1.

A 10 mL of K_(2)Cr_(2)O_(7) solution , liberated iodine from KI solution .The liberatediodine was titrated by 16 mL of M/25 sodium thiosulphate solution. Calculate the concentration of K_(2)Cr_(2)O_(7) solution per litre .

Answer»

Solution :`{:(Cr_(2)O_(7)^(2-) , +14H^(+)+6I^(-), =2Cr^(3+) +7H_(2)O+3I_(2)),(+ 12,,+6):}`
`{:(S_(2)O_(3)^(2-), +1/2 I_(2) = 1/2 S_(4)O_(6)^(2-) ,+I^(-1)),(+4,,+5 ):}`
EQ. wt of `K_(2)Cr_(2)O_(7) = (" MOL .wt")/( " change in ON per mole") `
` = (294.18)/6`
`= 49.03`
m.e of 10 mL of `K_(2)Cr_(2)O_(7)` solution = m.e of iodine
= m.e of sodium thiosulphate
` = 1/25xx 16 = 0.64`
` :. ` equivalent of 10 mL of `K_(2)Cr_(2)O_(7) ` solution ` = (0.64)/(1000) = 0.00064`
` :. ` weight per10 mL ` = 0.00064 xx 49.03`
` = 0.0313 `g
` :. ` concentration of `K_(2)Cr_(2)O_(7)` in grams per litre ` = 0.0313 xx 100`
` = 3.13 `g/L


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