A 6.85g sample of the hydrated ${S r (O H)}_{2} . {x H}_{2} O$ is dried in an oven to give 3.13 g of anhydrous ${S r (O H)}_{2}$ . What is the value of x?(atomic masses: Sr=87.60,O=16,H=1.0

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Solution

The correct option is A 8
$S r (O H)_{2} . x H_{2} O \to S r (O H)_{2} + x H_{2} O$

Molar mass of $S r (O H)_{2}$ = 122 g/mol

According to stoichiometry:
(122+18x) g of $S r (O H)_{2} . x H_{2} O$ decomposes to give 122 g of $S r (O H)_{2}$
Thus 6.85 g shall decompose to give $\frac{122}{122 + 18 x} \times 6.85 g$ of $S r (O H)_{2}$
As given :
$\frac{122}{122 + 18 x} \times 6.85 g = 3.13 g$

Thus the value of $x i s 8$ .
Option A is correct

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