jee main chemistry important questions | Electrochemistry jee main questions FREE

Q 11 :

Given below is the plot of the molar conductivity vs $\sqrt{concentration}$ for KCl in aqueous solution.

If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100 $Ω$ , then the resistance of the same cell with the dilute solution is $‘ x ’ Ω$ .

The value of $x$ is ______ (Nearest integer) [2025]

0%

(150)

$Λ_{m} = \frac{1000 \times κ}{C} \Rightarrow κ = \frac{Λ_{m} \times C}{1000}$

$Thus \frac{κ_{1}}{κ_{2}} = \frac{Λ_{m_{1}} \times C_{1}}{Λ_{m_{2}} \times C_{2}} = \frac{150 \times {(0.1)}^{2}}{100 \times {(0.15)}^{2}} = \frac{2}{3}$ ...(i)

$Also$

$R = ρ \frac{L}{a} = \frac{1}{κ} \times \frac{L}{a} \Rightarrow \frac{κ_{1}}{κ_{2}} = \frac{R_{2}}{R_{1}} = \frac{100}{x}$ ...(ii)

$Equating (i) and (ii) gives x = 150$

Q 12 :

0.2% (w/v) solution of NaOH is measured to have resistivity 870.0 m $Ω$ m. The molar conductivity of the solution will be ______ $\times 10^{2}$ mS ${dm}^{2}$ ${mol}^{- 1}$ . (Nearest integer) [2025]

0%

(23)

$Molarity of NaOH (C) = \frac{(% w / v) \times 10}{M_{NaOH}}$

$= \frac{0.2 \times 10}{40} = 0.05 M$

$ρ = 870 mΩ m = 870 \times 10^{- 3} Ω \times 100 cm = 87 Ω cm$

$κ = \frac{1}{ρ} = \frac{1}{87} {S cm}^{- 1}$

$Λ_{m} = \frac{1000 κ}{C} {S cm}^{2} {mol}^{- 1}$

$= \frac{1000}{87 \times 0.05} {S cm}^{2} {mol}^{- 1} = 229.885 {S cm}^{2} {mol}^{- 1}$

$= 229.885 \times 1000 {mS (10}^{- 1} {dm)}^{2} {mol}^{- 1}$

$= 22.99 \times 10^{2} {mS dm}^{2} {mol}^{- 1}$

Q 13 :

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S ${cm}^{2}$ ${mol}^{- 1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S ${cm}^{2}$ ${mol}^{- 1}$ , respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S ${cm}^{2}$ ${mol}^{- 1}$ , its degree of dissociation is given by $x \times 10^{- 1}$ .

The value of $x$ is _____. (Nearest integer) [2025]

0%

(3)

By Kohlrausch law:

$Λ_{{NH}_{4} OH}^{\circ} = λ_{{NH}_{4}^{+}}^{\circ} + λ_{{OH}^{-}}^{\circ}$

$= (λ_{{NH}_{4}^{+}}^{\circ} + λ_{{Cl}^{-}}^{\circ}) + λ_{{OH}^{-}}^{\circ} - λ_{{Cl}^{-}}^{\circ}$

$= Λ_{{NH}_{4} Cl}^{\circ} + λ_{{OH}^{-}}^{\circ} - λ_{{Cl}^{-}}^{\circ}$

$= (185 + 170 - 70) {S cm}^{2} {mol}^{- 1}$

$= 285 {S cm}^{2} {mol}^{- 1}$

$α = \frac{Λ_{{NH}_{4} OH}}{Λ_{{NH}_{4} OH}^{\circ}} = \frac{85.5}{285} = 0.3 = 3 \times 10^{- 1}$

Q 14 :

Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide. [2023]

1

2

3

4

(3)

$C_{6} H_{5} COOH + NaOH \to C_{6} H_{5} COONa + H_{2} O$

The conductance first drops rapidly due to loss of more conducting $H^{+}$ already in the solution.
However soon, due to common ion effect of $C_{6} H_{5} {COO}^{-}$ the free $H^{+}$ concentration in solution is almost negligible, and then the added NaOH only increases the number of ions by forming more of $C_{6} H_{5} COONa$ .
At the end point, the conductance rises much more rapidly due to addition of ${OH}^{-}$ which has higher conductivity.

Q 15 :

Following figure shows dependence of molar conductance of two electrolytes on concentration. $Λ_{m}^{°}$ is the limiting molar conductivity.

The number of incorrect statement(s) from the following is __________ . [2023]

(A) $Λ_{m}^{°}$ for electrolyte A is obtained by extrapolation.

(B) For electrolyte B, $Λ_{m}$ vs $\sqrt{c}$ graph is a straight line with intercept equal to $Λ_{m}^{°}$ .

(C) At infinite dilution, the value of degree of dissociation approaches zero for electrolyte B.

(D) $Λ_{m}^{°}$ for any electrolyte A or B can be calculated using $λ °$ for individual ions.

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(2)

Both (A) and (C) are incorrect.

Q 16 :

The resistivity of a 0.8 M solution of an electrolyte is $5 \times 10^{- 3} Ω cm$ . Its molar conductivity is ________ $\times 10^{4} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . (Nearest Integer) [2023]

0%

(25)

$Resistivity (ρ) = 5 \times 10^{- 3} Ω cm$

$Molarity (M) = 0.8 M$

$Molar conductivity (λ_{M}) = \frac{1}{ρ} \times \frac{1000}{M}$

$= \frac{1}{5 \times 10^{- 3}} \times \frac{1000}{0.8}$

$= 0.25 \times 10^{6}$

$= 25 \times 10^{4} {cm}^{2} {mol}^{- 1}$

Q 17 :

$1 \times 10^{- 5}$ M ${AgNO}_{3}$ is added to 1 L of saturated solution of AgBr. The conductivity of this solution at 298 K is _________ $\times 10^{- 8}$ S $m^{- 1}$ .

[Given: $K_{sp} (AgBr) = 4.9 \times 10^{- 13}$ at 298 K

$λ_{{Ag}^{+}}^{\circ} = 6 \times 10^{- 3} {S m}^{2} {mol}^{- 1}$

$λ_{{Br}^{-}}^{\circ} = 8 \times 10^{- 3} {S m}^{2} {mol}^{- 1}$

$λ_{{NO}_{3}^{-}}^{\circ} = 7 \times 10^{- 3} {S m}^{2} {mol}^{- 1}$ ] [2023]

0%

(13040)

$10^{- 5} {M AgNO}_{3}$ is added to saturate AgBr.

$K_{s p} = (s + 10^{- 5}) (s)$

$= s (10^{- 5}) = 4.9 \times 10^{- 13}$

$s = 4.9 \times 10^{- 8}$

Solution has

$[{Ag}^{+}] = 10^{- 5}, [{NO}_{3}^{-}] = 10^{- 5}$

$[{Br}^{-}] = 5 \times 10^{- 8}$

$λ = \frac{K \times 1000}{C}$

$K_{Total} = \frac{λ_{1} C_{1}}{1000} + \frac{λ_{2} C_{2}}{1000} + \frac{λ_{3} C_{3}}{1000}$

$\Rightarrow \frac{6 \times 10^{- 3} \times 10^{4} \times 10^{- 5}}{1000} + \frac{7 \times 10^{- 3} \times 10^{4} \times 10^{- 5}}{1000} + \frac{8 \times 10^{- 3} \times 10^{4} \times 5 \times 10^{- 8}}{1000}$

$\Rightarrow 6 \times 10^{- 7} + 7 \times 10^{- 7} + 40 \times 10^{- 10}$

$= 13.04 \times 10^{- 7} {S cm}^{- 1} = 13.04 \times 10^{- 5} {S m}^{- 1}$

$= 13040 \times 10^{- 8} {S m}^{- 1}$

$= 13040$

Q 18 :

The specific conductance of 0.0025 M acetic acid is $5 \times 10^{- 5} {S cm}^{- 1}$ at a certain temperature. The dissociation constant of acetic acid is ______ $\times 10^{- 7}$ . Consider limiting molar conductivity of ${CH}_{3} COOH$ as 400 S ${cm}^{2}$ ${mol}^{- 1}$ . [2023]

0%

(66)

$Λ_{m} = \frac{1000 \times K}{C} = \frac{1000 \times 5 \times 10^{- 5}}{0.0025} = 20 {S cm}^{2} {mol}^{- 1}$

$α = \frac{20}{400} = \frac{1}{20}$

$K_{a} = \frac{C α^{2}}{1 - α^{2}} = \frac{0.0025 \times \frac{1}{20} \times \frac{1}{20}}{1 - \frac{1}{20}} = \frac{0.0025}{19 \times 20} = 66 \times 10^{- 7}$

Q 19 :

The number of correct statements from the following is ________.

(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes.

(B) The number of ions per unit volume that carry current in a solution increases on dilution.

(C) Molar conductivity increases with decrease in concentration.

(D) The variation in molar conductivity is different for strong and weak electrolytes.

(E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation. [2023]

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(3)

(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes.

(B) The number of ions per unit volume that carry current in a solution decreases on dilution.

(C) Molar conductivity increases with decrease in concentration.

(D) The variation in molar conductivity is different for strong and weak electrolytes.

(E) For weak electrolytes, the change in the molar conductivity with dilution is due to increase in degree of dissociation.

The correct statements are (A), (C) and (D).

Topic Question Set

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Q 12 :

0.2% (w/v) solution of NaOH is measured to have resistivity 870.0 m $Ω$ m. The molar conductivity of the solution will be ______ $\times 10^{2}$ mS ${dm}^{2}$ ${mol}^{- 1}$ . (Nearest integer) [2025]

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Q 14 :

Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide. [2023]

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Q 16 :

The resistivity of a 0.8 M solution of an electrolyte is $5 \times 10^{- 3} Ω cm$ . Its molar conductivity is ________ $\times 10^{4} Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . (Nearest Integer) [2023]

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Q 18 :

The specific conductance of 0.0025 M acetic acid is $5 \times 10^{- 5} {S cm}^{- 1}$ at a certain temperature. The dissociation constant of acetic acid is ______ $\times 10^{- 7}$ . Consider limiting molar conductivity of ${CH}_{3} COOH$ as 400 S ${cm}^{2}$ ${mol}^{- 1}$ . [2023]

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