Question

Chile saltpetre is a natural source of $NaN{O}_3$ which also contains $NaI{O}_3$. The $NaI{O}_3$ can be used as a source of iodine produced in the following reactions:

${I{O}_3}^{-}+3{HS{O}_3}^{-}\to I^{-}+3H^{+}+3{S{O}_4}^{2-}$ ......(i)

$5I^{-}+{I{O}_3}^{-}+6H^{+}\to 3I_2\left(s\right)+3H_{2}O$ ....... (ii)

$1.0$ L of the starting solution which contains $5.80$ g $NaI{O}_3$ is treated with stoichiometric quantity of $NaHS{O}_3$. Further, a quantity of the starting solution is added to the reaction mixture to bring about the second reaction $[I=127,Na=23,S=32]$.

How many equivalents of ${I{O}_3}^{-}$ are used in the above reaction in step (ii)?

• A. $0.6$
• B. $0.5$
• C. $0.06$
• D. $0.035$

Answer 1

The correct option is D $0.035$

The molar mass of $NaI{O}_3$ $=23+127+3\left(16\right)=198$ g/mol

The number of moles of $NaI{O}_3$ is $\frac{1.16}{198}$ mol.

No. of equivalents of $NaI{O}_3$ = $n_{factor}\times no.ofmoles$

The number of equivalents of $NaI{O}_3$ is $\frac{6\times 1.16}{198}$ mol $=0.035$

Hence, the correct answer is $\left(D\right)$.