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On heating 1.763 g of hydrated $$\text{BaCl}_2$$ to dryness 1.505 g of anhydrous salt remained. Number of moles of $$\text{H}_2\text{O}$$ present in the mole of hydrated $$\text{BaCl}_2$$ is ____ (mol. wt. of anhydrous $$\text{BaCl}_2$$ is 208)
Correct Answer: 2
The water lost is $$1.763 - 1.505 = 0.258\text{ g}$$, i.e. $$\frac{0.258}{18} \approx 0.01433$$ mol. The anhydrous salt is $$\frac{1.505}{208} \approx 0.00724$$ mol. The ratio is $$\frac{0.01433}{0.00724} \approx 2$$, so the hydrate is $$\text{BaCl}_2 \cdot 2\text{H}_2\text{O}$$.
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