Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm$$^3$$ of SO$$_2$$ diffuses through the porous partition in 60 seconds. The volume of O$$_2$$ in dm$$^3$$ which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32 u):

According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. The rate can be expressed as the volume diffused per unit time. For two gases, SO₂ and O₂, under similar conditions, the relationship is given by:

$$\frac{\text{Rate}_{\text{SO}_2}}{\text{Rate}_{\text{O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{SO}_2}}}$$

where Rate = Volume / Time, and M is the molar mass.

Given:

• Volume of SO₂ diffused, $$ V_{\text{SO}_2} = 20 \text{dm}^3 $$
• Time for SO₂ diffusion, $$ t_{\text{SO}_2} = 60 \text{seconds} $$
• Time for O₂ diffusion, $$ t_{\text{O}_2} = 30 \text{seconds} $$
• Volume of O₂ diffused, $$ V_{\text{O}_2} = ? \text{dm}^3 $$ (to be found)

First, calculate the molar masses:

• Atomic mass of S = 32 u, O = 16 u
• Molar mass of SO₂, $$ M_{\text{SO}_2} = 32 + 2 \times 16 = 32 + 32 = 64 \text{u} $$
• Molar mass of O₂, $$ M_{\text{O}_2} = 2 \times 16 = 32 \text{u} $$

Using Graham's law:

$$\frac{\text{Rate}_{\text{SO}_2}}{\text{Rate}_{\text{O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{SO}_2}}}$$

Substitute the rate expressions:

$$\frac{\frac{V_{\text{SO}_2}}{t_{\text{SO}_2}}}{\frac{V_{\text{O}_2}}{t_{\text{O}_2}}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{SO}_2}}}$$

Plug in the known values:

$$\frac{\frac{20}{60}}{\frac{V_{\text{O}_2}}{30}} = \sqrt{\frac{32}{64}}$$

Simplify the left side:

$$\frac{20}{60} = \frac{1}{3}$$

So:

$$\frac{\frac{1}{3}}{\frac{V_{\text{O}_2}}{30}} = \frac{1}{3} \times \frac{30}{V_{\text{O}_2}} = \frac{30}{3V_{\text{O}_2}} = \frac{10}{V_{\text{O}_2}}$$

Simplify the right side:

$$\sqrt{\frac{32}{64}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

Now the equation is:

$$\frac{10}{V_{\text{O}_2}} = \frac{1}{\sqrt{2}}$$

Solve for $$ V_{\text{O}_2} $$:

Cross-multiply:

$$10 \times \sqrt{2} = V_{\text{O}_2}$$

So:

$$V_{\text{O}_2} = 10\sqrt{2}$$

Calculate the numerical value. $$ \sqrt{2} \approx 1.414 $$:

$$V_{\text{O}_2} = 10 \times 1.414 = 14.14 \approx 14.1 \text{dm}^3$$

Hence, the volume of O₂ that diffuses in 30 seconds is 14.1 dm³.

So, the answer is Option C.