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A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]
- a)– 0.18°C
- b)– 0.54°C
- c)– 0.36°C
- d)– 0.24°C
Correct answer is option 'D'. Can you explain this answer?
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A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for ...
To find the percent ionization of a weak acid, we need to use the equation:
% ionization = (concentration of ionized acid)/(initial concentration of weak acid) * 100
Given that the solution is 0.1 molal, we can assume that the initial concentration of the weak acid is also 0.1 molal.
Let x represent the concentration of ionized acid. Therefore, the concentration of the weak acid that remains unionized is (0.1 - x) molal.
The equation for the percent ionization can be written as:
30 = (x)/(0.1) * 100
Simplifying the equation, we have:
0.3 = x/0.1
x = 0.03 molal
Therefore, the concentration of the ionized acid is 0.03 molal.
Now, let's calculate the molality of water in the solution using the equation:
Kf = molality of water * boiling point elevation constant
Given that Kf for water is 1.86, we can rearrange the equation to solve for the molality of water:
molality of water = Kf / boiling point elevation constant
molality of water = 1.86 / 0.512 (boiling point elevation constant for water)
molality of water = 3.63 molal
Therefore, the molality of water in the solution is 3.63 molal.
% ionization = (concentration of ionized acid)/(initial concentration of weak acid) * 100
Given that the solution is 0.1 molal, we can assume that the initial concentration of the weak acid is also 0.1 molal.
Let x represent the concentration of ionized acid. Therefore, the concentration of the weak acid that remains unionized is (0.1 - x) molal.
The equation for the percent ionization can be written as:
30 = (x)/(0.1) * 100
Simplifying the equation, we have:
0.3 = x/0.1
x = 0.03 molal
Therefore, the concentration of the ionized acid is 0.03 molal.
Now, let's calculate the molality of water in the solution using the equation:
Kf = molality of water * boiling point elevation constant
Given that Kf for water is 1.86, we can rearrange the equation to solve for the molality of water:
molality of water = Kf / boiling point elevation constant
molality of water = 1.86 / 0.512 (boiling point elevation constant for water)
molality of water = 3.63 molal
Therefore, the molality of water in the solution is 3.63 molal.
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A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for ...
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A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]a)– 0.18°Cb)– 0.54°Cc)– 0.36°Cd)– 0.24°CCorrect answer is option 'D'. Can you explain this answer?
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A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]a)– 0.18°Cb)– 0.54°Cc)– 0.36°Cd)– 0.24°CCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]a)– 0.18°Cb)– 0.54°Cc)– 0.36°Cd)– 0.24°CCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]a)– 0.18°Cb)– 0.54°Cc)– 0.36°Cd)– 0.24°CCorrect answer is option 'D'. Can you explain this answer?.
A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]a)– 0.18°Cb)– 0.54°Cc)– 0.36°Cd)– 0.24°CCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]a)– 0.18°Cb)– 0.54°Cc)– 0.36°Cd)– 0.24°CCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]a)– 0.18°Cb)– 0.54°Cc)– 0.36°Cd)– 0.24°CCorrect answer is option 'D'. Can you explain this answer?.
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