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0.01 M solution of KCl and BaCl2 are prepared in water. The freezing point of KCl is found to be –20C. What is freezing point of BaCl2 solution when it is completely ionized?
- a)30C
- b)–30C
- c)–20C
- d)–40C
Correct answer is option 'B'. Can you explain this answer?
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0.01 M solution of KCl and BaCl2 are prepared in water. The freezing p...
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-1.86°C and the freezing point of BaCl2 is found to be -3.15°C. This is due to the fact that the freezing point depression of a solution is directly proportional to the concentration of solute particles in the solution.
To calculate the freezing point depression of the KCl solution, we can use the equation:
ΔTf = Kf * m
Where ΔTf is the freezing point depression, Kf is the freezing point depression constant for water (1.86°C/m), and m is the molality of the solution (moles of solute per kilogram of solvent).
Assuming we have 1 kilogram of water, the molality of a 0.01 M KCl solution would be:
molality = (0.01 moles KCl) / (1 kilogram water) = 0.01 m
Plugging this into the equation above, we get:
ΔTf = (1.86°C/m) * (0.01 m) = 0.0186°C
Therefore, the freezing point of the KCl solution would be:
-1.86°C - 0.0186°C = -1.8786°C
Similarly, we can calculate the freezing point depression of the BaCl2 solution:
molality = (0.01 moles BaCl2) / (1 kilogram water) = 0.02 m
ΔTf = (1.86°C/m) * (0.02 m) = 0.0372°C
Freezing point of the BaCl2 solution:
-3.15°C - 0.0372°C = -3.1872°C
To calculate the freezing point depression of the KCl solution, we can use the equation:
ΔTf = Kf * m
Where ΔTf is the freezing point depression, Kf is the freezing point depression constant for water (1.86°C/m), and m is the molality of the solution (moles of solute per kilogram of solvent).
Assuming we have 1 kilogram of water, the molality of a 0.01 M KCl solution would be:
molality = (0.01 moles KCl) / (1 kilogram water) = 0.01 m
Plugging this into the equation above, we get:
ΔTf = (1.86°C/m) * (0.01 m) = 0.0186°C
Therefore, the freezing point of the KCl solution would be:
-1.86°C - 0.0186°C = -1.8786°C
Similarly, we can calculate the freezing point depression of the BaCl2 solution:
molality = (0.01 moles BaCl2) / (1 kilogram water) = 0.02 m
ΔTf = (1.86°C/m) * (0.02 m) = 0.0372°C
Freezing point of the BaCl2 solution:
-3.15°C - 0.0372°C = -3.1872°C
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0.01 M solution of KCl and BaCl2 are prepared in water. The freezing point of KCl is found to be –20C. What is freezing point of BaCl2 solution when it is completely ionized?a)30Cb)–30Cc)–20Cd)–40CCorrect answer is option 'B'. Can you explain this answer?
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