$A$ $0.5\,M\,NaOH$ solution offers a resistance of $31.6\,\Omega$ in a conductivity cell at room temperature. What shall be the approximate molar conductance of this $NaOH$ solution if the cell constant of the cell is $0.367\,cm^{-1}$? (in $S\,cm^2\,mol^{-1}$)

For a saturated solution of $CaF_2$,choose the correct relationship-

Calculate $\Lambda _m^o$ for $CaCl_2$ and $MgSO_4$ from the data given in the table below:

Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$	Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$
$H^{+} : 349.6$	$OH^{-} : 199.1$
$Na^{+} : 50.1$	$Cl^{-} : 76.3$
$K^{+} : 73.5$	$Br^{-} : 78.1$
$Ca^{2+} : 119.0$	$CH_3COO^{-} : 40.9$
$Mg^{2+} : 106.0$	$SO_4^{2-} : 160.0$

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 $74.5 \; g \; mol^{-1}$.

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

The conductivity of sodium chloride at $298 \ K$ has been determined at different concentrations and the results are given below:

Concentration $/ M$	$0.001$	$0.010$	$0.020$	$0.050$	$0.100$
${10^2} \times \kappa / S \, m^{-1}$	$1.237$	$11.85$	$23.15$	$55.53$	$106.74$

Calculate ${\Lambda _m}$ for all concentrations and draw a plot between ${\Lambda _m}$ and $c^{1/2}$. Find the value of $\Lambda _m^o$.