25 ml of hydrogen peroxide solution were added to excess of acidified potassiutm iodide solution. The iodine so liberated required 20mL of 0.1 sodium this sulphate solution. Calculate the strenth in terms of normality, percentage and volume.

25 ml of hydrogen peroxide solution were added to excess of acidified

1.	25 ml of hydrogen peroxide solution were added to excess of acidified potassiutm iodide solution. The iodine so liberated required 20mL of 0.1 sodium this sulphate solution. Calculate the strenth in terms of normality, percentage and volume.
Answer» Solution :Let the molarity of `H_(2)O_(2)=N_(x)` 20mL of 0.1 N `Na_(2)SO_(3)-=20mL of 0.1 I_(2)=25mL.N_(x).H_(2)O_(2)` Normality of `H_(2)O_(2)(N_(x))=(20xx0.1)/(25)=0.08N` STRENGTH of `H_(2)O_(2)`=Normality`xx`Equivalent mass=`0.08xx17=1.36gL^(-1)` Percentage of `H_(2)O_(2)=(1.36)/(1000)xx100=0.136` This means that 0.136g of `H_(2)O_(2)` are PRESENT in 100mL solution or 0.00136g of `H_(2)O_(2)` are present in 1ML of the solution Now, 68g of `H_(2)O_(2)` give oxygen at N.T.P. =22400mL 0.00136g of `H_(2)O_(2)` given oxygen at N.T.P.`=(22400)/(68)xx0.00136=0.448mL` Thus. 1mL of `H_(2)O_(2)` givwes 0.048mL of oxgyen at N.T.P.