At the point of intersection of the two curves shown, the conc. of B is given by ...... for,$\mathrm{A}\to \mathrm{nB}$

At intersection point conc. reactant is equal to conc. product if x amount of B is formed at inter section point.

$\mathrm{A}\to \mathrm{nB}$

${\mathrm{A}}_0 0$

${\mathrm{A}}_0-\frac{\mathrm{x}}{\mathrm{n}}\mathrm{x}$

$\therefore {\mathrm{A}}_0-\frac{\mathrm{x}}{\mathrm{n}}=\mathrm{x}$

$\frac{{\mathrm{nA}}_{\mathrm{o}}-\mathrm{x}}{\mathrm{n}}=x$

${\mathrm{nA}}_{\mathrm{o}}-\mathrm{x}=\mathrm{nx}$

${\mathrm{nA}}_{\mathrm{o}}=\mathrm{x}+\mathrm{nx}$

${\mathrm{nA}}_{\mathrm{o}}=\mathrm{x}\left(\mathrm{n}+1\right)$

$\therefore \mathrm{x}=\frac{{\mathrm{nA}}_{\mathrm{o}}}{\mathrm{n}+1}$