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Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be
- a)5.08%
- b)7.01%
- c)4.08%
- d)6.05%
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Experimentally it was found that a metal oxide has formula M0.98O. Met...
Let the fraction o f metal which exists as M2+ be x. Then the fraction of metal as M2+ = (0.98 - x)
3x + 2( 0.98 - x) = 2
x + 1.96 = 2
x = 0.04
% of M3+ = 0.04/0.98 x 100
= 4.08%
3x + 2( 0.98 - x) = 2
x + 1.96 = 2
x = 0.04
% of M3+ = 0.04/0.98 x 100
= 4.08%
Most Upvoted Answer
Experimentally it was found that a metal oxide has formula M0.98O. Met...
Given: Metal oxide has formula M0.98O. Metal M, is present as M2 and M3 in its oxide.
To find: Fraction of the metal which exists as M3.
Solution:
Let the atomic weight of M be x.
M2+ ion will have (x-2) electrons and M3+ ion will have (x-3) electrons.
Let the fraction of M3+ ion be y.
Then the fraction of M2+ ion will be (1-y).
According to the given formula of metal oxide, the metal is present in the oxide in the ratio of 0.98:1.
Therefore, the mass of metal present in the oxide = 0.98x
From this, we can write the following equation:
(2*(1-y)*(x-2) + 3*y*(x-3))/x = 0.98
(2x - 4 - 2yx + 6y)/x = 0.98
2 - 2y + 6y/x = 0.98
2y/x = 0.02
y/x = 0.01
y = 0.01x
Therefore, the fraction of the metal which exists as M3 is 0.01 or 1%.
To convert this into percentage, we multiply by 100.
Therefore, the required fraction is 1% or 4.08% (option C).
Hence, option C is the correct answer.
Note: This type of problem can be solved by assuming the fraction of M3+ ion as y and then writing the equation based on the number of electrons in M2+ and M3+ ions. Then, solving the equation to find the value of y.
To find: Fraction of the metal which exists as M3.
Solution:
Let the atomic weight of M be x.
M2+ ion will have (x-2) electrons and M3+ ion will have (x-3) electrons.
Let the fraction of M3+ ion be y.
Then the fraction of M2+ ion will be (1-y).
According to the given formula of metal oxide, the metal is present in the oxide in the ratio of 0.98:1.
Therefore, the mass of metal present in the oxide = 0.98x
From this, we can write the following equation:
(2*(1-y)*(x-2) + 3*y*(x-3))/x = 0.98
(2x - 4 - 2yx + 6y)/x = 0.98
2 - 2y + 6y/x = 0.98
2y/x = 0.02
y/x = 0.01
y = 0.01x
Therefore, the fraction of the metal which exists as M3 is 0.01 or 1%.
To convert this into percentage, we multiply by 100.
Therefore, the required fraction is 1% or 4.08% (option C).
Hence, option C is the correct answer.
Note: This type of problem can be solved by assuming the fraction of M3+ ion as y and then writing the equation based on the number of electrons in M2+ and M3+ ions. Then, solving the equation to find the value of y.
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Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer?
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Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer?.
Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer?.
Solutions for Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+and M3+ in its oxide. Fraction of the metal which exists as M3+ would bea)5.08%b)7.01%c)4.08%d)6.05%Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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