6.85 g of hydrated Sr (OH)2 .XH2O was heated to give anhydrous mass of weight 3.13 g. The value of X is

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Solution

1. Calculate the mass and moles of H2O driven off:
6.85 g - 3.13 g = 3.72 g H2O / 18.0 g/mol = 0.207 moles H2O

2. Calculate moles Sr(OH)2 remaining:
3.13 g Sr(OH)2 / 121.6 g/mol = 0.0257 moles Sr(OH)2

3. Calculate the ratio of moles H2O / moles Sr(OH)2:
0.207 / 0.0257 = 8.0

So, your experimental determination of x is 8.0

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