Question

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)

a. BaCl₂​⋅2H₂​O
b. BaCl₂​⋅3H₂​O
c. BaCl₂​⋅4H₂​O
d. BaCl₂​⋅5H₂​O

Answer 1

To find the formula of the hydrated barium chloride, we calculate the mass of water lost (9 g) and determine the moles of water (0.5 moles) and anhydrous BaCl2 (0.25 moles). The ratio of moles of water to BaCl2 (2:1) reveals that the formula is BaCl2·2H2O. a. BaCl₂​⋅2H₂​O

Step-by-step explanation:

To determine the formula of the hydrated salt, we need to calculate the mass of water lost upon heating and relate it to the mole ratio between the anhydrous barium chloride and water. The initial mass of the hydrated barium chloride was 61 g, and upon drying, it weighs 52 g. Therefore, the mass of the water of hydration is 61 g - 52 g = 9 g. The molar mass of water (H2O) is approximately 18 g/mol, so the number of moles of water lost is 9 g / 18 g/mol = 0.5 moles.

Next, we need the molar mass of anhydrous BaCl2. The atomic masses for barium (Ba), chlorine (Cl), and hydrogen (H) are approximately 137 g/mol, 35.5 g/mol (each Cl), and 1 g/mol (each H), respectively. Thus, the molar mass of BaCl2 is 137 g/mol + 2(35.5 g/mol) = 208 g/mol. With 52 g of anhydrous BaCl2, we calculate the moles of BaCl2: 52 g / 208 g/mol = 0.25 moles.

To find the molar ratio of water to BaCl2, we divide the moles of water by the moles of BaCl2: 0.5 moles H2O / 0.25 moles BaCl2 = 2 moles H2O per mole of BaCl2. Thus, the formula of the hydrated salt is BaCl2·2H2O, which corresponds to option (a).

Answer 2

The mass of water removed from the hydrated barium chloride is 9 g. From the molar masses, we find there are 0.25 mol of BaCl2 and 0.5 mol of water, indicating a 1:2 ratio. The correct formula is option a. BaCl2 · 2H2O.

Step-by-step explanation:

To determine the formula of the hydrated salt of barium chloride, we need to first find the mass of water that has been removed. Since the hydrated sample weighs 61 g and the anhydrous (dried) sample weighs 52 g, the mass of the water is the difference, which is 9 g. Now, we need to calculate the molar mass of barium chloride (BaCl2) and water (H2O).

The molar mass of barium (Ba) is approximately 137.33 g/mol, chlorine (Cl) is approximately 35.45 g/mol, and water (H2O) is approximately 18.01 g/mol. The molar mass of BaCl2 is thus 137.33 g/mol + 2(35.45 g/mol) = 208.23 g/mol. The number of moles of BaCl2 in the dried sample is 52 g / 208.23 g/mol ≈ 0.25 mol. Next, the number of moles of water is 9 g / 18.01 g/mol ≈ 0.5 mol. Since the moles of water are twice the moles of BaCl2, the formula of the hydrated salt is BaCl2 · 2H2O, which corresponds to option (a). Therefore, the correct option in the final part is a. BaCl2 ·2H2O.