Question

$x$ mmol of ${XeF}_4$ quantitatively oxidized KI to $I_2$ and liberated $Xe$, along with formation of $KF$. This iodine required 20 ml of decinormal hypo solution for exact titration. The value of $x$ is:

• A. $0.5$
• B. $1.0$
• C. $2.0$
• D. $5.0$

Answer 1

Answer: (A)

$0.5$

$2N{a}_2{S}_2{O}_3+I_2=N{a}_2{S}_4{O}_6+2NaI$ (Balanced)

$20\times 0.1$ $10\times 0.1$

$mmoles$ $mmoles$

$\to 1$ $mmole$ of $I_2$ is formed in the below reaction.

$X_e{F}_4+4KI⟶Xe+2I_2+4KF$ (Balanced)

$1$ $mmole$

For $1$ $mmole$ of $Xe{F}_4$ $2$ $moles$ of $I_2$ are formed.

Therefore, moles of $Xe{F}_4=0.5$ $mmmole$