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A drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If the rate constant of disappearance of H+ is 1.0 x 10-7 mol litre-1 sec-1. How long would it take for H+ in drop to disappear?
- a)6 x 10-8 sec
- b)6 x 10-7
- c)6 x 10-9 sec
- d)6 x 10-10 sec
Correct answer is option 'C'. Can you explain this answer?
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A drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If...
Since rate constant = 1 x 10-7 mol litre-1 sec-1
∴ Zero order reaction
∴ Zero order reaction
0.05 mL has 3 x 10-6 mole of H+
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A drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If...
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A drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If...
To determine how long it would take for the H in the drop to disappear, we can use the first-order rate equation:
Rate = k[H]
Where:
- Rate is the rate of disappearance of H (in mol L-1 sec-1)
- k is the rate constant (in mol L-1 sec-1)
- [H] is the concentration of H (in mol L-1)
We are given the volume of the solution (0.05 mL) and the number of moles of H (3.0 x 10-6 mol). We can use these values to calculate the concentration of H:
[H] = moles of H / volume of solution
[H] = (3.0 x 10-6 mol) / (0.05 mL) = 6.0 x 10-5 mol L-1
Now we can substitute the values into the rate equation:
Rate = (1.0 x 10-7 mol L-1 sec-1)(6.0 x 10-5 mol L-1)
Rate = 6.0 x 10-12 mol sec-1
The rate of disappearance of H is 6.0 x 10-12 mol sec-1.
To find the time it would take for the H in the drop to disappear, we can rearrange the rate equation:
Rate = k[H]
t = 1 / (k[H])
t = 1 / ((1.0 x 10-7 mol L-1 sec-1)(6.0 x 10-5 mol L-1))
t = 1 / (6.0 x 10-12 mol sec-1)
t = 1.67 x 1011 sec
Converting this time to scientific notation, we get:
t ≈ 1.67 x 1011 sec = 1.67 x 1011 sec x 10-9 sec = 1.67 x 102 sec x 10-9 sec = 1.67 x 10-7 sec
Therefore, it would take approximately 1.67 x 10-7 seconds for the H in the drop to disappear. This is equivalent to 6 x 10-9 seconds, which matches option C.
Rate = k[H]
Where:
- Rate is the rate of disappearance of H (in mol L-1 sec-1)
- k is the rate constant (in mol L-1 sec-1)
- [H] is the concentration of H (in mol L-1)
We are given the volume of the solution (0.05 mL) and the number of moles of H (3.0 x 10-6 mol). We can use these values to calculate the concentration of H:
[H] = moles of H / volume of solution
[H] = (3.0 x 10-6 mol) / (0.05 mL) = 6.0 x 10-5 mol L-1
Now we can substitute the values into the rate equation:
Rate = (1.0 x 10-7 mol L-1 sec-1)(6.0 x 10-5 mol L-1)
Rate = 6.0 x 10-12 mol sec-1
The rate of disappearance of H is 6.0 x 10-12 mol sec-1.
To find the time it would take for the H in the drop to disappear, we can rearrange the rate equation:
Rate = k[H]
t = 1 / (k[H])
t = 1 / ((1.0 x 10-7 mol L-1 sec-1)(6.0 x 10-5 mol L-1))
t = 1 / (6.0 x 10-12 mol sec-1)
t = 1.67 x 1011 sec
Converting this time to scientific notation, we get:
t ≈ 1.67 x 1011 sec = 1.67 x 1011 sec x 10-9 sec = 1.67 x 102 sec x 10-9 sec = 1.67 x 10-7 sec
Therefore, it would take approximately 1.67 x 10-7 seconds for the H in the drop to disappear. This is equivalent to 6 x 10-9 seconds, which matches option C.
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A drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If the rate constant of disappearance of H+ is 1.0 x 10-7 mol litre-1 sec-1. How long would it take for H+ in drop to disappear?a) 6 x 10-8 sec b) 6 x 10-7 c) 6 x 10-9 sec d) 6 x 10-10 secCorrect answer is option 'C'. Can you explain this answer?
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A drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If the rate constant of disappearance of H+ is 1.0 x 10-7 mol litre-1 sec-1. How long would it take for H+ in drop to disappear?a) 6 x 10-8 sec b) 6 x 10-7 c) 6 x 10-9 sec d) 6 x 10-10 secCorrect answer is option 'C'. 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 drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If the rate constant of disappearance of H+ is 1.0 x 10-7 mol litre-1 sec-1. How long would it take for H+ in drop to disappear?a) 6 x 10-8 sec b) 6 x 10-7 c) 6 x 10-9 sec d) 6 x 10-10 secCorrect answer is option 'C'. 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 drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If the rate constant of disappearance of H+ is 1.0 x 10-7 mol litre-1 sec-1. How long would it take for H+ in drop to disappear?a) 6 x 10-8 sec b) 6 x 10-7 c) 6 x 10-9 sec d) 6 x 10-10 secCorrect answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice A drop of solution (volume 0.05 mL) contains 3.0 x 10-6 mole of H+. If the rate constant of disappearance of H+ is 1.0 x 10-7 mol litre-1 sec-1. How long would it take for H+ in drop to disappear?a) 6 x 10-8 sec b) 6 x 10-7 c) 6 x 10-9 sec d) 6 x 10-10 secCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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