# Question 769cb

Apr 21, 2017

$\text{Molarity} = 0.53 M$ $K C l {O}_{3}$

#### Explanation:

$\text{Molarity"=("moles of the solute")/"volume of the solution" = "moles"/"liters}$

We can begin by converting grams to moles
*The molar mass for $K C l {O}_{3}$ is $122.55 g$

$9.14 g$ x $\frac{1 \text{ mole}}{122.55 g}$

$9.14 \cancel{g}$ x (1 " mole")/(122.55cancel(g)) = 0.0746 " moles"

We have our moles and we have our volume but note: it's not in $L$ so we must convert $m L$ to $L$

We set up a conversion factor:

$\frac{0.001 L}{1 \cancel{m L}}$ x $\frac{140 \cancel{m L}}{1} = 0.14 L$

Finally, using the formula for molarity given earlier...

$\text{Molarity"="moles"/"liters}$

"Molarity" = (0.0746 " moles " KClO_3)/(0.14L)#

$\text{Molarity} = 0.53 M$ $K C l {O}_{3}$