Calculate the degree of dissociation of $0.01 \ M$ acetic acid at $25^{\circ} C$ given that the molar conductivity at concentration $c$ is $\Lambda_{c} = 15.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and the limiting molar conductivity is $\Lambda_0 = 300 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

For a salt $XY$, which is a strong electrolyte, the plot of $\Lambda_m$ versus $\sqrt{C}$ has a slope of $-90.0 \ S \ cm^2 \ mol^{3/2} \ L^{1/2}$ at $298 \ K$. At $0.01 \ M$ concentration of $XY$, the value of $\Lambda_m$ is $145.5 \ S \ cm^2 \ mol^{-1}$. The limiting molar conductivity of $Y^-$ ion ($\lambda^\circ_{Y^-}$, in $S \ cm^2 \ mol^{-1}$) at $298 \ K$ will be (Given $\lambda^\circ_{X^+} = 74.0 \ S \ cm^2 \ mol^{-1}$) (in $.0$)

Molar conductivities $\left(\Lambda_{m}^{\circ}\right)$ at infinite dilution of $NaCl$,$HCl$,and $CH_{3}COONa$ are $126.4$,$425.9$,and $91.0 \ S \ cm^{2} \ mol^{-1}$ respectively. $\Lambda_{m}^{\circ}$ for $CH_{3}COOH$ will be $:-$

The equivalent conductances of two strong electrolytes at infinite dilution in $H_2O$ (where ions move freely through a solution) at $25 ^\circ C$ are given below:
$\Lambda _{CH_3COONa}^o = 91.0 \ S \ cm^2 / equiv.$
$\Lambda _{HCl}^o = 426.2 \ S \ cm^2 / equiv.$
What additional information/quantity one needs to calculate $\Lambda ^o$ of an aqueous solution of acetic acid?

Assertion: Electrical conductivity of copper increases with increase in temperature.
Reason: The electrical conductivity of metals is due to the motion of electrons.

$A$ $5.0 \, mmol \, dm^{-3}$ aqueous solution of $KCl$ has a conductance of $0.55 \, mS$ when measured in a cell with a cell constant of $1.3 \, cm^{-1}$. The molar conductivity of this solution is ....... $mS \, m^{2} \, mol^{-1}$. (Round off to the Nearest Integer)