Question

${\mathrm{O}}_2$ And ${\mathrm{SO}}_2$ Gases Are Filled In The Ratio Of $1:3$ By Mass In A Closed Container Of $3\mathrm{L}$ At Temp Of $27°\mathrm{C}$ The Partial Pressure Of ${\mathrm{O}}_2$ Is $0.60\mathrm{atm}$ The Concentration Of ${\mathrm{SO}}_2$ Would Be

Answer 1

Step 1: Given data

Partial Pressure Of ${\mathrm{O}}_2$ Is $0.60\mathrm{atm}$

Volume of the container is $3\mathrm{L}$

Given temperature is $27°\mathrm{C}$

Ratio of ${\mathrm{O}}_2$ and ${\mathrm{SO}}_2$ is $1:3$

Step 2: Calculating total pressure

We have,

$$\begin{gathered} {\mathrm{P}}_{{\mathrm{O}}_2}={\mathrm{X}}_{{\mathrm{O}}_2}\times {\mathrm{P}}_{\mathrm{Total}} \\ \Rightarrow 0.60=\frac{{\mathrm{n}}_{{\mathrm{O}}_2}}{{\mathrm{n}}_{{\mathrm{O}}_2}+{\mathrm{n}}_{{\mathrm{SO}}_2}}\times {\mathrm{P}}_{\mathrm{Total}} \end{gathered}$$

Let us assume that the total number of moles of ${\mathrm{O}}_2$be $\mathrm{x}$

$$\begin{gathered} \therefore 0.60=\frac{{\displaystyle \frac{\mathrm{x}}{32}}}{\frac{\mathrm{x}}{32}+\frac{3\mathrm{x}}{64}}\times {\mathrm{P}}_{\mathrm{Total}} \\ \Rightarrow 0.60=0.4\times {\mathrm{P}}_{\mathrm{Total}} \\ \Rightarrow {\mathrm{P}}_{\mathrm{Total}}=\frac{0.60}{0.4} \\ \Rightarrow {\mathrm{P}}_{\mathrm{Total}}=1.5\mathrm{atm} \end{gathered}$$

Step 3: Finding the concentration of ${\mathbf{SO}}_{\mathbf{2}}$

As we know,

$$\begin{gathered} {\mathrm{P}}_{{\mathrm{O}}_2}+{\mathrm{P}}_{{\mathrm{SO}}_2}={\mathrm{P}}_{\mathrm{Total}} \\ \Rightarrow {\mathrm{P}}_{{\mathrm{SO}}_2}={\mathrm{P}}_{\mathrm{Total}}-{\mathrm{P}}_{{\mathrm{O}}_2} \\ \Rightarrow {\mathrm{P}}_{{\mathrm{SO}}_2}=1.5-0.60 \\ \Rightarrow {\mathrm{P}}_{{\mathrm{SO}}_2}=0.9\mathrm{atm} \end{gathered}$$

$\therefore$Concentration of ${\mathrm{SO}}_2$is:

$\frac{0.9}{0.0821\times 300}=0.036$

Therefore, the concentration of ${\mathrm{SO}}_2$ would be $0.036$