A sample of a 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, Ba = 137 amu, Cl = 35.5 amu)

The sample of the hydrate of barium chloride weighs 61 g. After heating, the dried sample, which is anhydrous barium chloride, weighs 52 g. The difference in mass is due to the removal of water. So, the mass of water removed is 61 g minus 52 g, which equals 9 g.

Now, we need to find the molar mass of anhydrous barium chloride, BaCl₂. The atomic mass of barium (Ba) is 137 amu, and chlorine (Cl) is 35.5 amu. Since there are two chlorine atoms, the molar mass of BaCl₂ is calculated as follows: 137 + 2 × 35.5 = 137 + 71 = 208 g/mol.

The mass of anhydrous BaCl₂ is 52 g. To find the number of moles, we use the formula: moles = mass / molar mass. So, moles of BaCl₂ = 52 g / 208 g/mol. Simplifying this fraction: 52 ÷ 208 = 52/208. Dividing both numerator and denominator by 52 gives 1/4, or 0.25 moles. Alternatively, 52 ÷ 208 = 0.25 moles.

The mass of water removed is 9 g. The molar mass of water (H₂O) is 18 g/mol (since hydrogen is 1 amu and oxygen is 16 amu, so 2 × 1 + 16 = 18 g/mol). Moles of water = mass / molar mass = 9 g / 18 g/mol = 0.5 moles.

We now have 0.25 moles of BaCl₂ and 0.5 moles of water. To find the ratio of water to BaCl₂, we divide the moles of water by the moles of BaCl₂: 0.5 / 0.25 = 2. This means there are 2 moles of water for every mole of BaCl₂.

Therefore, the formula of the hydrated salt is BaCl₂ · 2H₂O.

Alternatively, we can verify using the proportion method. Let the hydrate be BaCl₂ · nH₂O, where n is the number of water molecules. The molar mass of the hydrate is 208 + n × 18 g/mol. The mass ratio of BaCl₂ to the hydrate is 52 g / 61 g. This should equal the ratio of the molar mass of BaCl₂ to the molar mass of the hydrate: 208 / (208 + 18n) = 52 / 61.

Cross-multiplying: 208 × 61 = 52 × (208 + 18n). Calculating left side: 208 × 61 = 208 × 60 + 208 × 1 = 12480 + 208 = 12688. Right side: 52 × 208 = 52 × 200 + 52 × 8 = 10400 + 416 = 10816, and 52 × 18n = 936n. So, 12688 = 10816 + 936n.

Rearranging: 12688 - 10816 = 936n → 1872 = 936n. Solving for n: n = 1872 / 936. Simplifying: divide numerator and denominator by 936, 1872 ÷ 936 = 2. So, n = 2.

This confirms that the hydrate is BaCl₂ · 2H₂O.

Comparing with the options: A is BaCl₂ · H₂O, B is BaCl₂ · 3H₂O, C is BaCl₂ · 4H₂O, D is BaCl₂ · 2H₂O. Hence, the correct answer is Option D.