A solution was prepared by dissolving 0.800 g of sulfur, S g , in 100.0 g of acetic acid, HC 2 H 3 O 2 . Calculate the freezing point and boiling point of the solution.
A solution was prepared by dissolving 0.800 g of sulfur, S g , in 100.0 g of acetic acid, HC 2 H 3 O 2 . Calculate the freezing point and boiling point of the solution.
Author: Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
A solution was prepared by dissolving 0.800 g of sulfur, Sg, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and boiling point of the solution.
Expert Solution & Answer
Interpretation Introduction
Interpretation:
The boiling point and freezing point of 0.800 g Sulfur in 100.0 g of acetic acid have to be determined.
Concept Introduction:
Elevation in boiling point is a colligative property which refers to increase in boiling point of the solution due to the addition of non-volatile solute. It is expressed as,
ΔTb=Kb.cm
Where,
ΔTb= elevation in boiling pointKb= boiling point elevation constantcm= molal concentration
Depression in freezing point is a colligative property which refers to decrease in freezing point of the solution due to the addition of non-volatile solute. It is expressed as,
ΔTf=Kf.cm
Where,
ΔTf= depression in freezing pointKf= freezing point depression constantcm= molal concentration
Molality or molal concentration is one of the many parameters that is used to express concentration of a solution. It is expressed as,
Molality = number of molesof solutemass of solvent in kg
Answer to Problem 12.70QP
The boiling point of 0.800 g Sulfur in 100.0 g of acetic acid is determined as 118.6°C.
The freezing point of 0.800 g Sulfur in 100.0 g of acetic acid is determined as 16.49°C.
Explanation of Solution
Given that mass of solvent acetic acid is 100.0 g which is equivalent to 0.1000 kg. Mass of Sulfur is 0.800g
Number of moles of sulfur is,
No.of moles of sulfur= mass of Sulfurmolar mass of Sulfur=0.800 g256.52 g/mol= 0.003119 g/mol
Molality of the solution is,
Molality=0.003119 mol0.1000 kg= 0.03119 m ≈0.0312 m
From the data given in the table 12.3 in text book, for acetic acid, Kf= 3.59°C/m
Therefore, depression in freezing point is,
ΔTf = Kf.cm
Substitute the values,
ΔTf= 3.59°C/m × 0.0312 m= 0.1120°C
The depression in freezing point is equivalent to the difference between freezing temperature of pure solvent and solution. Hence freezing point of the solution is,
From the data given in the table 12.3 in text book, for Acetic acid, Kb= 3.08°C/m
Therefore, elevation in boiling point is,
ΔTb = Kb.cm
Substitute the values,
ΔTb= 3.08°C/m × 0.0312 m= 0.096096°C
The elevation in boiling point is equivalent to the difference between boiling temperature of and solution and pure solvent. Hence boiling point of the solution is,
Depression in freezing point and elevation in boiling point are colligative properties that depend upon the concentration of the solute in solution. The molal concentration of the solute, depression in freezing point and elevation in boiling point are the key parameters to determine freezing point and boiling point of the solution.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.