Tag: electrochemistry

Questions Related to electrochemistry

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

conductivity and its types electrochemistry

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

Show answer
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:

conductivity and its types electrochemistry

  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

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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 :

conductivity and its types electrochemistry

  1. $4\times { 10 }^{ -12 }$ 

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

  3. $2.5\times { 10 }^{ -13 }$ 

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

Show answer
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:

conductivity and its types electrochemistry

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

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

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

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

Show answer
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 :

conductivity and its types electrochemistry

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

Show answer
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?

conductivity and its types electrochemistry

  1. 45.9%

  2. 40.2%

  3. 60.4%

  4. 50.7%

Show answer
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:

conductivity and its types electrochemistry

  1. 0.005

  2. 0.043

  3. 0.01

  4. 0.02

Show answer
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}$

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

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

Which of the following is an insulator?

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. Wood

  2. Iron

  3. Graphite

  4. Silver

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Correct Option: A
Explanation:

Insulators donot conduct electricity due to the absence of ions or electrons.

Iron is metal therefore has pool of electrons on the surface and is a conductor.
Graphite has extensive $\pi$ electron cloud and is thus a conductor.
Silver is a metal and is a very good conductor of electricity.
Wood is made of organic molecule, thus its an insulator.
Option A is correct

Insulators:

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. conduct electricity.

  2. do not conduct electricity.

  3. conduct electricity only at low temperatures.

  4. conduct electricity at room temperature.

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Correct Option: B

What is [CN-] in a solution prepared by mixing 100ml 0.1M $KCN$ and 100ml 0.1M $HCl$. (Ka of $HCN$ is 510^-6 M)?

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. $5\times10^{-4} M$

  2. $2\times10^{-6} M$

  3. $2\times10^{-5} M$

  4. $4 \times10^{-6} M$

Show answer
Correct Option: C