10ml of a solution of $H_2{O}_2$  labeled 10 volume strength just decolourise 100ml of potassium permanganate solution acidified with dilute  $H_{2}S{O}_4$  calculate the amount of potassium permanganate in the given solution.

 $2H_2{O}_2\to 2H_{2}O+O_2$
 $22400mLof{O}_2\to 68g{H}_2{O}_2$
$10mLof{O}_2\to ?$ 
 $W{H}_2{O}_2=\frac{68\times 10}{22400}g$
1ml of  $H_2{O}_2$ contain  $\frac{0.0303}{34}mol=0.00178$ equivalent weight 10ml of  $H_2{O}_2$
Therefore 0.0178 equivalent will be present in 100ml of kmno4 solution 
Amount of kmno4 in given simple  $=\frac{158}{5}\times 0.0178=0.563g$