The equations involved are

$$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O] \times 2$$

$${H_2}{O_2} \to {O_2} + 2{H^ + } + 2{e^ - }] \times 5$$

$$2MnO_4^ - + 5{H_2}{O_2} + 6{H^ + } \to 2M{n^{2 + }} + 5{O_2} + 8{H_2}O$$

From this equation, we find that

Equivalent mass of $${H_2}{O_2},\,{M_{eq}} = {{Molar\,mass\,of\,{H_2}{O_2}} \over {2\,eq\,mo{l^{ - 1}}}} = {{34g\,mo{l^{ - 1}}} \over {2\,eq\,mo{l^{ - 1}}}} = 17\,g\,e{q^{ - 1}}$$

Mass of H₂O₂ in the given 1.0 g sample, $${m_{eq}} = {X \over {100}} \times 1.0\,g$$

Amount (in equivalents) of H₂O₂ in the given 1.0 g sample is

$${n_{eq}} = {m \over {{M_{eq}}}} = {{(X/100)g} \over {17\,g\,e{q^{ - 1}}}} = {X \over {17 \times 100}}eq$$

If N_KMnO4 is the normality of KMnO₄, the amount (in equivalents) of KMnO₄ consumed is

$$n{'_{eq}} = {V_{KMn{O_4}}}\,{N_{KMn{O_4}}} = \left( {{X \over {1000}}L} \right){N_{KMn{O_4}}}$$

Equating Eqs (1) and (2), we get

$$\left( {{X \over {1000}}L} \right){N_{KMn{O_4}}} = {X \over {17 \times 100}}eq$$

or, $${N_{KMn{O_4}}} = {{10} \over {17}}eq\,{L^{ - 1}} = 0.588\,eq\,{L^{ - 1}}$$