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 $.0$)

$\Lambda_m^0(NH_4OH)$ is equal to:

What is the molar conductivity of $0.02 \text{ M}$ $KI$ solution if its conductivity is $4.37 \times 10^{-4} \text{ } \Omega^{-1} \text{cm}^{-1}$?

Resistance of a conductivity cell (cell constant $129 \; m^{-1}$) filled with $74.5 \; ppm$ solution of $KCl$ is $100 \; \Omega$ (labelled as solution $1$). When the same cell is filled with $KCl$ solution of $149 \; ppm$,the resistance is $50 \; \Omega$ (labelled as solution $2$). The ratio of molar conductivity of solution $1$ and solution $2$ is i.e.,$\frac{\wedge_{1}}{\wedge_{2}} = x \times 10^{-3}$. The value of $x$ is (Nearest integer). Given,molar mass of $KCl$ is $74.5 \; g \; mol^{-1}$.

At $291 \ K$,a saturated solution of $BaSO_4$ was found to have a specific conductivity of $3.648 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and that of the water used is $1.25 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$. If the ionic conductances of $Ba^{2+}$ and $SO_4^{2-}$ are $110$ and $136.6 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively,the solubility of $BaSO_4$ at $291 \ K$ will be (Atomic masses: $Ba=137, S=32, O=16$)

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).