Question

A mixture of 0.02 mole of $KBr{O}_3$ and 0.01 mole of $KBr$ was treated with excess of $KI$ and acidified. The volume of 0.1 $MN{a}_2{S}_2{O}_3$ solution required to consume the liberated iodine will be:

• A. 1200 mL
• B. 800 mL
• C. 1000 mL
• D. 1500 mL

Answer 1

The correct option is A 1200 mL

Calculating moles of $l_2$ produced

When mixture of $KBr{O}_3$ and $KBr$ is treated with $KI$ following reaction takes place:

$Br{O}_3^{-}+I^{-}\to I_2+B{r}^{-}$
n-factor of $Br{O}_3^{-}=1\left(5\right(-1\left)\right)=6$
n-factor of $I_2=20(0-(-1\left)\right)=2$

According to this reaction:
Equivalents of $Br{O}_3^{-}=$ Equivalents of $I_2$
$(n_{Br{o}_3^{-}}\times n-factor)=(n_{I_2}\times n-factor)$
$0.02\times 6=n{I}_2\times 2$
$n_{I_2}=0.06$

Thus, 0.02 mol of $Br{O}_3^{-}$ will produce 0.06 mol of $I_2$
KBr does not react with KI.
So, 0.06 mol of $I_2$ will be liberated.

Calculating volume of $N{a}_2{S}_2{O}_3$ required to consume the liberated iodine

$I_2+2N{a}_2{S}_2{O}_3\to 2NaI+N{a}_2{S}_4{O}_6$
According to this reaction
1 mol of $I_2$ will react with 2 mol of $N{a}_2{S}_2{O}_3$
Thus, 0.06 mol of $I_2$ will react with $2\times 0.06$ i.e., 0.12 mol of $N{a}_2{S}_2{O}_3$

And molarity $=\frac{no.ofmoles}{Volume\left(L\right)}$
Molarity of $N{a}_2{S}_2{O}_3$ =0.1 M (given)

Volume of $N{a}_2{S}_2{O}_3$ solution $=\frac{no.ofmoles}{Molarity}$

$=\frac{0.12}{0.1}=1.2L=1200mL$

Hence, 1200 mL of $N{a}_2{S}_2{O}_3$ solution is required.

So, option (B) is the correct answer.