Answer: The partial pressure of sulfur dioxide gas in the flask is 160 mmHg
Explanation:
We are given:
Moles of sulfur dioxide gas = 0.25 moles
Moles of methane gas = 0.50 moles
Moles of oxygen gas = 0.50 moles
To calculate the mole fraction of sulfur dioxide, we use the equation:
[tex]\chi_{SO_2}=\frac{n_{SO_2}}{n_{SO_2}+n_{CH_4}+n_{O_2}}\\\\\chi_{SO_2}=\frac{0.25}{0.25+0.50+0.50}=0.2[/tex]
The partial pressure of a gas is given by Raoult's law, which is:
[tex]p_{SO_2}=p_T\times \chi_{SO_2}[/tex]
where,
[tex]p_{SO_2}[/tex] = partial pressure of sulfur dioxide gas
[tex]p_T[/tex] = total pressure = 800 mmHg
[tex]\chi_{SO_2}[/tex] = mole fraction of sulfur dioxide = 0.2
Putting values in above equation, we get:
[tex]p_{SO_2}=800\times 0.2=160mmHg[/tex]
Hence, the partial pressure of sulfur dioxide gas in the flask is 160 mmHg
