Given : Mass of compound taken = 0.50 g; Vol. of $\mathrm{H_2SO_4}$ = 50 mL Molarity of $\mathrm{H_2SO_4}$ = 0.5 M; Vol. of NaOH required = 60 mL Molarity of NaOH required = 0.5 M Method adopted : Kjeldahl’s method Formula used : % of N = $\frac{1.4 \times M \times 2\left[V - \frac{V_1}{2}\right]}{m}$ ... (i) M = 0.5 M (Molarity of $\mathrm{H_2SO_4}$), V = 50 mL, $V_1$ = 60 mL m = 0.5 g (mass of compound) By substituting the values in the formula, we get, % of N = $\frac{1.4 \times 0.5 \times 2\left(50 - \frac{60}{2}\right)}{0.5}$ = 56 ∴ % of N in the given compound = 56%