The volume strength of $$1.5\,N\,{H_2}{O_2}$$   solution is

Solution :
$$\eqalign{ & {\text{Normality}} = 1.5\,N \cr & {\text{Equivalent weight of}}\,{H_2}{O_2} = 17 \cr & {\text{So, strength of the solutions,}} \cr & S = E \times N \cr & \,\,\,\,\, = 17 \times 1.5 = 25.5 \cr & \,\,\,\,\,\,\,\,\,\,2{H_2}{O_2} \to 2{H_2}O + {O_2} \cr & \,\,\,\,\, = 2 \times 34 \cr & \,\,\,\,\, = 68\,g \cr} $$
$$\because \,\,68\,g$$   of $${H_2}{O_2}$$  produce $${O_2}$$  at $$NTP = 22.4\,L$$
$$\therefore 25.5\,g$$   of $${H_2}{O_2}$$  will produce
$$\eqalign{ & = \frac{{22.4}}{{68}} \times 25.5 \cr & = 8.4\,L\,\,{\text{of}}\,{O_2} \cr} $$