What would be the formula of a barium iodide hydrate that contains $0.0243 \cdot m o l$ $0.0243mol$ $B a I_{2}$ $BaI_2$ , and $0.75 \cdot m o l$ $0.75mol$ water?

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What would be the formula of a barium iodide hydrate that contains $0.0243 \cdot m o l$ $0.0243mol$ $B a I_{2}$ $BaI_2$ , and $0.75 \cdot m o l$ $0.75mol$ water?

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Answer 1 · 2017-06-10T15:16:57

Well it would be $B a I_{2} \cdot 30 H_{2} O$ $BaI_2*30H_2O$

Explanation:

And how did I get this?

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

And thus we gets........ $\frac{0.75 \cdot m o l \cdot H_{2} O}{0.0243 \cdot m o l \cdot B a I_{2}}$ $(0.75*mol*H_2O)/(0.0243*mol*BaI_2)$

I think that you have quoted the wrong units; barium iodide is typically supplied as the dihydrate, i.e. $B a I_{2} \cdot 2 H_{2} O$ $BaI_2*2H_2O$ ; well, that's what the (old!) bottle on the shelf says.

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What would be the formula of a barium iodide hydrate that contains $0.0243 \cdot m o l$ $0.0243mol$ $B a I_{2}$ $BaI_2$ , and $0.75 \cdot m o l$ $0.75mol$ water?

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What would be the formula of a barium iodide hydrate that contains 0.0243⋅mol0.0243*mol BaI2BaI_2, and 0.75⋅mol0.75*mol water?

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What would be the formula of a barium iodide hydrate that contains $0.0243 \cdot m o l$ $0.0243mol$ $B a I_{2}$ $BaI_2$ , and $0.75 \cdot m o l$ $0.75mol$ water?