Tag: electrochemistry

Quizzes (9) Questions (132)

Questions Related to electrochemistry

Multiple choice conductivity and its types electrochemistry

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

  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. $1.3 \times {10^3}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{mo}}{{\text{l}}^{ - 1}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}}$

Multiple choice conductivity and its types electrochemistry

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice conductivity and its types electrochemistry

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. $4\times { 10 }^{ -6 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice conductivity and its types electrochemistry

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. $\dfrac{C}{1000R}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $=\dfrac{k\propto 1000}{c}$
                                         $=\dfrac{1000}{RC}$

Multiple choice conductivity and its types electrochemistry

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. $625\times10^{-4}\ S\ m^{2} mol^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Molar conductivity is calculated as (kappa * 1000) / M. First, find the cell constant (G*) using the 0.2 M solution: G* = kappa * R = 1.3 * 50 = 65 m^-1. For the 0.4 M solution, kappa = G* / R = 65 / 260 = 0.25 S/m. Molar conductivity = kappa / concentration = 0.25 / 400 (converting 0.4 M to 400 mol/m^3) = 6.25 * 10^-4 S m^2 mol^-1.

Multiple choice conductivity and its types electrochemistry

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%
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $D$

Multiple choice conductivity and its types electrochemistry

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$

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

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

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Which of the following is an insulator?

  1. Wood

  2. Iron

  3. Graphite

  4. Silver
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Insulators:

  1. conduct electricity.

  2. do not conduct electricity.

  3. conduct electricity only at low temperatures.

  4. conduct electricity at room temperature.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Insulators are materials that prevent the flow of electric current because they have very few free electrons or ions to carry charge. Conductors, by contrast, allow electricity to pass through them easily.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

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)?

  1. $5\times10^{-4} M$
  2. $2\times10^{-6} M$
  3. $2\times10^{-5} M$
  4. $4 \times10^{-6} M$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mixing 100 ml of 0.1M KCN and 100 ml of 0.1M HCl results in a total volume of 200 ml containing 0.01 moles of HCN and neutralized chloride ions. The concentration of HCN is 0.01 mol / 0.2 L = 0.05 M. Since HCN is a weak acid, [H+] = sqrt(Ka * C) = sqrt(5 * 10^-6 * 0.05) = sqrt(2.5 * 10^-7) = 5 * 10^-4 M.