# What would be the formula of a barium iodide hydrate that contains 0.0243*mol BaI_2, and 0.75*mol water?

Jun 10, 2017

Well it would be $B a {I}_{2} \cdot 30 {H}_{2} O$

#### Explanation:

And how did I get this?

I took the quotient $\text{Moles of water"/"Moles of barium iodide}$.

And thus we gets........$\frac{0.75 \cdot m o l \cdot {H}_{2} O}{0.0243 \cdot m o l \cdot B a {I}_{2}}$

I think that you have quoted the wrong units; barium iodide is typically supplied as the dihydrate, i.e. $B a {I}_{2} \cdot 2 {H}_{2} O$; well, that's what the (old!) bottle on the shelf says.