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A mixture of methane and ethane in the mole ratio of x : y has a mean molecular mass equal to 20. What would be the mean molecular mass of a mixture of the same gases present in the ratio y : x?
- a)20
- b)22
- c)24
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A mixture of methane and ethane in the mole ratio of x : y has a mean ...
Molar ratio of methane : ethene = a:b
Average molecular mass of the mixture = 16a + 28b / a+b
Average molecular mass of the mixture = 16a + 28b / a+b
Now,
16a + 28b / a+b = 20
On solving,
a/b = 2/1
On reversing b/a = 1/2
so, 1:2 = (16 x1 + 28x 2)/3 = 24
Most Upvoted Answer
A mixture of methane and ethane in the mole ratio of x : y has a mean ...
Mean molecular mass of a mixture of gases is calculated by taking the weighted average of the individual molecular masses of the gases present in the mixture.
Given:
Mole ratio of methane to ethane = x : y
Mean molecular mass of the mixture = 20
To find:
Mean molecular mass of the mixture in the ratio y : x
Let's assume the molecular masses of methane and ethane are M1 and M2 respectively.
Weighted average of the molecular masses of the mixture can be calculated using the following formula:
Mean molecular mass = (x * M1 + y * M2) / (x + y)
According to the given information, the mean molecular mass of the mixture is 20:
20 = (x * M1 + y * M2) / (x + y) ........(1)
We need to find the mean molecular mass of the mixture in the ratio y : x. Let's represent it as MM:
MM = (y * M1 + x * M2) / (y + x) ........(2)
We can find the relationship between MM and 20 by rearranging equation (1) as follows:
20 = (x * M1 + y * M2) / (x + y)
20 * (x + y) = x * M1 + y * M2
20x + 20y = x * M1 + y * M2 ........(3)
Now, let's rearrange equation (2) to match the form of equation (3):
MM = (y * M1 + x * M2) / (y + x)
MM * (y + x) = y * M1 + x * M2
MMy + MMx = y * M1 + x * M2 ........(4)
Comparing equations (3) and (4), we can see that the left-hand side of both equations is the same (20x + 20y = MMy + MMx). However, the right-hand side of both equations is different.
Since the right-hand sides of equations (3) and (4) are not equal, the mean molecular mass of the mixture in the ratio y : x is not equal to 20. Therefore, the correct answer is option D: None of these.
Given:
Mole ratio of methane to ethane = x : y
Mean molecular mass of the mixture = 20
To find:
Mean molecular mass of the mixture in the ratio y : x
Let's assume the molecular masses of methane and ethane are M1 and M2 respectively.
Weighted average of the molecular masses of the mixture can be calculated using the following formula:
Mean molecular mass = (x * M1 + y * M2) / (x + y)
According to the given information, the mean molecular mass of the mixture is 20:
20 = (x * M1 + y * M2) / (x + y) ........(1)
We need to find the mean molecular mass of the mixture in the ratio y : x. Let's represent it as MM:
MM = (y * M1 + x * M2) / (y + x) ........(2)
We can find the relationship between MM and 20 by rearranging equation (1) as follows:
20 = (x * M1 + y * M2) / (x + y)
20 * (x + y) = x * M1 + y * M2
20x + 20y = x * M1 + y * M2 ........(3)
Now, let's rearrange equation (2) to match the form of equation (3):
MM = (y * M1 + x * M2) / (y + x)
MM * (y + x) = y * M1 + x * M2
MMy + MMx = y * M1 + x * M2 ........(4)
Comparing equations (3) and (4), we can see that the left-hand side of both equations is the same (20x + 20y = MMy + MMx). However, the right-hand side of both equations is different.
Since the right-hand sides of equations (3) and (4) are not equal, the mean molecular mass of the mixture in the ratio y : x is not equal to 20. Therefore, the correct answer is option D: None of these.
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