16 ml of oxygen diffused through a porous partition in t seconds . what volume of carbon dioxide will diffuse in same time by volume

From Graham's law of diffusion of gases, rate of diffusion is inversely proportional to molar mass of gas.

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

Rate = $\frac{Volume of gas}{time}$

Rate is directly proportional to the volume of gas.

Suppose r₁ is the rate of diffusion for O₂ and r₂ is the rate of diffusion for CO₂.
V₁ is the volume of diffusion of O₂
V₂ is the volume of diffusion of CO₂.

M₁, molar mass of O₂ = 32 g mol-1
​M₂, molar mass of CO₂ = 44 g mol-1
​As the time of diffusion of two gases are same and from the above two equations we can write

$\frac{V_1}{V_2} = \sqrt{\frac{M_2}{M_1}}$

V₁ = 16 mL

$$\begin{gathered} \frac{16}{V_2}=\sqrt{\frac{44}{32}} \\ \frac{16}{V_2}= 1.173 \\ {V}_2 = \frac{16}{1.173} \end{gathered}$$

V₂ = 13.64 mL.

Hence, volume of CO₂ diffused is 18.76 mL.