Question

The average molar mass of a mixture of methane and ethene present in the ratio $a:b$ is found to be $20g\,mo{{l}^{-1}}$. If the ratio were reversed, what would be the molar mass of the mixture?

Answer 1

Use the formula for finding out the average with the number of moles and find the ratio, then reverse it and use the same formula of average again.

Complete answer:
In order to answer the question, we need to learn about the concept of molarity and molality. Both of the terms are used for describing the concentration of a solute in a particular solvent or solution.
Now, molality is used when a particular mole of a solute is present in 1kg of the solvent, whereas in the case of molarity, the latter part is taken to be 1000 mL of the solution itself. Now, let us come to the solution. We have been given with two compounds and the ratio they are present in. Now, we know that:
$no\,of\,moles=\dfrac{given\,mass}{molar\,mass}$
Now, the molar mass of methane is 16 and that of ethene is 20 grams. As the average molar mass of mixture is $30g\,mo{{l}^{-1}}$, so we use the formula for calculating the average and we get:

& \dfrac{16a+30b}{a+b}=20 \\\ & \Rightarrow 16a+30b=20a+20 \\\ & \Rightarrow 10b=4a \\\ & \Rightarrow \dfrac{a}{b}=\dfrac{5}{2} \\\ \end{aligned}$$ Now we have obtained the ratio as 2.5, with the help of this we are going to calculate when the ratio is reversed. Now, both of the compounds are present but in the ratio of $b:a$. Again, by the using the formula for calculating the average, we have: $$\dfrac{16b+30a}{a+b}=\dfrac{16+30\dfrac{a}{b}}{1+\dfrac{a}{b}}=\dfrac{16+30(\dfrac{5}{2})}{1+(\dfrac{5}{2})}=\dfrac{16+75}{\dfrac{5}{2}}=26$$ So, when the ratio by which the compounds are reversed, then we get the average molar mass of the mixture as $26g\,mo{{l}^{-1}}$, which is the required answer for our question. **Note:** It is to be noted that the new average molar mass depends on the molar mass of the compounds as well as the initial ratio of the compounds.