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

Question

Smart Achievers · Accepted Answer

(4)

${BaCl}_{2} \cdot {xH}_{2} O \to {BaCl}_{2} + {xH}_{2} O,$ $(137 + 2 \times 35.5 + 18 x) = (208 + 18 x) g / mole$

$\frac{208 + 18 x}{208} = \frac{61}{52},$ $52 (208 + 18 x) = 61 \times 208,$

$10816 + 936 x = 12688,$ $936 x = 1872,$ $x = 2,$

Formula is ${BaCl}_{2} \cdot 2 H_{2} O$

Solution

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

1 ${BaCl}_{2} \cdot H_{2} O$

2 ${BaCl}_{2} \cdot 3 H_{2} O$

3 ${BaCl}_{2} \cdot 4 H_{2} O$

4 ${BaCl}_{2} \cdot 2 H_{2} O$

Ans.
(4)

${BaCl}_{2} \cdot {xH}_{2} O \to {BaCl}_{2} + {xH}_{2} O,$ $(137 + 2 \times 35.5 + 18 x) = (208 + 18 x) g / mole$

$\frac{208 + 18 x}{208} = \frac{61}{52},$ $52 (208 + 18 x) = 61 \times 208,$

$10816 + 936 x = 12688,$ $936 x = 1872,$ $x = 2,$

Formula is ${BaCl}_{2} \cdot 2 H_{2} O$

Want Us to Email you About Special Offers & Updates?

Solution

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

1 BaCl2·H2O

2 BaCl2·3H2O

3 BaCl2·4H2O

4 BaCl2·2H2O

Ans. (4) BaCl2·xH2O → BaCl2+xH2O, (137+2×35.5+18x)=(208+18x) g/mole 208+18x208=6152, 52(208+18x)=61×208, 10816+936x=12688, 936x=1872, x=2, Formula is BaCl2·2H2O