Question

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

Answer 1

Answer:

● 50% helium and 5% methane by volume

= Molar and volume ratio are same

$= \frac{50}{50} = \frac{1}{1} \: volume \: ratio$

$molar \: ratio \ = \frac{1}{1} = \frac{helium}{methane}$

= 1 mole of helium = 4g

= 1 mole of methane = 16g

= total = 4 + 16 = 20g

= %by weight of methane in the mixture

$= \frac{12}{16} \times 100 = 80\%$

Explanation:

$\huge\mathfrak\orange{verified \: answer}$