Question

A gas mixture contains SO 2 SO2 ( molar mass = 64 g/mol ) (molar mass=64 g/mol) and no other source of sulfur. If the mixture is 10 % 10% sulfur by mass, what is the percentage (by mass) of SO 2 SO2 in the mixture?

Answer 1

Answer : The percentage (by mass) of $SO_2$ in the mixture is 20 %

Explanation :

As we are given that, 10 % sulfur by mass that means 10 grams of sulfur present in 100 grams of mixture.

Mass of sulfur = 10 g

Mass of mixture = 100 g

Now we have to calculate the mass of $SO_2$.

As, 32 grams of sulfur present in 64 grams of $SO_2$

So, 10 grams of sulfur present in $\frac{10}{32}\times 64=20$ grams of $SO_2$

Thus, the mass of $SO_2$ is 20 grams.

Now we have to calculate the percentage (by mass) of $SO_2$ in the mixture.

$\% \text{ of }SO_2=\frac{\text{Mass of }SO_2}{\text{Mass of mixture}}\times 100$

$\% \text{ of }SO_2=\frac{20g}{100g}\times 100=20\%$

Therefore, the percentage (by mass) of $SO_2$ in the mixture is 20 %