(D) According to Kohlrausch's law of independent migration of ions:$\Lambda_{m}^{\infty}(BaSO_{4}) = \lambda_{m}^{\infty}(Ba^{2+}) + \lambda_{m}^{\inf

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(D) According to Kohlrausch's law of independent migration of ions:$\Lambda_{m}^{\infty}(BaSO_{4}) = \lambda_{m}^{\infty}(Ba^{2+}) + \lambda_{m}^{\infty}(SO_{4}^{2-})$We can express this in terms of the given electrolytes:$\Lambda_{m}^{\infty}(BaSO_{4}) = \Lambda_{m}^{\infty}(BaCl_{2}) + \Lambda_{m}^{\infty}(H_{2}SO_{4}) - 2\Lambda_{m}^{\infty}(HCl)$Substituting the given values:$\Lambda_{m}^{\infty}(BaSO_{4}) = 280 + 860 - 2(426)$$= 1140 - 852$$= 288 \, S \, cm^{2} \, mol^{-1}$

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Medium

The molar conductivities at infinite dilution of barium chloride,sulphuric acid and hydrochloric acid are $280$,$860$ and $426 \, S \, cm^{2} \, mol^{-1}$ respectively. The molar conductivity at infinite dilution of barium sulphate is $...... \, S \, cm^{2} \, mol^{-1}$ ($Round$ off to the Nearest Integer).

JEE MAIN 2021 All JEE MAIN

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$102$
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$328$
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$288$

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Conductor and Conductance and Cell constant

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The limiting molar conductivities of the given electrolytes at $298 \ K$ follow the order: $\lambda^{\circ}_{(K^{+})} = 73.5, \lambda^{\circ}_{(Cl^{-})} = 76.3, \lambda^{\circ}_{(Ca^{2+})} = 119.0, \lambda^{\circ}_{(SO_{4}^{2-})} = 160.0 \ S \ cm^{2} \ mol^{-1}$.

MediumKVPY 2017

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Resistance of $0.1\, M\, KCl$ solution in a conductance cell is $300\, \Omega$ and conductivity is $0.013\, S\, cm^{-1}$. The value of cell constant is

Medium

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If the conductivity of mercury at $0^{\circ} \, C$ is $1.07 \times 10^{6} \, S \, m^{-1}$ and the resistance of a cell containing mercury is $0.243 \, \Omega$,then the cell constant of the cell is $x \times 10^{4} \, m^{-1}$. The value of $x$ is ...... (Nearest integer).

DifficultJEE MAIN 2021

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Calculate $\Lambda _{HOAc}^{\infty }$ using appropriate molar conductances of the electrolytes listed below at infinite dilution in $H_2O$ at $25\ ^oC$.

Electrolyte $\Lambda ^{\infty } (S\ cm^2\ mol^{-1})$

$KCl$ $149.9$

$KNO_3$ $145.0$

$HCl$ $426.2$

$NaOAc$ $91.0$

$NaCl$ $126.5$

Electrolyte	$\Lambda ^{\infty } (S\ cm^2\ mol^{-1})$
$KCl$	$149.9$
$KNO_3$	$145.0$
$HCl$	$426.2$
$NaOAc$	$91.0$
$NaCl$	$126.5$

MediumAIEEE 2005

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Resistance of $0.05 \ M$ electrolytic solution at $298 \ K$ temperature is $30.0 \ \Omega$. The cross-sectional area of the conductivity cell having $Pt$ electrodes is $3.8 \ cm^{2}$ and the distance between the two electrodes is $1.5 \ cm$. What is the molar conductivity of the electrolytic solution?

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The limiting molar conductivities of the given electrolytes at $298 \ K$ follow the order: $\lambda^{\circ}_{(K^{+})} = 73.5, \lambda^{\circ}_{(Cl^{-})} = 76.3, \lambda^{\circ}_{(Ca^{2+})} = 119.0, \lambda^{\circ}_{(SO_{4}^{2-})} = 160.0 \ S \ cm^{2} \ mol^{-1}$.

Resistance of $0.1\, M\, KCl$ solution in a conductance cell is $300\, \Omega$ and conductivity is $0.013\, S\, cm^{-1}$. The value of cell constant is

If the conductivity of mercury at $0^{\circ} \, C$ is $1.07 \times 10^{6} \, S \, m^{-1}$ and the resistance of a cell containing mercury is $0.243 \, \Omega$,then the cell constant of the cell is $x \times 10^{4} \, m^{-1}$. The value of $x$ is ...... (Nearest integer).

Calculate $\Lambda _{HOAc}^{\infty }$ using appropriate molar conductances of the electrolytes listed below at infinite dilution in $H_2O$ at $25\ ^oC$. Electrolyte $\Lambda ^{\infty } (S\ cm^2\ mol^{-1})$ $KCl$ $149.9$ $KNO_3$ $145.0$ $HCl$ $426.2$ $NaOAc$ $91.0$ $NaCl$ $126.5$

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Calculate $\Lambda _{HOAc}^{\infty }$ using appropriate molar conductances of the electrolytes listed below at infinite dilution in $H_2O$ at $25\ ^oC$.

Electrolyte $\Lambda ^{\infty } (S\ cm^2\ mol^{-1})$

$KCl$ $149.9$

$KNO_3$ $145.0$

$HCl$ $426.2$

$NaOAc$ $91.0$

$NaCl$ $126.5$