Question

The concentration of $Na_{2}S_{2}O_{3}.5H_{2}O$ solution in grams per litre is $y$ , $10mL$ of which just de-colourised $15mL$ of $\frac{N}{20}$ iodine solution, calculate $\frac{y}{4 . 65}$ value?

Answer 1

$2Na_{2}S_{2}O_{3}+I_{2} \rightarrow Na_{2}S_{4}O_{6}+2Nal$
$n$ -factor of $I_{2}=2$
Therefore, number of $mmoles$ of $I_{2}=15\times \frac{M}{40}=\frac{15}{40}$
Number of $mmoles$ of $Na_{2}S_{2}O_{3}=\frac{15}{40}\times 2=\frac{15}{20}$ in $10mL$ solution.
Weight of $Na_{2}S_{2}O_{3}\cdot 5H_{2}O$ (in $g$ ) in $10mL$ solution $=\frac{15}{20}\times 10^{- 3}\times 248=0.186g$
Therefore, concentration of $Na_{2}S_{2}O_{3}\cdot 5H_{2}O=18.6g/L$
$\frac{y}{4 . 65}=\frac{18 . 6}{4 . 65}=4$