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The average molar mass of a mixture of methane and ethene present in the ratio a:b is found to be 20 gmol . If the ratio were reversed, what would be the molar mass of the mixture.

Answer
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Hint: Average molecular mass can be written as n1m1+n2m2+n3m3.....n1+n2+n3.... where ‘n’ is the no of moles of compound and ‘m’ is the molecular mass of compound.

Complete step by step answer:
In the question it is given that the mixture of methane and ethane are present in the ratio a:b and the average molar mass of the mixture is given to be 20 g/mol .
Molar mass of Methane (CH4)=(12×1)+(1×4)=16g/mol
Molar mass of ethene (C2H4)=(12×1)+(1×4)=28g/mol
Average molar mass of the mixture = 20 g/mol
We know,
Number of moles = Weight (g) / Molecular mass
Weight (g) = Number of moles * Molecular mass
There are ‘a’ mole of methane and ‘b’ moles of ethene in the mixture.
16a+28b=20a+20b4a=8b
So, solving the equation we get: ab=84=21
So, a:b=2:1
If the ratio is reversed, i.e. Methane and ethene present in the mixture in the ratio of b:a,
Then,
16b+28aa+b=16+28ab1+ab
16+28(21)1+21
(16+56)3
723=24

So, if the ratio is reversed, the molar mass of the mixture will be 24g/mol.


Note: Molar mass is a physical quantity defined as the mass of 6.022×1023 atoms of an element. Also the higher the molar mass the higher the density of a gas and vice versa. Average molecular weight is defined for mixture of atoms or molecules. The molar ratios are very important for quantitative chemistry calculations.
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