What is the concentration, in grams per liter, of a solution prepared by dissolving 0.00040 mol HCl in 2.2 L H2O? Assume that th
2 answers:
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<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C
<u>Explanation:</u>
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute
= 0.800 g
= Molar mass of solute
= 256.52 g/mol
= Mass of solvent (acetic acid) = 100.0 g
Putting values in above equation, we get:
- <u>Calculation for freezing point of solution:</u>
Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.
To calculate the depression in freezing point, we use the equation:
or,
where,
Freezing point of acetic acid = 16.6°C
i = Vant hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant = 3.59°C/m
m = molality of solution = 0.0312 m
Putting values in above equation, we get:
Hence, the freezing point of solution is 16.5°C
- <u>Calculation for boiling point of solution:</u>
Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.
The equation used to calculate elevation in boiling point follows:
To calculate the elevation in boiling point, we use the equation:
or,
where,
Boiling point of acetic acid = 118.1°C
i = Vant hoff factor = 1 (for non-electrolyte)
= molal boiling point elevation constant = 3.08°C/m
m = molality of solution = 0.0312 m
Putting values in above equation, we get:
Hence, the boiling point of solution is 118.2°C