# Question bbc92

##### 1 Answer
Aug 14, 2016

Let w gram of $K I {O}_{3}$ be dissolved in water to make a solution V L.

Now molar mass of $K I {O}_{3} = 39 + 127 + 3 \cdot 16 = 214 \frac{g}{\text{mol}}$

So w g KIO_3="w g"/(214g/"mol")=w/214mol#

So the strength of this solution will be
$= \text{no of moles of solute"/"Volume} \times 1000 M$

$= \frac{w}{214 V} \times 1000 M = S \left(M\right)$

The no of moles of solute present xL solution will be $x \times S \text{ mol}$

As the solute $K I {O}_{3}$ is a strong electrolyte which dissociates in water as follows , the solution will contain same no. of moles of $I {O}_{3}^{-}$ ions as it contains no.of moles of $K I {O}_{3}$

$K I {O}_{3} r i g h t \le f t h a r p \infty n s {K}^{+} + I {O}_{3}^{-}$