0.0444 g of BaCl2 * nH2O was placed in a flask, which was immersed in a melt of lead (melt = 327 C)

| May 29, 2021 | education

0.0444 g of BaCl2 * nH2O was placed in a flask, which was immersed in a melt of lead (melt = 327 C), an overpressure of 15.3 mm Hg was created in it. Art. The volume of the flask is 0.888 l. Determine the formula of the crystalline hydrate.

1. We translate mm Hg into kPa P = 15.3 * 0.133 = 2.0349 kPa
2. We translate C into K. T = 273 + 327 = 600 K
3. We use the Mendeleev-Clapeyron equation, P * V = n * R * T, whence nH2O = P * V / (R * T) = 2.0349 * 0.888 / (8.314 * 600) = 0.00036224 mol of water is contained in the flask.
4. Find the total mass of water. 18 * 0.00036224 = 0.00652032 grams of water
5. We are looking for the mass of barium chloride, subtracting it from the total mass of the crystalline hydrate. 0.0444 – 0.00652032 = 0.03787968 g BaCl2.
6.find the number of mol of BaCl 0.03787968 / 208 = 0.000182114 mol
Water and salt refer to each other as 0.00036224: 0.000182114, i.e. 2 to 1
Accordingly, for 1 mol of salt there is 2 mol of water. BaCl2 * 2H2O



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