1.

In a 50 - mL solution of H_(2)O_(2)an excess of KI and dilute H_(2)SO_(4)were added The I_(2)so liberated required 20 mL of 0.1 N Na_(2)S_(2)O_(3)for complete reaction . Caculate the strength of H_(2)O_(2) in grams per litre .

Answer»

Solution :`2KL +H_(2)SO_(4) +H_(2)O_(2) to K_(2)SO_(4) +2H_(2)O + I_(2)`
`2Na_(2)S_(2)O_(3) +I_(2) to Na_(2)S_(4)O_(6) +2NaI `
m.e of `H_(2)O_(2)`in 5 mL = m.e of `I_(2)` = m.e of `Na_(2)S_(2)O_(3)`
` :. ` m.e of `H_(2)O_(2)` in 50 mL = `0.1 xx20 = 2 ` ...(EQN.1)
` :. ` m.e of `H_(2)O_(2)` in 1000 mL = `2/50 xx1000 = 40`
Equivalents per litre` = 2/50 xx1000 = 40 `
Equivalents per litre ` = 40/1000`..(Eqn . 3)
Grams per litre of `H_(2)O_(2) = 40/1000 xx 17` ...(Eqn .4)
`(" EQ. wt . of "H_(2)O_(2) = 34/2 = 17 )`


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