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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:

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

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

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

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

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

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

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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 12 : 0.2% (w/v) solution of NaOH is measured to have resistivity 870.0 mΩ m. The molar conductivity of the solution will be ______ ×102 mS dm2 mol-1. (Nearest integer) [2025]

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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]