-$$^1$

Hence the correct option is 'B'.

What are the units of equivalent conductivity of a solution?

  1. $mho\ cm^{-1}$

  2. $ohm\ cm^{-1} g\ equiv^{-1}$

  3. $mho\ cm^{-2}g\ equiv^{-1}$

  4. $mho\ cm^{2}g\ equiv^{-1}$


Correct Option: A
Explanation:
$\rightarrow$ If 1 $m^{3}$ of a solution of an electrolyte placed between two large electrodes 1m apart. The cross-sectional area of the solution will be $1m^{2}$.
$\rightarrow$ The conductance of solution will evidently be its specific conductance because we have a one-meter cube of the solution.
$\rightarrow$ If $1m^{3}$ of the solution contains 1 gram equivalent of the electrolyte.
Hence, $Conductance(C)=Specific Conductance(K)=Equivalent Conductance(\wedge)$
$\rightarrow$ As unit of Specific conductance is $Sm^{-1}$ or $Scm^{-1}$, so the unit of Equivalent Conductance is
$Scm^{-1}$ or $ mhocm^{-1}$.
$\rightarrow$ $mho$ is reciprocal of $ohm$ & reciprocal of $ohm$ i.e. $ohm^{-1}$ is also called $mho$ or $Siemen(S)$.

The equivalent conductances of $NaCl$ at concentration $c$ and at infinite dilution are $\lambda _{c}$ and $\lambda _{\infty}$ respectively. The correct relationship between $\lambda _{c}$ and $\lambda _{\infty}$ is given as: (where the constant $b$ is positive).

  1. $\lambda _{c} = \lambda _{\infty} - b\sqrt {c}$

  2. $\lambda _{c} = \lambda _{\infty} + b\sqrt {c}$

  3. $\lambda _{c} = \lambda _{\infty} + bc$

  4. $\lambda _{c} = \lambda _{\infty} - bc$


Correct Option: A
Explanation:

According to debye huckel onsager equation we have,

equivalent conductance of NaCl is given by
$\lambda _C$ = $\lambda _{\infty}$ - $b C^{1/2}$
where b is positive.

The equivalent conductances at infinite dilution of $HCl$ and $NaCl$ are $426.15$ and $126.15\ mho\ cm^{2}g\ eq^{-1}$ respectively. If can be said that the mobility of:

  1. $H^{+}$ ions is much more than that of $Cl^{-}$ ions

  2. $Cl^{-}$ ions is much more than that of $H^{+}$ ions

  3. $H^{+}$ ions is much more than that of $Na^{+}$ ions

  4. $Na^{+}$ ions is much more than that of $H^{+}$ ions


Correct Option: C
Explanation:
$\rightarrow$ The equivalent conductance of $HCl$ is more than $NaCl$ , it is because of difference in speed of ions.
$\rightarrow\,HCl$ has conductance at infinite dilution more than three times as high as that of $NaCl$.
$\rightarrow$ As Chlorine ion is common in both of them, it seems that speed of $H^{+}$ ion is much more than that of $Na^{+}$ ions.

The resistance of $0.01\ N$ solution at $25^{\circ}$ is $200\ ohm$. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

  1. $200\ ohm^{-1}cm^{2} eq^{-1}$.

  2. $300\ ohm^{-1}cm^{2} eq^{-1}$.

  3. $400\ ohm^{-1}cm^{2} eq^{-1}$.

  4. $500\ ohm^{-1}cm^{2} eq^{-1}$.


Correct Option: D
Explanation:

since, we have

$conductance * cell constant$ =  specific conductance
cell consyant = 1
conductance = specific conductance= $\dfrac{1}{200}$

equivalent conductance= $\dfrac{K*1000}{N}$ $Scm^{2}eq^{-1}$

equivalent conductance= $\dfrac{1*1000}{200*0.01}$

equivalent conductance= $500$ $ Scm^{2}eq^{-1}$
 

The specific conductance $(K)$ of an electrolyte of $0.1\ N$ concentration is related to equivalent conductance $(\wedge _{e})$ by the following formula.

  1. $\wedge _{e} = K$

  2. $\wedge _{e} = 10 K$

  3. $\wedge _{e} = 100 K$

  4. $\wedge _{e} = 10000 K$


Correct Option: D
Explanation:

we have,

equivalent conductance = $\dfrac{K*1000}{N}$

equivalent conductance= $\dfrac{K*1000}{0.1 N}$

equivalent conductance= $10000K$

The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was

  1. methyl isocyanate

  2. methylamine

  3. ammonia

  4. phosgene


Correct Option: A

The specific conductivity of $0.1$ $N$ $KCl$ solution at $20^0$C is $0.0212$ $ohm^{-1} cm^{-1}$. The solution was found to offer resistance of $55$ ohms. Find the cell constant of the conductivity cell.

  1. $2.25$

  2. $1.166$

  3. $1.936$

  4. none of these


Correct Option: B
Explanation:
Specific conductance $K=0.0212 \Omega ^{-1}cm^{-1}$

Normality = $0.1 N$, Resistance = $55\Omega $

Cell constant $G^*$ = conductivity $\times $ Resistance

$G^*=K\times R$

$=0.0212\times 55$

$=1.166 cm^{-1}$

Option B

What would be the equivalent conductivity of a cell in which $0.5$ N salt solution offers a resistance of $40$ ohm whose electrodes are $2$ cm apart and $5$ $cm^{2}$ in area?

  1. $10$ $ohm^{-1}$ $cm^2$ $eq^{-1}$

  2. $20$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$

  3. $30$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$

  4. $25$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$


Correct Option: B
Explanation:
Given,
$R=40 ohm ,l=2cm,A=2cm^{2}$

We know the relation,

$\kappa=\dfrac{1}{R}\times\dfrac{l}{A}=\dfrac{1}{40}\times\dfrac{2}{5}=\dfrac{1}{100} ohm^{-1} m^{-1}$


Now we also know the relation,

$\Lambda _{eq}=\dfrac{\kappa\times1000}{N}=\dfrac{1}{100}\times\dfrac{1000}{0.5}=20 ohm^{-1} cm^{2} eq^{-1}$

Hence, option B is correct.

The equivalent conductivity of $N/10$ solution of acetic acid at $25^o$C is $14.3$ $ohm^{-1}$ $cm^2$ $equiv^{-1}$. What will be the degree of dissociation of acetic acid?
$(\Lambda _{\infty CH _3COOH}=390.71$ $ohm^{-1}$ $cm^2$ $equiv^{-1}$).

  1. $3.66\%$

  2. $3.9\%$

  3. $2.12\%$

  4. $0.008\%$


Correct Option: A
Explanation:

The equivalent conductivity of an electrolyte is defined as the conductivity of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solution.
Given,
$\Lambda^{\infty} _{m(CH _3COOH)}=390.71 ohm^{-1} cm^{2} eq^{-1}$
$\Lambda _{m(CH _3COOH)}=14.3 ohm^{-1} cm^{2} eq^{-1}$
Degree of dissociation is given by,
$\alpha=\dfrac{\Lambda _m}{\Lambda^{0} _m}=\dfrac{14.3}{390.71}=0.0366 \implies$3.66%


Which of the following statements is true?

  1. When an aqueous solution of NaCl is electrolysed, sodium metal is deposited at cathode

  2. There is no difference between specific conductivity and molar conductivity

  3. Silver nitrate solution can be stored in a copper container

  4. The addition of liquid bromine to iodide solution turns it violet


Correct Option: D
Explanation:

When an aqueous solution of NaCl is electrolysed, hydrogen is liberated at cathode. Specific conductivity and molar conductivity are different terms. Silver nitrate solution cannot be stored in a copper container as silver will get precipitated because of high reactivity of Cu than Ag.
The addition of liquid bromine to iodide solution turns it violet.

$Br _{2(l)}+2I^{-} _{(aq)}\rightarrow 2Br^{-} _{(aq)}+I _{2(aq)}$
Here, $Br _{2(l)}$ is Reddish-brown in color and $I _{2(aq)}$is violet in color

The ionic conductivity of $B{a^{2 + }}$ and $C{l^ - }$ at infinite dilution are 127 and 76 respectively. The equivalent conductivity of $BaC{l _2}$ at infinite dilution (in $oh{m^ - }\,c{m^2}\,e{q^{ - 1}}$) would be:

  1. 279

  2. 280

  3. 139.5

  4. 102


Correct Option: C
Explanation:

Given data,


$\lambda (Ba^{2+}) = 127$


$\lambda(Cl) = 76$

$\wedge _m^{\infty} = \lambda (Ba^{2+}) + 2\lambda(Cl)$

$= 127 + 2\times 76$

$\wedge _m^{\infty} = 279$

We know that-

$\wedge _{eq}^{\infty} = \dfrac{\lambda _m^{\infty}(Ba^{2+})}{2} + \lambda _m^{\infty}(Cl^-)$ 

$= \dfrac{127}{2} + 76$

$\wedge _{eq}^{\infty} = 139.5\ {ohm}^{-1}{cm}^2$

Hence, option C is correct.

The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is:

  1. LiCl $>$ NaCl $>$ KCl

  2. KCl $>$ NaCl $>$ LiCl

  3. NaCl $>$ KCl $>$ LiCl

  4. LiCl $>$ KCl $>$ NaCl


Correct Option: B
Explanation:

Since, the anion are same, comparing the size of cation.

$Li^+<Na^+<K^+$
but due to hydration of cation, the radii order is inverted.
$Li^+>Na^+>K^+$
$Conductance \propto\cfrac{1}{radii}$
$\therefore$ The order is $KCl>NaCl>LiCl$

The equivalent conductivity conductivity of 1M ${{\text{H}} _{\text{2}}}{\text{S}}{{\text{O}} _{\text{4}}}$ solution would be if specific conductance is ${\text{26}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{S}}\,{\text{c}}{{\text{m}}^{ - 1}}$.

  1. $1.3 \times {10^2}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{e}}{{\text{q}}^{ - 1}}$

  2. $1.6 \times {10^2}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,$

  3. $13\,{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

  4. $1.3 \times {10^3}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{mo}}{{\text{l}}^{ - 1}}$


Correct Option: A
Explanation:

Normality = Molarity $ \times 2\,{\text{factor}}$ 

$ = 1 \times 2 = 2{\text{N}}$     $2{{\text{H}}^ + } + S{\text{c}}{{\text{m}}^2}$
$\Delta eq = \dfrac{{\kappa  \times 1000}}{N} = \dfrac{{2.6 \times {{10}^{ - 2}} \times 1000Sc{m^{ - 12 = 2}}}}{2}$  $1Lt = {10^3}c{m^3}$
$ = 1.3 \times 10/{10^3}c{m^3}$
$ = 1.3 \times {10^{ - 1 + 3}}Sc{m^2} = 1.3 \times {10^2}Sc{m^2}e{q^{ - 1}}$

Equivalent conductance at infinite dilution for weak electrolyte HF:

  1. can be determined by measurement of equivalent conductance at infinite dilution for dilute solution of $HCL, \; HBr$ and $HI$

  2. can be determined by measurement of equivalent  conductance at infinite dilution for very dilute $HF$ solutions

  3. can best be determined from measurements on dilute

  4. can not be calculated


Correct Option: B
Explanation:
For weak electrolytes, equivalent conductance at infinite dilution can be determined by their very dilute solutions. Hence, for $HF$. equivalent conductance at infinite dilution can be determined by measurements on very dilute $HF$ solutions.

The conductivity of a saturated solution of $Ba{ SO } _{ 4 }$ is $306\times { 10 }^{ -6 }{ ohm }^{ -1 }{ cm }^{ -1 }$ and its equivalent conductance is $1.53 \ { ohm }^{ -1 }{ cm }^{ 2 }{ equiv }^{ -1 }$. 


The ${ K } _{ sp }$ for ${ BaSO } _{ 4 }$ will be :

  1. $4\times { 10 }^{ -12 }$ 

  2. $2.5\times { 10 }^{ -9 }$

  3. $2.5\times { 10 }^{ -13 }$ 

  4. $4\times { 10 }^{ -6 }$


Correct Option: D
Explanation:
Given : $\wedge eq=1.53 \Omega^{-1}eq^{-1}$

$\wedge = 3.06\times 10^{-6}\Omega^{-1}$

Solubilicty, $s=\dfrac{\wedge\times 1000}{\wedge eq}=\dfrac{3.06\times 10^{-6}\times 10^{3}}{1.53}=2\times 10^{-3}M$

$K _{\wedge p}=[Ba^{2+}][SO _{4}^{2-}]=S^{2}$

$K _{\wedge p}=(2\times 10^{-3})^{2}=4\times 10^{-6}M^{2}$

Therefore, the correct option is D.

If the specific resistance of a solution of concentration C g equivalent/litre is R, then its equivalent conductance is:

  1. $\dfrac{100R}{C}$

  2. $\dfrac{RC}{1000}$

  3. $\dfrac{1000}{RC}$

  4. $\dfrac{C}{1000R}$


Correct Option: C
Explanation:

Specific resistance for $C _g eq/lt=R$

Dont know the meaning of conductance it must be conductance
Conductance of solution $=k=\dfrac{1}{k}$
Equivalent conductance $=\dfrac{k\propto 1000}{c}$
                                         $=\dfrac{1000}{RC}$

The resistance of $0.2\ M$ solution of an electrolyte is $50\ \Omega$.The specific conductance of the solution is $1.3\ S\ m^{-1}$. If the resistance of the $0.4\ M$ solution of the same electrolyte is $260\ \Omega$, its molar conductivity is :

  1. $62.5\ S\ m^{2} mol^{-1}$

  2. $6250\ S\ m^{2} mol^{-1}$

  3. $6.25\ \times10^{-4}S\ m^{2} mol^{-1}$

  4. $625\times10^{-4}\ S\ m^{2} mol^{-1}$


Correct Option: C

The equivalent conductivity of monobasic acid at infinite dilution is 348 $ohm^{-1}$ $cm^2$ $eq^{-1}$. If the resistivity of the solution containing 15 g acid (molar mass 49) in 1 litre is 18.5 ohm cm, what is the degree of dissociation of acid?

  1. 45.9%

  2. 40.2%

  3. 60.4%

  4. 50.7%


Correct Option: D
Explanation:

equivalent conductivity of monobar'c acid at infinite dilution $\wedge^{\circ} m = 348 \Omega^{-1} cm^2 eq^{-1}$

Amount of acid $= 15 g$
Molar mass $= 49$
Molarity = $\dfrac{15}{49} / 1 \, litre = 0.306 M$
Resistivity = $18.5$ ohm cm
conductivity = $\dfrac{1}{18.5} = 0.054$
Molar conductivity = $\dfrac{0.054}{0.306 \times 10^{-3}}$
$\wedge _m  = 176.64$
dissociation constant $\alpha = \dfrac{\wedge _m}{\wedge _m^{\circ}} = \dfrac{176.64}{348}$
$= 0.507$
$50.7 \%$
option $D$

The conductivities at infinite dilution of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{Cl,NaOH}}$ and $\text{NaCl}$ are 130, 218, 120 ${\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$. If equivalent conductance of N/100 solution of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{OH}}$ is 10, then degree of dissociation of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{OH}}$ at this dilution is:

  1. 0.005

  2. 0.043

  3. 0.01

  4. 0.02


Correct Option: B
Explanation:

$NH _4Cl+NaOH\longrightarrow NH _4OH+NaCl$


$\therefore \wedge _m^{\infty}$ $ _{NH _4OH}=\wedge _m^{\infty}$ 

$ _{NH _4Cl}+\wedge _m^{\infty}$ $NaOH-\wedge _m^{\infty}$ $ _{NaCl}$

$\wedge _m^{\infty}=130+218-120$
$\implies \wedge _m^{\infty}=228$ $scm^2eq^2$

$\alpha=\cfrac{\wedge _m}{\wedge _{m^{\infty}}}=\cfrac{10}{228}=0.0438$

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