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Q.

A 6.85 g sample of the hydrate Sr(OH)2.xH2O is dried in an oven to give 3.13 g of anhydrous Sr(OH)2. what is the value of x ? (Atomic weights : Sr = 87.60, O = 16.0, H = 1.0)

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a

8

b

12

c

10

d

6

answer is A.

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Detailed Solution

Sr(OH)26.85 g⋅xH2O⟶Sr(OH)23.13 g+xH2O3.72 g% of water =3.726.85×100=18x121+18x×100  ⇒x=8
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