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)

Question

Toppr Admin · Accepted Answer

Let the formula of the hydrated salt be $B a C l_{2} . x H_{2} O$
$B a C l_{2} . x H_{2} O Δ - \to B a C l_{2} . x H_{2} O$ .

$B a C l_{2} . x H_{2} O$ $B a C l_{2}$
Molar mass (g/mol)
208+18x
208
Mass (g)
61
52
Thus $208 = \frac{52}{61} \times (208 + 18 x)$
$244 = 208 + 18 x$
$36 = 18 x$
$x = 2$
Hence, the formula of the hydrated salt is $B a C l_{2} .2 H_{2} O$ .

Mass of crucible and cover	31.623 grams
Mass of crucible, cover, and sample prior to heating	33.632 grams
Mass of crucible, cover, and sample after heating	33.376 grams

Mass of crucible and cover	31.623 grams
Mass of crucible, cover, and sample prior to heating	33.632 grams
Mass of crucible, cover, and sample after heating	33.376 grams

	$B a C l_{2} . x H_{2} O$	$B a C l_{2}$
Molar mass (g/mol)	208+18x	208
Mass (g)	61	52

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) BaCl2.H2OBaCl2.2H2OBaCl2.4H2OBaCl2.3H2O

Let the formula of the hydrated salt be BaCl2.xH2OBaCl2.xH2OΔ−→BaCl2.xH2O.BaCl2.xH2OBaCl2Molar mass (g/mol)208+18x208Mass (g)6152Thus 208=5261×(208+18x)244=208+18x36=18xx=2Hence, the formula of the hydrated salt is BaCl2.2H2O.