Q. A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is: (atomic mass, Ba = 137 amu, Cl = 35.5 amu)

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A

$BaCl_{2}⋅H_{2}O$

B

$BaCl_{2}⋅3 H_{2}O$

C

$BaCl_{2}⋅4 H_{2}O$

D

$BaCl_{2}⋅2H_{2}O$

Solution:

$BaCl_{2}⋅xH_{2}O \rightarrow BaCl_{2}+x \, H_{2}O \, $
$\left(\right.137+2\times 35.5+18x\left.\right)$
$=\left(208 + 18 x\right) \, $ g/mole
$\frac{208 \, + \, 18 \, x}{208}=\frac{61}{52}$
$10816+936 \, x=12688$
$936 \, x=1872$
$x=2$
Formula is $BaCl_{2}⋅2H_{2}O$

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