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1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:
- a)3.015 × 102 m2
- b)5 × 10–19 m2
- c)2.5 × 10–19 m2
- d)1.25 × 10–19 m2
Correct answer is option 'B'. Can you explain this answer?
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1g charcoal show chemical adsorption of some acetic acid on coming in ...
The initial number of moles of acetic acid in the 100 mL solution can be calculated using the formula:
moles = volume (L) x molarity
Converting 100 mL to L:
volume = 100 mL / 1000 mL/L = 0.1 L
Calculating the initial number of moles:
moles = 0.1 L x 0.50 M = 0.05 moles
After the adsorption of some acetic acid onto the charcoal, the molarity of the remaining acetic acid in the solution is 0.49 M. We can calculate the final number of moles of acetic acid using the same formula:
moles = volume (L) x molarity
Calculating the final volume of the solution:
volume = moles / molarity
moles = 0.05 - x (where x is the number of moles of acetic acid adsorbed onto the charcoal)
volume = (0.05 - x) / 0.49
Since the final volume of the solution is still 0.1 L, we can set up the equation:
(0.05 - x) / 0.49 = 0.1
Simplifying the equation:
0.05 - x = 0.1 x 0.49
0.05 - x = 0.049
x = 0.05 - 0.049
x = 0.001
Therefore, the number of moles of acetic acid adsorbed onto the charcoal is 0.001 moles.
To calculate the molarity of acetic acid after adsorption, we need to subtract the moles of acetic acid adsorbed from the initial moles and divide by the final volume:
moles = 0.05 - 0.001 = 0.049 moles
molarity = moles / volume
molarity = 0.049 moles / 0.1 L = 0.49 M
Therefore, the molarity of acetic acid after adsorption is 0.49 M.
moles = volume (L) x molarity
Converting 100 mL to L:
volume = 100 mL / 1000 mL/L = 0.1 L
Calculating the initial number of moles:
moles = 0.1 L x 0.50 M = 0.05 moles
After the adsorption of some acetic acid onto the charcoal, the molarity of the remaining acetic acid in the solution is 0.49 M. We can calculate the final number of moles of acetic acid using the same formula:
moles = volume (L) x molarity
Calculating the final volume of the solution:
volume = moles / molarity
moles = 0.05 - x (where x is the number of moles of acetic acid adsorbed onto the charcoal)
volume = (0.05 - x) / 0.49
Since the final volume of the solution is still 0.1 L, we can set up the equation:
(0.05 - x) / 0.49 = 0.1
Simplifying the equation:
0.05 - x = 0.1 x 0.49
0.05 - x = 0.049
x = 0.05 - 0.049
x = 0.001
Therefore, the number of moles of acetic acid adsorbed onto the charcoal is 0.001 moles.
To calculate the molarity of acetic acid after adsorption, we need to subtract the moles of acetic acid adsorbed from the initial moles and divide by the final volume:
moles = 0.05 - 0.001 = 0.049 moles
molarity = moles / volume
molarity = 0.049 moles / 0.1 L = 0.49 M
Therefore, the molarity of acetic acid after adsorption is 0.49 M.
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1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer?
Question Description
1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer?.
1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer?.
Solutions for 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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ample number of questions to practice 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Chemistry tests.
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