Question

Question #769cb

Answer 1

`"Molarity" = 0.53M` `KClO_3`

Explanation:

`"Molarity"=("moles of the solute")/"volume of the solution" = "moles"/"liters"`

We can begin by converting grams to moles
*The molar mass for `KClO_3` is `122.55g`

`9.14g` x `(1 " mole")/(122.55g)`

`9.14cancel(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 `mL` to `L`

We set up a conversion factor:

`(0.001L)/(1cancel(mL))` x `(140cancel(mL))/(1)= 0.14L`

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

`"Molarity"="moles"/"liters"`

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

`"Molarity" = 0.53M` `KClO_3`