Question

The concentration of $N{a}_2{S}_2{O}_3.5H_{2}O$ 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

Answer: 4

$2N{a}_2{S}_2{O}_3+I_2\to N{a}_2{S}_4{O}_6+2Nal$
$n$ -factor of $I_2=2$
Therefore, number of $mmoles$ of $I_2=15\times \frac{M}{40}=\frac{15}{40}$
Number of $mmoles$ of $N{a}_2{S}_2{O}_3=\frac{15}{40}\times 2=\frac{15}{20}$ in $10mL$ solution.
Weight of $N{a}_2{S}_2{O}_3\cdot 5H_{2}O$ (in $g$ ) in $10mL$ solution $=\frac{15}{20}\times 10^{-3}\times 248=0.186g$
Therefore, concentration of $N{a}_2{S}_2{O}_3\cdot 5H_{2}O=18.6g/L$
$\frac{y}{4.65}=\frac{18.6}{4.65}=4$