At $298 \ K$ the molar conductivities at infinite dilution $(\Lambda_m^{\circ})$ of $NH_4Cl, KOH$ and $KCl$ are $152.8, 272.6$ and $149.8 \ S \ cm^2 \ mol^{-1}$ respectively. The $\Lambda_m^{\circ}$ of $NH_4OH$ in $S \ cm^2 \ mol^{-1}$ and $\%$ dissociation of $0.01 \ M \ NH_4OH$ with $\Lambda_m = 25.1 \ S \ cm^2 \ mol^{-1}$ at the same temperature are
- A$275.6, 0.91$
- B$275.6, 9.1$
- C$266.6, 9.6$
- D$30, 84$
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Similar Questions
Assertion : On increasing dilution,the specific conductance keeps on increasing.
Reason : On increasing dilution,the degree of ionization of a weak electrolyte increases and the mobility of ions also increases.
Reason : On increasing dilution,the degree of ionization of a weak electrolyte increases and the mobility of ions also increases.
MediumAIIMS 2015
View SolutionConductivity of $0.001 \ M$ aqueous solution of $Na_2SO_4$ is found to be $2.6 \times 10^{-3} \ S \ cm^{-1}$ at $25 \ ^oC$. If limiting molar conductance of $Na^+$ is $50 \ S \ cm^2 \ mol^{-1}$,then limiting molar conductance of $SO_4^{2-}$ will be .............. $S \ cm^2 \ mol^{-1}$ (neglect conductivity of water).
Medium
View SolutionThe molar conductivity of a $0.04 \text{ M}$ $AB_2$ type salt solution at $300 \text{ K}$ is $200 \text{ } \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$. Find the conductivity.
DifficultMHT CET 2026
View SolutionThe molar conductivity of $KCl$ solutions at different concentrations at $298 \, K$ are given below:
$c^{1/2} / (mol \, L^{-1})^{1/2}$ $\Lambda_m / S \, cm^2 \, mol^{-1}$ $0.000198$ $148.61$ $0.000309$ $148.29$ $0.000521$ $147.81$ $0.000989$ $147.09$
Show that a plot between $\Lambda_m$ and $c^{1/2}$ is a straight line. Determine the values of $\Lambda_m^o$ and $A$ for $KCl$.
| $c^{1/2} / (mol \, L^{-1})^{1/2}$ | $\Lambda_m / S \, cm^2 \, mol^{-1}$ |
| $0.000198$ | $148.61$ |
| $0.000309$ | $148.29$ |
| $0.000521$ | $147.81$ |
| $0.000989$ | $147.09$ |
Show that a plot between $\Lambda_m$ and $c^{1/2}$ is a straight line. Determine the values of $\Lambda_m^o$ and $A$ for $KCl$.
Medium
View SolutionThe resistance of a $0.1 \ N$ solution of acetic acid is $250 \ \Omega$ and the cell constant is $1.15 \ cm^{-1}$. The equivalent conductance of $0.1 \ N$ acetic acid in $\Omega^{-1} \ cm^2 \ eq^{-1}$ is:
Easy
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