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Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fraction of the m etal which exists as M3+ would be
- a)7.01%
- b)4.08%
- c)6.05%
- d)5.08%
Correct answer is option 'B'. Can you explain this answer?
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Experimentally, it was found that a metal oxide has formula M 0 9B0 . ...
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Experimentally, it was found that a metal oxide has formula M 0 9B0 . ...
Given:
Metal oxide has formula M0.98O.
Metal M is present as M2 and M3 in its oxide.
To Find:
The fraction of the metal which exists as M3.
Solution:
Step 1:
Let's assume the total number of moles of the metal in the oxide is x.
Step 2:
According to the given information, the formula of the metal oxide is M0.98O.
This means that for every 1 mole of oxygen, there are 0.98 moles of the metal.
So, the number of moles of oxygen in the oxide is 0.98x.
Step 3:
Since the metal is present as M2 and M3 in the oxide, the total number of moles of metal can be represented as:
2y + 3z = x
where y is the number of moles of M2 and z is the number of moles of M3.
Step 4:
The number of moles of oxygen is given as 0.98x.
Therefore, the total number of moles in the oxide is:
x + 0.98x = 2.98x
Step 5:
Since the total number of moles in the oxide is equal to the sum of moles of M2 and M3, we can write:
2y + 3z = 2.98x
Step 6:
The fraction of metal present as M3 can be calculated by dividing the number of moles of M3 by the total number of moles of metal.
So, the fraction of metal present as M3 is given by:
z/x = (2.98x - 2y)/x
Step 7:
Now, let's substitute the values of x and y from Step 3 into the above equation:
z/x = (2.98(2y + 3z) - 2y)/x
Step 8:
Simplifying the equation:
z/x = (5.96y + 8.94z - 2y)/x
z/x = (3.96y + 8.94z)/x
Step 9:
Since we need to find the fraction of metal present as M3, we can divide the equation by z:
(z/x)/(z/z) = ((3.96y + 8.94z)/x)/(z/z)
(z/x)/(1) = (3.96y + 8.94z)/(z)
z/x = (3.96y + 8.94z)/z
z/x = 3.96y/z + 8.94
Step 10:
Since y and z are both variables representing the number of moles, we can assume any value for y and solve for z.
Let's assume y = 1, which means there is 1 mole of M2.
Substituting this value into the equation:
z/x = 3.96(1)/z + 8.94
Step 11:
Simplifying the equation:
z/x = 3.96/z + 8.94
Metal oxide has formula M0.98O.
Metal M is present as M2 and M3 in its oxide.
To Find:
The fraction of the metal which exists as M3.
Solution:
Step 1:
Let's assume the total number of moles of the metal in the oxide is x.
Step 2:
According to the given information, the formula of the metal oxide is M0.98O.
This means that for every 1 mole of oxygen, there are 0.98 moles of the metal.
So, the number of moles of oxygen in the oxide is 0.98x.
Step 3:
Since the metal is present as M2 and M3 in the oxide, the total number of moles of metal can be represented as:
2y + 3z = x
where y is the number of moles of M2 and z is the number of moles of M3.
Step 4:
The number of moles of oxygen is given as 0.98x.
Therefore, the total number of moles in the oxide is:
x + 0.98x = 2.98x
Step 5:
Since the total number of moles in the oxide is equal to the sum of moles of M2 and M3, we can write:
2y + 3z = 2.98x
Step 6:
The fraction of metal present as M3 can be calculated by dividing the number of moles of M3 by the total number of moles of metal.
So, the fraction of metal present as M3 is given by:
z/x = (2.98x - 2y)/x
Step 7:
Now, let's substitute the values of x and y from Step 3 into the above equation:
z/x = (2.98(2y + 3z) - 2y)/x
Step 8:
Simplifying the equation:
z/x = (5.96y + 8.94z - 2y)/x
z/x = (3.96y + 8.94z)/x
Step 9:
Since we need to find the fraction of metal present as M3, we can divide the equation by z:
(z/x)/(z/z) = ((3.96y + 8.94z)/x)/(z/z)
(z/x)/(1) = (3.96y + 8.94z)/(z)
z/x = (3.96y + 8.94z)/z
z/x = 3.96y/z + 8.94
Step 10:
Since y and z are both variables representing the number of moles, we can assume any value for y and solve for z.
Let's assume y = 1, which means there is 1 mole of M2.
Substituting this value into the equation:
z/x = 3.96(1)/z + 8.94
Step 11:
Simplifying the equation:
z/x = 3.96/z + 8.94
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Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fractionof the m etal which exists as M3+ would bea)7.01%b)4.08%c)6.05%d)5.08%Correct answer is option 'B'. Can you explain this answer?
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Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fractionof the m etal which exists as M3+ would bea)7.01%b)4.08%c)6.05%d)5.08%Correct answer is option 'B'. 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 Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fractionof the m etal which exists as M3+ would bea)7.01%b)4.08%c)6.05%d)5.08%Correct answer is option 'B'. 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 Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fractionof the m etal which exists as M3+ would bea)7.01%b)4.08%c)6.05%d)5.08%Correct answer is option 'B'. Can you explain this answer?.
Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fractionof the m etal which exists as M3+ would bea)7.01%b)4.08%c)6.05%d)5.08%Correct answer is option 'B'. 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 Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fractionof the m etal which exists as M3+ would bea)7.01%b)4.08%c)6.05%d)5.08%Correct answer is option 'B'. 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 Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fractionof the m etal which exists as M3+ would bea)7.01%b)4.08%c)6.05%d)5.08%Correct answer is option 'B'. Can you explain this answer?.
Solutions for Experimentally, it was found that a metal oxide has formula M 0 9B0 . Metal M is present as M2+ and M3+ in its oxide. Fractionof the m etal which exists as M3+ would bea)7.01%b)4.08%c)6.05%d)5.08%Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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