The slope of $\mathrm{\pi }$ vs $\mathrm{C}$ was made against insulin concentration $\left(\mathrm{C}\right)$ in $\mathrm{g} {\mathrm{mL}}^{-1}$ and temperature $27°\mathrm{C}$. If $\mathrm{C}$ is in $\mathrm{g} {\mathrm{mL}}^{-1}$  and the slope of line obtained is $4.65\times {10}^{-3} \mathrm{g}/\mathrm{mL}$, which of the following are not the molar mass (in $\mathrm{g} {\mathrm{mol}}^{-1}$) of insulin?

A graph showing the variation of osmotic pressure $\left(\mathrm{\pi }\right)$ versus molar concentration $\left(\mathrm{C}\right)$ of an aqueous solution at temperature $\mathrm{T}$ is given below.

The slope of the line doesn't represent

A dilute solution contains $\text{'}\mathrm{m}\text{'}$ mol of solute $\mathrm{A}$ in $1 \mathrm{kg}$ of a solvent with molal elevation constant ${\mathrm{K}}_{\mathrm{b}},$ the solute dimerises in solution as $2\mathrm{A}\rightleftharpoons {\mathrm{A}}_2$.

Show that the equilibrium constant $\mathrm{K}$ for this dimer formation is $\mathrm{K}=\frac{{\mathrm{K}}_{\mathrm{b}}\left({\mathrm{K}}_{\mathrm{b}}\mathrm{m}-{\mathrm{\Delta T}}_{\mathrm{b}}\right)}{{\left(2{\mathrm{\Delta T}}_{\mathrm{b}}-{\mathrm{K}}_{\mathrm{b}}\mathrm{m}\right)}^2}$ , where ${\mathrm{\Delta T}}_{\mathrm{b}}$ is the elevation in boiling point for the given solution.