Plotting 1 / Lambda_{m} against c Lambda_{m} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a weak acid,the degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$.The dissociation constant $K_a$ is given by $K_a = \frac

Vedclass · Accepted Answer

$K_{a} \Lambda_{m}^{\circ}$

Vedclass · Answer

(A) For a weak acid,the degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$.The dissociation constant $K_a$ is given by $K_a = \frac{c \alpha^2}{1 - \alpha}$.Substituting $\alpha$,we get $K_a = \frac{c (\Lambda_m / \Lambda_m^{\circ})^2}{1 - (\Lambda_m / \Lambda_m^{\circ})} = \frac{c \Lambda_m^2}{\Lambda_m^{\circ}(\Lambda_m^{\circ} - \Lambda_m)}$.Rearranging gives $K_a \Lambda_m^{\circ} (\Lambda_m^{\circ} - \Lambda_m) = c \Lambda_m^2$.Dividing by $K_a \Lambda_m \Lambda_m^{\circ}$,we get $\frac{\Lambda_m^{\circ} - \Lambda_m}{\Lambda_m} = \frac{c \Lambda_m}{K_a \Lambda_m^{\circ}}$,which simplifies to $\frac{\Lambda_m^{\circ}}{\Lambda_m} - 1 = \frac{c \Lambda_m}{K_a \Lambda_m^{\circ}}$.Dividing by $\Lambda_m^{\circ}$,we get $\frac{1}{\Lambda_m} = \frac{c \Lambda_m}{K_a (\Lambda_m^{\circ})^2} + \frac{1}{\Lambda_m^{\circ}}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = 1 / \Lambda_m$ and $x = c \Lambda_m$:Slope $S = \frac{1}{K_a (\Lambda_m^{\circ})^2}$ and $y$-intercept $P = \frac{1}{\Lambda_m^{\circ}}$.Therefore,the ratio $P / S = \frac{1 / \Lambda_m^{\circ}}{1 / (K_a (\Lambda_m^{\circ})^2)} = K_a \Lambda_m^{\circ}$.

List-$I$	List-$II$
$A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$	$I$. Primary battery
$B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$	$II$. Discharging of secondary battery
$C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$	$III$. Fuel cell
$D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$	$IV$. Charging of secondary battery

$A$. Leclanche cell	$1$. Converts energy of combustion into electrical energy
$B$. Fuel cell	$2$. Rechargeable cell
$C$. Ni-Cd cell	$3$. At anode,$Zn \longrightarrow Zn^{2+} + 2e^{-}$

Similar Questions

At $300 \ K$,the conductivity of $0.01 \ mol \ dm^{-3}$ aqueous solution of acetic acid is $19.5 \times 10^{-5} \ S \ cm^{-1}$ and the limiting molar conductivity of acetic acid at the same temperature is $390 \ S \ cm^2 \ mol^{-1}$. The degree of dissociation of acetic acid is:

Which of the following statements is not correct?

Match the entries from Column $I$ and Column $II$ and choose the correct order.
$A$. Leclanche cell $1$. Converts energy of combustion into electrical energy
$B$. Fuel cell $2$. Rechargeable cell
$C$. Ni-Cd cell $3$. At anode,$Zn \longrightarrow Zn^{2+} + 2e^{-}$

Given the values of $ᴧ^{0}$ for $KCl, KNO_3, HCl, NaOAc,$ and $NaCl$ are $149.9, 146.0, 426.2, 91.0,$ and $226.5 \, S \, cm^2 \, eq^{-1}$ respectively,calculate the value of $ᴧ^{0}$ for $HOAc$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module