Question

A gas mixture contains 50% helium and 50% methane by volume. What is the percent by weight of methane in the mixture?

A. 88.89%
B. 20%
C. 71.43%
D. 80%

Answer 1

Using the % volume of helium and methane gas calculate the volume ratio of helium to methane gas. Using the relation of the volume of gas with moles of gas at constant temperature and pressure determine the mole ratio of helium to methane gas. Finally, using the weight of methane gas and weight of mixture calculate the percent by weight of methane in the mixture.

Formula Used:
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{n_1}}}{{{n_2}}}$
${\% \text{ Weight = }}\dfrac{{{\text{Weight of gas}}}}{{{\text{Total weight of the mixture}}}} \times 100\%$

Complete step by step answer:
Calculate the volume ratio of helium to methane gas as follows:
A gas mixture contains 50% helium and 50% methane by volume.
So,
$\dfrac{{{\% \text{ Volume of He}}}}{{{\%\text{ Volume C}}{{\text{H}}_{\text{4}}}}} = \dfrac{{50\% }}{{50\% }} = \dfrac{1}{1}$
Avogadro’s law states that at constant pressure and temperature, the volume of gas is directly proportional to the number of moles of gas.
Thus, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{n_1}}}{{{n_2}}}$
As $\dfrac{{{\%\text{ Volume of He}}}}{{{\%\text{ Volume C}}{{\text{H}}_{\text{4}}}}} = \dfrac{1}{1}$
So, $\dfrac{{{\text{moles of He}}}}{{{\text{moles C}}{{\text{H}}_{\text{4}}}}} = \dfrac{1}{1}$
Calculate the percent by weight of methane in the mixture as follows:
Mass of 1 mole of ${\text{C}}{{\text{H}}_{\text{4}}}$ = 16g
Mass of 1 mole of $He$ = 4g
The total mass of the mixture of gas = 16g+4g =20g
Now, we have a mass of methane gas and a mass of the mixture of gas.
So, ${\% \text{ Weight = }}\dfrac{{{\text{Weight of gas}}}}{{{\text{Total weight of the mixture}}}} \times 100\%$
Substitute 16g for the weight of methane gas and 20 g for the weight of the mixture of gas and calculate the percent weight of methane gas in the mixture.
$\Rightarrow {\% \text{ Weight C}}{{\text{H}}_{\text{4}}}{\text{ = }}\dfrac{{{\text{16g}}}}{{20}} \times 100\%$
$\Rightarrow {\% \text{ Weight C}}{{\text{H}}_{\text{4}}}{\text{ = 80}\% }$
Thus, the percent by weight of methane in the mixture is 80%.

**Hence, the correct option is (D) 80%

Note: **
According to Avogadro’s law, the constant temperature and pressure ratio of the volume of gases is equal to the ratio of moles of gases. The molecular weight of a substance is a mass of 1 mole of the substance.