The boiling point of water is 100.00 °C at 1 atmosphere. A student dissolves 10.58 grams of barium sulfide, BaS (169.4 g/mol), in 185.8 grams of water. Whats the molality(m) and boiling point of the solution in Celsius?
The elevation in boiling point is given by
ΔTb= i * Kb * molality
Kb for water = 0.512 °C/m
Number of moles of solute BaS =mass/mol.wt. = 10.58 g/(169.4 g/mol) = 0.062456 mol
Mass of water (solvent) = 185.8 g = 0.1858 kg
Molality = moles/mass of solvent in kg = 0.062456 mol/ 0.1858 kg = 0.336145 m = 0.3361 m
For BaS, van’t Hoff factor = i =2, as it dissociates to Ba+2 and S–2 ions
ΔTb = i * Kb * molality = 2 * 0.512 °C/m * 0.336145 m = 0.344212 °C = 0.34 °C
ΔTb = Ts – T0 = 0.34 °C
Ts – 100 °C = 0.34 °C
Ts = 100 + 0.34 = 100.34 °C
Molality of the solution = 0.3361 m
Boiling point of the solution = 100.34 °C
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