# How many moles of Ba(OH)_2 are present in 225 mL of 0.800 M Ba(OH)_2?

Jul 7, 2016

0.180 moles of $B a {\left(O H\right)}_{2}$ are present.

#### Explanation:

Molarity is represented by the equation below:

We know the molarity and the volume of solution. The volume does not have good units since it is given in terms of milliliters instead of liters.

225 mL can be converted into liters by using the conversion factor 1000mL = 1L.

When 225 mL is divided by 1000 mL/L, you obtain a volume of 0.225 L.

Now we can rearrange the equation to solve for the number of moles. This can be accomplished by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:

Moles of solute = $\text{liters of solution} \times M o l a r i t y$

Moles of solute = $0.225 L \times 0.800 M$

$0.180 m o l$