Given,

$\mathrm{k}=1.237\times {10}^{-2}\mathrm{S} {\mathrm{m}}^{-1}, \mathrm{c}=0.001 \mathrm{M}$

Then, $\mathrm{k}=1.237\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}, {\mathrm{c}}^{1/2}=0.0316 {\mathrm{M}}^{1/2}$

$\therefore {\mathrm{\Lambda }}_{\mathrm{m}}=\frac{\mathrm{k}}{\mathrm{c}}$

$=\frac{1.237\times {10}^{-1} \mathrm{S} {\mathrm{cm}}^{-1}}{0.001 \mathrm{mol} {\mathrm{L}}^{-1}}\times \frac{1000 {\mathrm{cm}}^3}{\mathrm{L}}$

$=123.7 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$ Given,

$\mathrm{k}=11.85\times {10}^{-2} \mathrm{S} {\mathrm{m}}^{-1}, \mathrm{c}=0.010 \mathrm{M}$

Then, $\mathrm{k}=11.85\times {10}^{-4} \mathrm{S} {\mathrm{m}}^{-1}, {\mathrm{c}}^{1/2}=0.1 {\mathrm{M}}^{1/2}$

$\therefore {\mathrm{\Lambda }}_{\mathrm{m}}=\frac{\mathrm{k}}{\mathrm{c}}$

$=\frac{11.85\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}}{0.010 \mathrm{mol} {\mathrm{L}}^{-1}}\times \frac{1000 {\mathrm{cm}}^3}{\mathrm{L}}$

$=118.5 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$ Given,

$\mathrm{k}=23.15\times {10}^{-2} \mathrm{S} {\mathrm{m}}^{-1}, \mathrm{c}=0.020 \mathrm{M}$

Then, $\mathrm{k}=23.15\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}, {\mathrm{c}}^{1/2}=0.1414 {\mathrm{M}}^{1/2}$

$\therefore {\mathrm{\Lambda }}_{\mathrm{m}}=\frac{\mathrm{k}}{\mathrm{c}}$

$=\frac{23.15\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}}{0.020 \mathrm{mol} {\mathrm{L}}^{-1}}\times \frac{1000 {\mathrm{cm}}^3}{\mathrm{L}}$

$=115.8 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$Given,

$\mathrm{k}=55.53\times {10}^{-2} \mathrm{S} {\mathrm{m}}^{-1}, \mathrm{c}=0.050 \mathrm{M}$

Then, $\mathrm{k}=55.53\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}, {\mathrm{c}}^{1/2}=0.2236 {\mathrm{M}}^{1/2}$

$\therefore \mathrm{k}=\frac{\mathrm{k}}{\mathrm{c}}$
$=\frac{55.53\times {10}^{-4}\mathrm{S} {\mathrm{cm}}^{-1}}{0.050 \mathrm{mol} {\mathrm{L}}^{-1}}\times \frac{1000 {\mathrm{cm}}^3}{\mathrm{L}}$

$=111.1 1 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$ Given,

$\mathrm{k}=106.74\times {10}^{-2} \mathrm{S} {\mathrm{m}}^{-1}, \mathrm{c}=0.100 \mathrm{M}$

Then, $\mathrm{k}=1.06.74\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}, {\mathrm{c}}^{1/2}=0.3162 {\mathrm{M}}^{1/2}$

$\therefore {\mathrm{\Lambda }}_{\mathrm{m}}=\frac{\mathrm{k}}{\mathrm{c}}$
$=\frac{106.74\times {10}^{-4} \mathrm{S} {\mathrm{cm}}^{-1}}{0.100 \mathrm{mol} {\mathrm{L}}^{-1}}\times \frac{1000 {\mathrm{cm}}^3}{\mathrm{L}}$

$=106.74 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$ Now, we have the following data:

Since the line interrupts ${\mathrm{\Lambda }}_{\mathrm{m}}$ at $124.0 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}, {\mathrm{\Lambda }}_{\mathrm{m}}^0=124.0 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$