A sample of a hydrate of barium chloride weighing 61 g was heated...

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)

Options

A $BaC{l}_{2} ⋅ {H}_{2}O$
B $BaC{l}_{2} ⋅ {3H}_{2}O$
C $BaC{l}_{2} ⋅ {4H}_{2}O$
D $BaC{l}_{2} ⋅ 2{H}_{2}O$

Correct answer

D. $BaC{l}_{2} ⋅ 2{H}_{2}O$

Step-by-step solution

BaC{l}_{2} &sdot; x{H}_{2}O \to BaC{l}_{2} + x{H}_{2}O

( 137 + 2 \times 35.5 + 18x )

=\left(208 + 18x\right)

g/mole<br/><br/>

\dfrac{208 + 18x}{208} = \dfrac{61}{52}

<br/><br/>

10816 + 936x = 12688

<br/><br/>

936x = 1872

<br/><br/>

x = 2

<br/><br/> Formula is

BaC{l}_{2} ⋅ 2{H}_{2}O

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