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Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. c. 5.5% NaNO3 by mass (in water)

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The freezing point and boiling point of a solution can be calculated using the following formula:

Freezing Point = Kf x molality

Boiling Point = Kb x molality

where Kf and Kb are the freezing and boiling point constants respectively.

The molality of a solution is the number of moles of solute (NaNO3) in 1 kg of solvent (water). For this solution, the mass of the solute (NaNO3) is 5.5%, and the mass of the solvent (water) is 94.5%.

Therefore, the mass of NaNO3 = 5.5g/100g = 0.055g

The molar mass of NaNO3 = 85.0 g/mol

Therefore, the number of moles of NaNO3 = 0.055g/85.0 g/mol = 0.000647 mol

The mass of water = 94.5g/100g = 0.945g

Therefore, the molality of the solution = 0.000647 mol/0.945 kg = 0.000682 mol/kg

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