Titration of `l_(2)` produced from `0.1045` g of primary standard `KlO_(3)` required `30.72` mL of sodium thiosulphate as shown below :
`lO_(3)^(-) + 5l^(-) + 6H^(+) to 3l_(2) + 3H_(2) O `
` l_(2) + 2S_(2)O_(3)^(2-) to 2l^(-) + S_(4)O_(6)^(2-)`
The molarity of sodium thiosulphate ion is :

Options

A. `0.079`
B. `0.095`
C. `0.084`
D. `0.064`

Answer

B

Explanation

`{:(lO_(3)^(-) + 5l^(-) + 6H^(+) to 3l_(2) + 3H_(2) O ),(ul(3l_(2) + 6S_(2)O_(3)^(2-) to 6l^(-) + 3S_(4)O_(6)^(2-))),(lO_(3)^(-) + 6S_(2)O_(3)^(2-) + 6H^(+) to l^(-) + 3S_(4)O_(6)^(2-)):}`
Mole of `KlO_(3) -= 6 " moles of " S_(2)O_(3)^(2-)`
Mole of `KlO_(3) = (0.1045)/(21.4) = 4.88 xx 10^(-4)`
Mole `S_(2)O_(3)^(2-) " used " = 4.88 xx 10^(-4) xx 6`
` = 2.93 xx 10^(-3)`
` (M xx 30.72)/1000 = 2.93 xx 10^(-3)`
` :. M = 0.095`