Question

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

Answer 1

Hint: Average molecular mass can be written as $\dfrac{{n_1m_1 + n_2m_2 + n_3m_3.....}}{{n_1 + n_2 + n_3....}}$ 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 $(C{H_4}) = (12 \times 1) + (1 \times 4) = 16g/mol$
Molar mass of ethene $({C_2}{H_4}) = (12 \times 1) + (1 \times 4) = 28g/mol$
Average molar mass of the mixture = 20 $g/mol$
We know,
Number of moles = Weight (g) / Molecular mass
$\Rightarrow$ Weight (g) = Number of moles * Molecular mass
There are ‘a’ mole of methane and ‘b’ moles of ethene in the mixture.
\[\begin{array}{*{20}{l}}
\Rightarrow {16a + 28b = 20a + 20b} \\
\Rightarrow {4a = 8b}
\end{array}\]
So, solving the equation we get: $\dfrac{a}{b} = \dfrac{8}{4} = \dfrac{2}{1}$
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,
$\Rightarrow$ $\dfrac{{16b + 28a}}{{a + b}} = \dfrac{{16 + 28\dfrac{a}{b}}}{{1 + \dfrac{a}{b}}}$
$\Rightarrow$ $\dfrac{{16 + 28\left( {\dfrac{2}{1}} \right)}}{{1 + \dfrac{2}{1}}}$
$\Rightarrow$ $\dfrac{{(16 + 56)}}{3}$
$\therefore$ $\dfrac{{72}}{3} = 24$

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 $6.022\times 10^{23}$ 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.