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Calculation the normality of 30 volume solution of hydrogen peroxide |
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Answer» SOLUTION :Calculation of MASS of `H_(2)O_(2)` per litre of the solution The decomposition of `H_(2)O_(2)` takes PLACE as follows: `2H_(2)O_(2)to 2H_(2)O+O_(2) ` `{:(2(2xx1+2xx16),22400ml),(=68g,"at N.T.P):}` From the equation 22400mL of `O_(2)` at N.T.P are obtained from `H_(2)O_(2)` from `H_(2)O_(2)` =68g 30ML of `O_(2)` at N.T.P are obtained from `H_(2)O_(2)=(68)/(22400)xx30=0.091g` Now, by defination 1mL of hydrogen peroxide solution weights =0.091g 1000ML of hydrogen perooxide solution will weigh=`0.091xx1000=91.0g` The strength of the solution=91//g STEP-II Calculation of equivalent mass of `H_(2)O_(2)` Consider the equation: `underset("68 parts by mass")(2H_(2)O_(2))to 2H_(2)O+underset("32 parts by mass")(O_(2)) ` Now, 32 parts by mass of `O_(2)` are envolved from `H_(2)O_(2)` =68parts 8 parts by mass of `O_(2)` are evolved from `H_(2)O_(2)` `=(68)/(32)xx8=17` parts Thus equivalent mass of `H_(2)O_(2)`=17 or 17g/equiv STEP-III Calculation of normality of the solution `"Normality of solution"=("Strength of solution")/("Equivalent mass")=(91g//L)/(17g//"equiv")=5.35"equiv"L^(-1)=5.35N.` |
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