A sample of a hydrate of barium chloride weighing 61 g has heated until all the hydration is removed. The dried sample weighed 52 g. The formula of hydrated salt is (atomic mass and Ba = 137 and Cl = 35.5)
1. BaCl2.H2O
2. BaCl2.2H2O
3. BaCl2.3H2O
4. BaCl2.4H2O
Dear Student,
Please find below the solution to the asked query:
On heating the water of crystallisation loses from the sample leaving behind the anhydrous salt, hence
Mass of water left = 61 - 52 = 9 g
Moles of water = 9 g / 18 g mol-1 = 0.5 moles
Similarly, the moles of anhydrous BaCl2 will be = 52g / 208 g mol-1 = 0.25 moles
Now, by dividing moles of water by moles of BaCl2 we will have the value of water of crystallization = 0.5 / 0.25 = 2
Hence, the formula is = BaCl2.2H2O i.e. option (2).
Hope this information will clear your doubts about Some Basic Concepts of Chemistry.
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Please find below the solution to the asked query:
On heating the water of crystallisation loses from the sample leaving behind the anhydrous salt, hence
Mass of water left = 61 - 52 = 9 g
Moles of water = 9 g / 18 g mol-1 = 0.5 moles
Similarly, the moles of anhydrous BaCl2 will be = 52g / 208 g mol-1 = 0.25 moles
Now, by dividing moles of water by moles of BaCl2 we will have the value of water of crystallization = 0.5 / 0.25 = 2
Hence, the formula is = BaCl2.2H2O i.e. option (2).
Hope this information will clear your doubts about Some Basic Concepts of Chemistry.
If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Best Wishes !